Tr1-30(3)

7/23/2019 Tr1-30(3)

http://slidepdf.com/reader/full/tr1-303 1/30

i

DOXUANHUNG

Gi o

vign

chuy^n luy^n

thi DH-CD H6a

( o/UH'V'Aijy <k

Pharyldng

phuangphap

giai

7/23/2019 Tr1-30(3)


PHAN DANG

VA PHLfCJNG

PHAP

GIAI HOA HQC 1 1 H Q U Cd

D

6

XUAN

HUNG

C h j u

trach n h i ^ m xu^t ban

N G U Y E N

T H I

T H A N H H l / O f N G

Bientap :

H A I

A U

Sufa

ban in :

H 6 N G

H A I

Tnnh bay : C6ng ty

K H A N G VI : T

Bia : C6ng ty

K H A N G

VlfeX

NHA

XUAT

BAN T

5

N G HOP TP. H6 CHf

MINH

NHA

SACH

T 6 N G H0P

62 N g u y i n Th j

M i n h

Khai,

Q

. l

D T :

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-

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Fax: 84.8.38222726

E m a i l : [email protected]

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Tong

phdt hanh

C NG

TY T N M T V

D|CH

vg

V N H

6

KHANG

VI|T

r D j a c h f : 71 Dinh TiSn Ho^ng - P.Da Kao - Q.I MP.HCM

Di in thoai: 08. 39115694 -

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www.nhasachkhangvlet.vn

In lln

tha

1. So la<?ng 2.000 cuon, kho 16x24cm.

T?i: Cong ty

TNHH

MTV in ^n M A I T H j N H

DLfC '

Dja ch l: 71 , Kha V^n CSn, P.HiOp Binh

Ch^nh

Q.Thii Dufc, Tp.HCM.

So

DKKHXB:

1482-12/CXB/12-181/THTPHCM

ngSy

06/12/2012.

Quy^t djnh xuat bkn s6:

1815/QD-THTPHCM-2012

do NXB

Tong

h(?p

Thanh

ph 6 Chi

Minh

cS'p ng^y

28/12/2012.

In xong nOp lUU chi û quy' I nanri 201

3.

uMtta

4

p

D A I

CI CJNG V E H A

H p C

HtfU CCJ

A. TOM TAT Lf

THUY^T

M d D A U V E H b A

HOC

H i r U C

( ] f V A H 0 P

CHAT H0UCCf

1 .

H^p

chaft

hffu ccf la hdp chat cua cacbon (trilf CO,

CO2,

muoi cacbonat xianua,

cacbua,...)

2.

PhSn

loaii

h^p

chSt

hf fu

ccf .

9 ^: : J^'l • ,

p cha't hQu CO

Hidrocacbon

^ ^ ' Di n xuSt cua hidrocacbon

Hidrocacbon

Hidrocacbon Hidrocacbon

no

khOng no

thorn

Din Ancol

Andehit

Amin Axit

HOP

xuSt Phenol Xeton

nitro

Este

chat

halogen

Ete

tap

chile

polime

3. Danh phap h^p

chS't hf fu

cd

a) Te n thong thiidng (nguon goc flm ra chat)

^ - - ^ M s g

« * a v * f «

b )

Ten

h^

thong (lUPAC) '

: i , 1 ^ / ^ ( r y i

Ten goc chiJc = Ten phan goc + T6n phan dinh chiJc

Ten thay the = Ten phan thay the + Ten mach C chinh + Ten phan dinh

I I . P H A N T I C H

NGUYEN

TO

1 . Ph3n tichdjnh

t i n h

f'''' /''-;'

* Xdc

dinh

C : Dot chdy chat hS u cd CO2 CaCOjl

* Xac

dinh

H : Dot chdy chat

h i h i

cd

H2O ^CuS04khan CUSO4.5H2O

'•• - (xanh)

X ic

dinh

nitd

: C^HyCN, (NH4)2S04 + .j

V- v* ?

n . f •

* «

^ • • *

( N H , ) 2 S 0 4

+ 2NaOH >Na2S04 + 2NH3t + 2H2O

* Xac

dinh

halogen : CxHyO,N,

CO2 H2O + HCl 14 ;

H C l

+ AgNOj

> A g C l i

+

HNO

vi ,

2.

PhSn t i ch d jnh I r f^ng

(A la chat

hffu

cd c6 H I A

chffc

m c = 1 2

. - i 4

^ ( g )

44

- H

=

2

. i ^ ( g )

= i^.100

=> H=-2^.100

= > N = i ^ . l O O

n i

7/23/2019 Tr1-30(3)


Phan djing \ik phuong ph p g Ai H6a hpc 11 HJu co

B

Xuan Hung

=> mo = ni A - ( m c +

m H

+ I H N )

h a y % 0 = 1 0 0 %

- ( C +

H +

N ) ; ,

I I I

C O N G

T H L T C PHAN T l T H O P

C H A T

H 0 U C O f

^ i M j ^ J

1. C6ng t h t f c defn

g i a n

n h S ' t Y i t M i * t ,vN ( .

C 6 n g

t h u- c p h a n

tuT

(CTPT) :

CxHyO,Ni

v,

j

C o n g t h u -c d d n g i a n n h a t (CTDGN): (CpH,OrN,)„ ( n = 1,2, 3,...) ,|

* C a c h t h i e t l a p : T i m t i 1 (C,HyO,)

so,..^^.,^

m r

mH

niQ ^ %C

H

% 0 , ;

x : y : z =— ^ :— ^ :— ^

hay

x : y : z =

: : = p : d : r

12 1 16 12 1 16

2 .

C d n g t h t f c p h a n

tuT

( C T P T )

^

/ • ^ ^ j , ; . - . • i s / V i ' — •• "^ n: ir::.xM\i

* C a c h t h i e t l a p :

- Xdc d i n h k h o i l u W n g m o l p h a n tu T: M A =

M e. d A / B

h o a c

M A =

29.dA /k k • i t i i

'

- D i f a vac % k h o i l u ' d n g cac

n g u y e n

to :

j

CxHyO, x C + y H + z O ' ' '

I „ 12x y 16z M

T a c d t i l ? : = - ^ = = => x, y, z

C H % 0 100 % • • , J;

- D u r a v ao c o n g t h i J c d d n g i a n n h a ' t : iî^Y j

j

C, HyO, = (CpH, 0, )„ = M = ^ n = ^ „ ..... i ^ ' 0 : ^

1 2 p + q + 1 6 r

,/>a

F .

- T i n h t n i c t i e p

t h e o

k h o ' i I t f d n g s a n p h a m d o 't

c h d y :

-H

-

I

CxHA

+

^ y z

x + - - -

4 2

02

^ x C O ^ + ^ H ^ O '

•"':i f <(T}iD 00 2 H2O

1 p f f

' 1 '*

T a c d t i i e : — =

— =

^ x , y ; t i r M A t i n h d i r d c z . ^ - ^^^^^ ^ '

t

A

00 2

2 n, , 2 0

i

I V . C A U T R U C

P H A N

T l f H O P C H A T H 0 U C O i i i *

1. C6ng t h i t c c f i u t a o js

* C o n g t h i J c c a u t a o k h a i t r i e n ;|

.>fiX

* C o n g t h i J c c a ' u t a o t h u g o n

2 .

D d n g d a n g

N h f f n g h d p c h a t

c6

th a n h p h a n p h a n

tuT

h d n k

^ m

n h a u

mpt

h a y n h i e u

n h o m CH2

n h i f t i g c6 t i n h c h a t h o a hpc t i i d n g t i f n h a u .

"V' ' ' ;^;

,4^

3 .

D 6 n g p h a n

. . .: „ ,;„., , , . „ . , . j , . j ^ w ' ^ ^

N h i J n g h d p

c h a t k h d c

n h a u n h u ^ g c6

c C ln g c 6 n g

t h i Jc p h a n t i l . iti

4 .

L i d n k d 't

t r o n g p h a n

tuT

hflTp

chS ' t h f l u cc f

L i e n ket d d n , l i e n ket

do i

va h e n

ke t

b a . '* JifJK hn-

A

V .

P H A N

0 N G H O U C O

1 . P h a n t fng the : CH4 + C I 2 C H 3 C I + HCl HO

- : J

r

2 . P h a n uTng

c O n g :

C2H4 + Br2 >C 2 \ I ^ B T 2 '

3. P h a n u r n g t a c h : C H 3 - C H 2 - O H - S ^ ^ CH2=CH2 + H2O H

D a c diem cua p h a n

iJng

h 6 a hpc t r o n g hoa hoc hf fu cd : - 1

- Phan

i J n g t h i f d n g

xay ra c h a m . \'t } -\{^ An

-

Phan

iJn g t h u d n g s in h

ra hon hdp san

p h a m .

; '. ^

B.

PHAN LOAI

VA PH JCJNG

PHAP GIAI

CA C

DANG

BAI TAP

- (Vm f i h a n

l i n g ^ hau c ^ r i r i O ' - v H ^ ^ - T H ; } , H I 4 . . - , H

; : H H D - ; . i i : ) : - , - i - ' '

•

- < J j h a i i l o a i f d i t u t i t i i g . h i t i i eet

' t

- (Vi eMiq. I h i i e e a i i t im , c t m t g ^ p h d n ^ d d t i g ^ d q t t Q ^ ede e h d t

BAI TAP

M A U

VA BAI TAP NANG CAO

i?

B a i 1 . Viet

p h i f d n g

t r i n h hoa hoc

c i i a

cac

p h a n i J ng

xay ra theo sd do sau f.

CH^CH CH2=CH2

CH3-CH2-OH CH3CH2-Br Q j | j

Trong

cac

p h a n i J ng t r e n , p h a n i J n g

nao

t h u p c l o a i p h a n t fn g

t h e ,

p h a n iJ n g

cong,

p h a n i J n g t d c h ? ,,,,

. ^

i a i

( 1 ) CH^CH +

H2 ^'^''^^^^^) CH2=CH2

:

p h a n i J n g cpng

( 2 )

CH2=CH2 + H2O '^ °> CH3- CH2

-OH

:

p h a n iJng

cpng

1,,,

?

. . M r t r ;

3 ) CH3-CH2-OH + HBr CH3-CH2-Br + H2O : p h d n

t fng thg*

' '

( 4 ) 3CH=CH ^•^°°°^) ( Q ) : p h a n u-ng cpng

B a i 2. Co ba c h a t hCu cd c6 cong

thiJc

Ian l i Tdt la : C4Hii)0, C4H9CI, C4H11N, Cho

b i e t c h a t nao c6

n h i e u

dong

p h a n

h d n ?

i d i

* C4H|()0

: „., 'J , '(.,, .

- ,M

>•• ^-.;j;y--.

CH3-CH2-CH2-CH2-OH ; CH3-CH2-CH2-O-CH3

CH3

-CH2

-C H-CH3 ; CH3-CH2-O-CH2-CH3

7/23/2019 Tr1-30(3)


Phan

djing va phUdng ph^p g ia i H6a hpc 11 HOu cO 5 XuSn

HiJng

O H

^ , ..:Jt

CH 3-C H-CH2-OH ; CH3 -CH-O -CH3 ; CH3 -C

-CH3

i

CH3 CH3

CH3

*

C4H9CI: i

,:u ;.

- C H 3 C H C H C H - C I ; C H 3 - C -C H3 • „

C H 3 - C H 2 - C H - C H 3

;

C H 3 - C H - C H 2 - C I

C H 3

- C H 2 - C H 2 - C H 2 - N H 2 ;

CH 3

-CH2-CH2-NH-CH3 uhi^t^ya:f ••••

C H 3 - C H 2 - C H - N H 2

; CH3

-CH2-NH-CH2

-CH 3 . „

C H

^

, ,»:v A,«;

tv i f i^ i^ ; nvA ^ ihiyii, I l- iM mhh, \ ' v t;^5 * •••

C H 3 - C H - C H 2 - N H 2

;

C H 3 - C H - N H - C H 3

C H 3 - C - C H 3 ^ '

.HDsHD v ;^;

C H 3 - C H 2 - N - C H 3 ,m--m

^ i ^ ^ .

f i C t e l t

I 1

CH3 CH3 ( i i j

V ay c ha t

cd

n h i e u d o n g p h a n nha t

Ik

C iH^N r o i

den

C 4 H , „ 0 ,

C 4 H y C l .

a i

3 .

Cho ba hdp

c h a t

: A (cau tao tit C, H); B (cau tao

tu f

C, H, N) ; C (cau tao tit

C, H, O).

Chaft

nao

thupc loa i h idrocacb on, cha ' t

nao

thuQ l o a i

dan

x u a t

cua

h id r oc ac b on .

. ,

*

A la

h id r oc ac b on c h i

cSu tao gom

ha i ng uy e n

to

C.

H.

1;,)

j i l

*

B va C la dan

x u a t

cua

h id r oc ac b on

vi

n g o a i

hai

ng uy e n

to C, H cdc hcJp

c ha t

B,

C con

chiJa

cdc

n g u y e n

to

k h a c

(N va O).

a)

T r o n g

cac

hdp chaft

sau.

h

(p

c ha t

nô la hffu cd ,

hcJp cha't

nao la v 6 cd CH4,

CHCI3. C2H7N, HCN, C H j C O O N a , C,2H220|,, f C2H3Cl}-„, AI4C3.

b)

C h a t

nao sau day la

h id r oc ac b on

? la dan

x u a t

cua

h id r oc ac b on

: CH2O,

C2H5Br,

CH2O2,

CfiHsBr, CfiHfi,

CH3COOH.

i d i

a)

H d p c h a t h i ? u c d : C H 4 , C H C l 3 , C 2 H 7 N . C H 3 C O O N a , C , 2 H 2 2 0 , i , { C 2 H 3 C l } „ .

*

H d p c h a t v 6 cd : H C N , AI4C3.

b) H id r oc ac b o n : C fiH fi . frt )

D a n xua 't c u a h id roc ac b on

: CH2O,

C2H5Br ,

CH2O2,

C f iH jBr ,

CH3COOH.

a i 5. Cho 6 hdp cha t hi?u cd ddn chiJc, mach hcl cd cung cong thiJc phan tuT 1

C4HXO2. Vi et cdc cong thiJc ca u tao thu gpn cua cdc chat dd . • , • ,

CH3-CH2-CH2 -COOH ; CH3-COO-CH2-CH3 ; CH3 -CH -COO H

r CH3

HCOO-CH2-CH2-CH3 ; CH3-CH2-COOCH3 ; HCOO-CH -CH3.

C H

a i 6. Cac

p h a n

uTng sau day thuOc

l o a i p han i Jng

nho ?

( p han i Jng the , p han i Jng

c ong , p han i Jng ta c h ) ?

a)

C2H6 + CI2 ^ C2H5CI

+ HCl b

C4HX + H2O C4H

,oO

c) CzH s C l )

C2H4

+ HC l d) 2 C 2 H 5 0 H ^ ^ C 2 H 5 0 C 2 H 5 + H20

e)

CH2=CH2 + H2O CH3 -CH 2

-O H f CfiHft + Br j Cf iHsBr + HBr

g) C4HH, + 5F2

> 4C

+

l O H F

h) CH3OH + HCl CH3CI + H2O

a) Phan iJng the

b

Phan iJng

c p n g c)

P han t fng t a c h

^

d) Phan iJng tach e) Phan iJng

c p n g

f Phan iJng the

g)

P han

i f n g

t a c h

h)

Phan u-ng the .

a i 7.

Dt fa

vao t i n

c ha t

hoa hoc cua

C H 2 = C H 2

v a

C H ^ C H

(da hoc

Idp

9 .

Hay

Viet

p h i T d n g

trinh hoa hoc

k h i

ch o C H 3 - C H = C H - C H 3

va

C H 3 - C ^ - C H 3 l a c d u n g

vdi

Br2, H2 va cho b i e t

nhufng nhdm nguyen

tu T nao t r ong

phan tuf

cua hai

hd p cha ' t

t r e n

da gay nen cac

phan u 'ng dd . , . . , : (

; .

*

CH 3

-CH=CH

-CH3 + Br2

CH 3

-CH-CH

-CH3

-mwaot/rh sn6:; ?iv .

^ ; H ^ Br Br 'hi: , ' '

CH 3-CH=CH-CH 3 + H2 CH3- CH2-C H2-CH 3 - .

. • r i . V H , ?

„

» n i ; >

;

y

. > t 3 f f i . ; * ) : .

=> Nh dm n guye n tuf

gay nen

p han i Jng

: -C =C . 0;| 0ktxi-nUym¥'' ''

B r Br ....,:.„„..; ,

* CH 3 -CHC-CH3 +

2Br2

—

CH3 -C—C

-CH3

| . . ^ . l ^ :

CH 3-C^C-CH3 + 2H2 CH3- CH2-C H2-CH

3

=> N hd m ng uy e n tuT gay nen pha n uTng 1^

:

-C = C - .

a i

8.

V ie t phufdng

t r inh

hda hoc cua cdc

p han i Jng

xay ra

t r ong

cac

tr iTdng hdp

sau

va

cho

b i e t

cac

p h a n

u ng dd

thuQC loai phdn uTng

nô

(ph an vJng

the,

ph an uTng

c p ng , p han u 'ng t a c h ) .

7/23/2019 Tr1-30(3)


Phan dgng v i phuong phap giSi H6a hgc 11 HDu co - 8 XuSn Hung

a ) E t i l en t ac d un g vd i h i d ro c6 Ni l am x u c t d c va d un n 6 ng .

b ) D un n6 ng ax e t i l en d 6 0 0 " C v d i b g t t han l am x u c t ac t hu diTdc

b e n z e n .

^f

c ) D ung d i ch anco l e t y l i c d^ l a u n g o S i k ho ng k h i chuy ^n t hanh d ung d i ch ax i t

a x e t i c ( g i a m

an).

• t i >

. i f . . i J

; W;>

; /

:i•0-.-, K;;»

,•

' a) CH2=CH2 +

H2 CH3-CH3

( p han i J ng cOng)

,li::).:.:H':;).-..H':) i :X ) »

^

^

(Cf iHfi) [Q] (phan iJng c p n g )

) 3 C H = C H

600°

C

c ) C2H5OH + 02(kk)

^ ^

>

CH3COOH

+

H2O ^

p han i f ng o x i h6 a

, , , r ; .^ <.': , , , . k ho n g ho an t o an .

B a i 9 . Cho cac chat sau, nhf fng chat nao la dong d in g cua nhau ? NhCng cha 't nao la

d d ng p han cua nhau ?

,

CH3-CH3

• ill

( 1 )

CH3 -CH -CH3 Ui + j K ' ( 2 ) : } VJ {.7

C H ,

ÎM;;; +

Q I . ^

-

C H .

C H ,

C H 3 - C H 2 - C H 2 - O H

( 4 )

•>1

3

C H 3 - C H = C H 2 (5)

C H 3 - C H 2 - O - C H 3 (7)

v a ) C H 3 - C H 2 - C H 2 - C H 2 - O H ( 9 )

C H 3 - C H 2 - C H 2 - C H 3

C H 2 C I - C H 2 - C H 3

(6 ) i i i

t )

J

I d I i.i i

I , /

411

X

^m )h W f m --\0-

: H ' : ) fjv

f i ' 0 - - i idi .rf

A n r

^ i i J ' ] i f h

Iq i H u

f£;Cic c h a t

la

d o n g d i n g : ( 1 )

va

( 6 ) ; ( 4 )

va

( 9 ) . rt rirt 4 t

\ i ' •^ 'c f f,.

II }H

* C dc chaf t d o n g p h a n c u a n h a u : (2 ) va ( 8 ) ; ( 4 ) ( 7 ) , ( 3 ) (5). . < h .

B a i 1 0 .

H a y v i e t c o n g thu fc e lec t ro n va c o n g thiJc cau tao c ^c p han tuT sau : > ~ . , r ) i } f ' 3 *

C H 3 C I

C H 4 O C H 2 O

CH5N.

b ) Hay vie't c o n g thufc cau tao k ha i t r i en va c o n g thiJc cau ta o thu gon n h a t cula c a c

h d p c h a t s a u :C 3H 6 C H 3 C H O C H 3 C O O C 2 H 5 C H 3 C N b i e t r i n g t r ong p h a n

tiJf

c u a c h i i n g d i u c l i e n k e t b p i . : ' ' r f "'v • ^ 'H^;-^'' ijik I'-'iy-- '•^• •.i-r., ,

id i

a )

If/

C T P T

C T e l e c t r o n

C T C T

H M ...

V ,

C H 3 C I

H : C : C 1

' H - c - c i • • ^ -v: ^ 5 -«•

/- ii^gj-..4;

H::)---?:HD +

H

C H 4 O

H : C J O : H

H - C - O - H

' '

1

H H

b )

C H 2 0

H : C : : 0 :

i i

H - C = 0

H

H

H

1

CH5N

H : C : N : H

H H

1 H- ( j : -N- H

H H

C T P T

C T C T kha i t r i en

C T C T t hu g o n n ha t

H

1 C H 3 - C H = C H 2

C 3 H 6

H - C - C = C - H

A A A

h o a c

C H 3 C H O

•? /°

H - C - C

A

h o a c

C H 3 - C H O

H - c - c

H 0 - C - C - H

1

1

H H

C H 3 C O O C H 2 - C H 3

C H 3 C O O C 2 H 5

H - c - c

H 0 - C - C - H

1

1

H H

h o a c

H

1

C H 3 C N

C H 3 C N

H - C - C = N

ho Sc

= N

B a i 1 1 . D o n g p h a n la gi ? The nao la nh f f ng cha t d o ng d i ng ? T r o n g so nhuTng cha t

diTdi day, nhCTng chat nao la d o ng p han cua nhau ? Nhf fng chat nao la d o n g d i n g

cua nhau

?

a) C H 3 - C H 2 - C H = C H 2

b)

C H 3 - C H = C H 2

c) I I ^

CH2

—CH2

..i,..J

^ p 'f':''f(\f . ' / i i V ; ' lyi- V /• y . i •

U

H

CH3

d C

=CH2

C H ,

7/23/2019 Tr1-30(3)


^han dang phuong

phiip giai

H6a hgc 11 HOu co - D8 Xufln

Ht ng

•)) V i e l c o n g t h u c c a ' u t a o c 6 t h e c d c u a c a c d o n g

d a n g

c u a a n c o l

e t y l i c

c d c o n g

t h u c

p h a n

tuT

C s H x O

v a C 4 H|„0 .

I * C 2 H 6 O : C H 3 - C H 2 - O H ;

C H 3 - O - C H 3

* C 3 H 6 O : C H 3 - C H 2

- C H O ;

C H 2

= C H

- C H 2

- O H ;

C H 2

= C H - 0 - C H 3 ;

O H

C H 3

- C

- C H 3

; C H ; ^ \ P 2 - C H 2 -

* C 4 H : C H 3 - C H 2 - C H 2 - C H 3 ; C H 3

- C H

- C H 3

IB

. ' IV

M

I . . ,

C H ,

*

C H 2 C I 2 :

H - c - c i ;

H

I

O

* C 2 H 4 O 2 :

H - C - C

H ^ 0 - H

h a y C H 3 C O O H . i

,

H - ^ - O - C - H

h a y

H C O O C H 3

O H

.

H -O-C-C

5 f s | . ; y ' | a ) ( j l

sS wi?

H H

^ , H H . .

*

C 2 H 4 C I 2 :

H - C - C - C I

H

CI ^..

H H

• i A ' ' V A ; « ( T H i , v . H - C - C - H

I I

C I C I

h a y C H g - C H O ^>|ri

i ^ q

i j

1

i . i a i i K u H f i i d q s n o b i i i

u V i f i

} « b

j j r f M f i i f t r f j

h a y C H i C H C l :

1 ir }

h a y C H 2 C I - C H 2 C I .

.) * C j H s O :

C H 3 - C H 2 - C H 2 O H

; C H ^ - C H - C H g

i n

* C 4 H , o O ;

C H 3 - C H 2 - C H 2 - C H 2 O H ; C H , - C H , - C H - C H

3

—v^xa2

^~^ -3

O H

OH

C H g - C

- C H 3 ; C H 3

- C H

- C H 2 O H .

• AO. *

KH Jg G V I E

B a i 1 3 NhCng

cha't nao sau day la d o n g d i n g c u a n h au , d o n g phan cua nhau ?

a)

C H 3 - C H = C H - C H 3

b )

C H 2 = C H - C H 2 - C H 3

c) C H 3 - C H 2 - C H 2 - C H 2 - C H 3 d) C H 2 = C H - C H 3

e) C H 3 - C H = C H - C H 2 - C H 3 g ) C H g = C H - C H

- C H 3 -)

C H 3

Si I

lfU.

_ n —

i

^ n . o .

I

h )

C H 3 - C H 2 - C H 2 - C H 2 - C H 2 - C H 3

i )

C H 3 - C H 2

- C H

- C H 3 .

G i d i

* C a c cha ' t l a d o n g d a n g c u a n h a u :

d , a v a e ; d , a v a g ; d , b v a e ; d , b v a g ; i v a h ; e h . . ,

* C a c c h a t l a d o n g

p h a n

c u a

n h a u

: a va b; e va g; c va i . - •

B a i 1 4

a ) H a y v i e ' t c o n g t h u ' c p h o i c a n h c u a C H 3 O H ( m e t a n o l ) v a c i i a C H C I 3 ( c i o r o f o m ) .

b ) H a y v i e t c o n g th iJ c

p h o i

c a n h c u a

C 2 H 6

( e t a n ) v a

C 2 H 5 O H

( e t a n o l ) .

. . . , . ' , G id i . ' ' ,

a ) *

C H 3 O H

H ^ V , , . * C H C l 3 : H

H O ^ " C I

b ) * C 2 H , : H H * C 2 H 5 0 H : » H

\« . \

H ' H i l I C — C - * H

r

H . . i i i | C - C - * H

V \1 t : : . i t h , < / - : Hut

4 \

H

H H O H

,,, r

<

W r

c u i a 2 . . , ,

-

X i d e

(t i i i h i t / e d m a t l u / m j e u t a , t h a n h p l i d i t e d e i i t f i i i f j e i i t o t r m t q

c h u t h f f l l et f

i ) -f;>i.-vJ.(}v--

. 4\ j l i ^ f , )

ym

C.

I J , O f n i

r; i V ) , ( J -~ -

-

£ g f i e t n i f f t h i i e i t t f i i q l u n i i h a l .

, , ) , , . , „,,,,,v,y,) . > . 4 , j

j . ^ j ; , ^ ; ^

. ^ B A I T A P M A U V A B A I T A P N NG C A O

B a i 1

Ox i h 6 a h o ^ n toan 0,6g hdp cha t hi?u cd A th u

duTdc

0,672

l it

C O 2

dktc)

v j

0,72g

H 2 O .

T in h than h phan % k ho i liTdng cua cac

n g u y e n

to tron g ph an tuf cha't A .

G i d i

Kh o i l i f d n g c a c

n g u y e n

to : viul^h^

n j o r , -

m c = 1 2 . n e o , ^ 1 2 . ^

=

0.36 g)^

'

= 2 . n H , o = 2 . ^ = 0,08 g) "

i ' ^

= > m o = 0,6 - (0 ,36 + 0,08) = 0,16 (g )

7/23/2019 Tr1-30(3)


Phan d?ng

va

phudng

ph p giS H6a hpc 11 HQu co -

D

8

XuSn ung

Thanh phan % khoi lifdng cic nguyen 16' trong A : d, i jy.

- f ir J i' ' 'J

^rNw

% C = • ^ x l O O % = 6 0 % j r i H . / ,

0.6

• • • : >

; i f ^ ; r . ^ : j f : > . . . . f t - : .

% H = 100% = 13,33% ^ % 0 = 26,67%. £ >-

i H ^ * ^

> ^^

V 0,6 ' •

Bai 2. T i n h

k h o i

l i f d ng mol phSn tuf cua cdc chat sau : j j , ; . , ,,,,,

a) Chat

A

c5 ti k h o i hdi so vdi khong khi bkng 2,07.

b)

The t ic h hdi cua 3,3g chat X b^ng the t ic h cua l,76g kh i oxi (do cf ciang dieu

k i f n

ve nhiet do, ap sua't).

Gidi

•^ ^ ^ ^

*

a) d

k k

= ^ = 2,07 ^

M A

= 29.2.07 = 60

(dvC).

, ^g ,, .^ .^ .

b

Ta c6:

no2

= ^ ^ = 0,055 m ol =>

n

= = 0,055

m o l = > M A

= = 60

(g/mol).

32 0,055

Bai

3. Dot chay hoan toan 0,282g hdp chat hiJu cd

A,

san pham sinh ra cho qua

binh

dtrng

C a C l khan va

binh

dUng

NaOH,

thay

binh diTng

CaCla tSng them 0,194g,

binh

dufng NaOH tSng them 0,8g. Mat khac dot chay liTdng chat

A

tren thu difdc

22,4ml khi n i t d

(dktc).

Xac

dinh

thanh phan %

k h o i

liTdng cac nguyen to trong A.

Gidi

Cho san pham chay qua

binh dtfng

CaCb : mi = 0,194g

=>mH20

=0'194g

= Qua

binh dirng

NaOH : m = 0,8g=> m co , = 0.8g

K ho' i l i f d ng cac nguyen to :

m c = 1 2 . M = o,22(g); mH

= 2. ^ = 0,02 (g) . ^ - W i ^

44

18

0 0224

' . V . . J' 1.

t

N z

= = 0.001 (mol) mw = 0,001 x 28 = 0,028 (g) . | |

Thanh

phin

% khî l i W n g cac nguydn ttf f * « *»*'*^ '

% C = - ^ ^ x

100% =

78,01%;

%H = - ^ ^ x 100% = 7,09%

..- 'f

0,282

- * v . > M ; - i i . , vi^ uuu

0,282

fu,.//

iu;.'>fi

ix O , | ijiitt

% N = ° i ^ x 100% = 9,93% : ^ %0 = 4,97%. ^ f^^

0,282

ai 4. Oxi h6a hoan toan 4,92mg mot hdp chat A chiJa C, H, N O ro i cho san

pham Ian liTdt qua

binh

chiJa

H S O 4

dam dac,

binh

chiJa KO H thi thay k h o i

l iTdng binh chiJa H S O 4 dSc tang them 1,8

I m g ,

binh chiJa K OH tSng them

10,56mg.

d

thi nghie

m

khac, k hi nung 6,15mg hdp chat

A

do vdi CuO thi thu

difdc

0,55ml (dktc) khi n i t d . Hay xdc

djnh

h^m liTdng phan trSm cua C, H, O va

N d

hdp chat

A.

Gidi

*

Kho i lifdng binh chiJaH2S04 tang them 1,8Img chinh la mH20=>mH20 =1.81m

*

Kho i

liTdng

binh chiJa

KO H

tang them 10,56mg chinh la mco, =>

"^C02

^0,56 mi

Kho i

l i f d ng cac nguyen

to :

10,56

1,81

m c = 1 2 . ^ ^ ^ = 2,88(mg);

mH

= 2 . ^ = 0,2 (mg)

44 18 • I

0 55 4 92

mN(4,92.g)

=

2 8 . ^ x - i — =

0,55(mg) > • '> mmiiy. j : } . , H „ p : A nr v

=> Thanh phan %

k h o i

l i f d ng

% H =

2,88

4,92

0.2

4.92

^ x l O O %

= 58.54% , ^ ,„ £,... s fM ) i xSk ^ ^ ^ ^ ^

x 100% = 4.065%

% N =

^ X

100% = 11,18% - ^ ^ ' ^ - ' ^ ^ o ^ ' ' ^ ^ ' ' ^

r:> % 0 = 100% - %C -

% H

- % N = 26.215%.

Bai 5.

Hai chat

hffu

cd

A

va

B

c i in g

chtJa cac nguyen to C, H,

O.

Khi dot chay moi

chat deu phai dung mOt l i f d ng oxi b^ng 8 Ian l i f d ng oxi c6 trong moi chat va thu

difdc

l i f d ng khi C O va H O theo ti le k h o i l i f d ng C O so vdi k h o i l i f d ng nuTdc =

— . Xac

dinh

cong thiJc ddn gian nhat cua moi chat ? ' ' ' ' -

• CUM

Theo dieu

ki?n

de bai thi

A, B la

hai chat dong phan cua nhau. •

Dat CTTQ cua A, B la

CxHyO^.

. i ,, ,, ,

Mrira?- ^ .-YV .^ „

Phifdng t r i n h dot chay : , . , i

CxH y O , +

^

y

x + ^^

4 2

O > X C O + - H j O ,

Theo de : 32

y z

x + - - -

. 4 2)

= 8x l6 z <:>4x + y= 18 z (1)

' I K ' ' -

m c o , 22 44x 22 44x 22 .

^ =— o

= — <=> = — =>y = 2x

m 9 y 9 9y 9

The y = 2xvao(;i) => x = 3z=> z = - '

T a c 6 t i l $

x : y : z = 3 z :6 z : z = 3 : 6 : 1

V a y cong thiJc ddn gian nhat cua hai chat la CsHeO. ' ' ' '

1

7/23/2019 Tr1-30(3)


Phan

d j ing

va phuong ph^p

gi i

H6a hpc 11 HOu

co

- Dfi Xuan Hung

Ba i

6.

Vitamin

A (retinol) c6 c6ng thiJc phan tuf

C 2

( ) H 3 ( ) 0 ,

vitamin

C c6 cong thiJc

phan tur la CfiHsOfi.

a)

Vie't

cong thiJc ddn gian nhat ciia moi chat. 'li,, ,.

b)

Tinh t i 1? % ve kho'i li/dng va t i ie % so' nguyen tur cua cac nguyen to d vitamin

A

va vitamin C.

Gidi ,

a) Cong thiJc ddn gia n nhat cua cdc cha't : ^

Vitamin A : C 2 ( ) H 3 ( , 0 ; Vitamin C : C 3 H 4 O 3 . ,0 • 7. ,

b) * T i 1

Q

% ve khoi liTdng cdc nguyen to :

+ Vitamin

A :

C a j H j d O

(M = 286) ' '' '' ' '

% m c = - ^ ^ ^ x 100% =83.92% - ' * ^ > ; ' : ' - ,

286

mH

— x l O O % = 10,49% ^ ' ' ^ '

286

=> %mo = 100% - 83,92% - 10,49% = 5,59%. ' - ,

+ Vitamin

C :

QHsOfi

(M = 176)

6x12 •

%mc = X 100% = 40,9% , , , ,

176 ' I f , ^ i .1' ' i V lih

m H = — x l 0 0 %

= 4,55%=^ %mo = 54,55%. ' ^ ' ^ ' "^ ' '

176 ^.1 . | w ' i

* T i 1? so nguyen tuf cdc nguyen to: , / j

+ Vitamin A : C 2

( i H 3 „ 0

(51 nguyen tuf)

20

n c = — X 100% =39,22%

51 ' \ b t i i - . ^ . i r f i h - i c ^ i T

n H

—

x l O O % =58,82% ^ %no =

1,96%.*'

' ' ' '

+ Vitamin

C :

C6Hx06

(20 nguyen

tijf)

; v ;

% n c = - - x l 0 0 %

=30 % '

20 - ^ ^ . .

n H

—

x l00%

=4 0% => %no = 30%. f' •

B k i 7 . Chat A chiJa C, H . O c6 ti 1$

khoi li/dng

mc : mo = 3 : 2 kh i dot chdy h6't A thu

diTdc

C O 2 va hdi

niTdc

theo u ' I 9 the tich V c o V H J O =4 : 3 (cdc the tich do

d

cung

dieu ki^n

nhi§t

dp, ip suat). Ti m cong thiJc ddn gian nha't cua A. ^,

(TrichTSDHSuphamKythu^tTP.HCM)

Z^:

Gidi

D$tCTTQcuaA:C,HyO , . ,^^, 'v. ^ ^ . f r s ^^ i r ^ : ' : -tf

| i • • • • ;

14

PhiTdng

trinh dot chay : C x H y O +

T h e o d I : ^

3 12x

•

= — <=>

mo 2 16z

y z

X —

. 4

2)

X

=>

z = —

2

X C O 2 + y H j O

(1)

Ta

C O :

V,

C O 2 _ 4

^

JL,^1

H20

y

3

2

3

X

y =

3x

(2) ,

Ta c6 t i le x : y : z = x : ^ x : ^ = 2 : 3 : 1 .r^:^^^^'": M

2 2 { ••••.i-n

Va y cong Ihu'c ddn gian nhat cua A : C 2 H 3 O . ^ = ' '

Ba i 8. Hay thict lap cong thiJc ddn gian nhat tCf cac .so

lieu

phan

tich

sau :

a) 70,94%C; 6,4%H; 6,9%N; con la i la o x i . i r i o n ' . .r ;o 5 '

b) 65,92%C; 7,75%H; t , ; , c o n lai la oxi.

l f :> H : \ tP kv j, ;t v |: :, . ',

'

., ,;• X^:,, ,,,, Gidi

. •iîl i orf anf

giiW utkj I.

a) Than h phan % cua oxi • 'jiinod:?^'^

n i ^ q di

%0 = 100% - 70,94% - 6,4% - 6,9% = 15,76%

Dat

CTTQ cua chat hi lu cd :

C,HyO,N,.

f ; ^ ^

%C %H %0 %N

Ta

c6

ti

1^ X : y : z :

t =

12 1 16 14

70,94% 6,4% 15,76%. 6,9%

12 1 16 14

= 5,91:6,4:0,985:0,49= 12:13:2:1

Vay cong Ihu'c ddn gian nhat la C 1 2 H 1 3 O 2 N . , „ ;

b)

Thanh phan % cOa o x i : H .

X : y : z

%0 = 100% - 65,92% - 7,75% = 26,33%

Dat CT TQ cua chat hffu cd : C x H y O . Ta c6

ti

1?

^ 6 5 ^ . 7 J 5 % ^ 2 M 3 % ^ 5,49:7,75:1,64 = 10:14:3

12 1 16

Va y cong thu-c ddn gian nhat la C 1 0 H 1 4 O 3 .

Bki 9. Dot chdy ho^n loan 4,6g mpt hdp chat huTu cd A thu diTdc 3,6g H 2 O . Dan khi

C O 2 sinh ra vao dung dich ni/dc voi trong 0,1M thi difdc 8g ket tua v^ dung dich

Y . nho tiep dung dich NaOH vô dung dich Y thi difdc 3,5g ket tua nffa. Xac

dinh cong Ihu'c ddn gian nhat cua X.

Gidi

T a c 6 : m H

= 2 . — =

0,4g

18

CO2

+ Ca(0H

)2

CaCOj i +

H2O

0,08 mo l 0,08 mo l

7/23/2019 Tr1-30(3)


Phan jing

va

phudng phip

giSi H6a

hpc

11 HQu

co

D

8 Xuan Hifng

2 C O 2 C a O H ) 2 C a ( H C 0 3) 2 ^

? 0,07 mol 0,035 mol ' ' « ;

Ca(HC03)2 + 2NaOH C a C O j i Na2C03 + 2 H 2 O

0,035mol

' • ' '

0,035mol

yi ' ; i; i,^

SomolC02: nco2 =0>08 + 0,07 =0,15 (mo l) .

r=>mc=

12x0,15= 1,8 (g) ' ^ ^

^ , , f f V

=> mo = 4,6 - 0,4 - 1,8 = 2,4 (g) 1'

Dat CTTQ

cua A la

C.HyO,.

Ta c6

ti

le :

, £ , , . ,

,

v

x : y : z = ; : .

M M ; M =

o,15 : 0,4 : 0,15 = 3 : 8 : 3

12 1 16 12 1 16

yv

V a y

cong thtfc ddn gian

nhat

cua X la

CsHgOs. •> iah- ~iSi

i i l r f i

y \M

J

B ai 10.

KM cho 5,3 gam hon hdp gom etanol

C 2 H 5 O H ; H ' ' i ? i ' , f l ; C i v ^ , i i '

/ j ;

va propan-l-ol

C H 3 C H 2 C H 2 O H

tac dung vdi natri drft hu diTdc 1,12 lit khi (dktc) .

a)

Vie't

phifdng

trinh

hoa hoc cua cac

phan iJng

xay ra.

b )

Tinh

thanh phan

% khoi lifdng cua

moi

chat

trong hon hdp.

S' d i n M T

{a

a)

C2H50H

+ Na

C2H50Na

+-H2 ;

X

mol ,

—mol

,:'S*C,d, ,S'dt<ci_, :ki2,

C H 3 C H 2 C H 2 O H

+ Na

>

CHsCHzCHzONa + ^ H ,

2

S^molH2:

n H , = - ^

= 0,05 =

+ y

'

V v : - . .v i v . c . * 0 ( ) ( . . X ) ^ '

, ,, ^

^ 22,4 2 2 ^ ^ _ , , , , i i i , 3 , j t o . t f O r { i i M 3 M p ^

Ta

c6

h? phudng

trlnh

:

-

+ = 0,05 fx = 0,05 (mol) .•

s ; y • K

2 2 => <

<cn C O y = 0,05 (mo l)

46x + 60y = 5,3

. t , * veV

b )

Thanh phan %

khoi

lOdng moi

chat:

r ^ ^ - i m

3^,* a. ,wsFi i b i i J(? M «

%C2H30H=Q' ^^^^

100% = 43.39% :

=>%CH3CH2CH20H

= 56,61%.

:>tU^

J l f l n f t? . ., , ^ . ^ .,. .fltfc^ :

.

t.;.,,iV

- : 7

- . , S '

* j i m '

V ; - & . r f i T

(DoMig 3. <£Afi. eS^ntf,

thi te

fthdit tilt :

<T)t/a

ocLO cang, thiie

fteiti Qtatt

nhdi

i

<

'

<T)ita oda khai liMtiq. hoae % khoi lii&iig. ede nqiujjen t»

î/a i9dn the tieh (phiMuj fthdfi khi nhien ke)

B A I T A P M A U -v^

B ai 1.

Dot chay ho^n toan 5,6 lit

chat

hiJu cd A d t h i

tich

kh i thu difdc 16,8 li t

C O 2

va 13,5g,

H 2 O .

Xac

dinh cong thiJc phan tuT

cua A

biet

t i

khoi

so

vdi hidro

la 21

(cac khi do

a

dktc). , , ,

Gidi

5 g y,C:VO£l A f i t i p • ^ ^ ) ; -

So mol

chat

hu-u

cd

A:

n = — ^ =

0,25 (mol)

>, „ \

22,4 :>iA,..,.f= ,r|.v.o.H;:>

Khoi

liTdng

mol phan

tuT

A: M = 21 x 2 = 42 =>

m

= 0,25 x 42 = 10,5 (g)

16,8

22,4

,tyr

Khoi lifdng

cac nguyen to' : mc = 12. ^^ = 9 (g) j

. 3 5

06/

m H = 2 . - ^ = 1,5 (g)=:> mo =

10 ,5-9-1 ,5

= 0 n l " , . i , , ^

J 1 I'M

11* 1

V a y

trong

A

khong

C O

oxi.

, • 1

Dat CTTQ cua A :

CJiy.

Ta c6 ti

I9

: x : y = — : — =

0,75:1,5

= 1:2

,

i •

12 1

^

Cong thu-c A c6

dang C H 2 ) „ .

,..

. l

„, „ .,,

Taco: 14n = 42 ^ n = 3. V$y CTPT A :

C 3 H . . - ^^- ^-

>

^

B ai 2.

Limonen

la mot chat c6 mdi

thdm diu difdc tdch

tilf

tinh

dau

chanh. Ke't

qua

phSn

tich

nguyen

to cho

thay limonen diTdc

cau tao tiJ hai

nguyen

to C va H,

trong do C chiem 88,235% ve

khoi

l i f d n g . Ti

khoi

hdi cua limonen so vdi khong

k h i

gan

b^ng 4,69. Lap cong thiJc phan

tuT

cua

hmonen.

, '

Gidi

Thanh phan %

kh6i li/dng

nguyen to hidro :

% H = 100% - 88,235% = 11,765%

' ' '^^'^

^

Khoi

liTdng

mol phan tur: M = 4,69x29= 136 ,,

Dat CTTQ cua limonen :

C x H y .

Ta c(5 ti 1§ : . ,

%C %H 88,235% 11,765%

'^'^'''^^'^^'^'''^^'^^^^

n

I 12

1

= 7,353 :

11,765

= 5 :8

=>

Cong thtfc limonen

c6 dang (C5H8)„. ' • V,t> - '

Tacd : (5 x 12 + 8)n = 136=> n = 2 . v i ,

V $ y

CTPT cua limonen

m C j p H i i .

THLT VIEN TMH

BSNH THU N

-zr::^

'

A 9

17

7/23/2019 Tr1-30(3)


Phan dgng va phuong phap giSi H6a hoc 11 Hau co - P i Xuan Hung

B a i 3. Dot chay 100ml hdi cha't hffu cd A chtfa ba nguyen to C, H, O vdi 450ml

O 2

la 'y dff trong khi nhien ke. Sau phan ffng the tich hon hdp khi va hdi thu difdc

650ml. Sau khi cho hdi nffdc ngffng tu hon hdp chi con 350ml va sau khi cho Igi

tiep qua NaOH chi con 50ml. Xac dinh cong thiJc phan tuf cua A .

Gidi

The

tIch cic

chat:

ui r

= 50ml

=>

Vo2(ptf)= 4 50- 50 = 400ml

Vco2

= 35 0- 50 = 300ml

V H J O =6 50 -3 50 = 300ml

Dat C T T Q cua A la C x H y O ^ .

t

i 1' I rt ? •

C,H„0, +

T a c6:

100ml

1

y z

x

+ — - -

4 2

400ml

O , X C O 2 + ^H j O

300ml 300ml

;-.,.X

,<'):H

100 300

_ L = _ y

100

1

2.300

4~ 2

x = 3

y

= 6

« x + i - - = 4; The x = 3,y = 6 = > z = 1

4 2

00 400

V a y CTPT cua A la C j H f i O . . ,• ; _ , ^

B a i 4. Hay thie't lap cong thiJc phan tuf hdp chat

hffu

cd trong moi trifclng hdp sau :

a Dot chdy ho^n toan lO.OOmg hdp chat hCu cd Y sinh ra 33,85mg

CO2

6,95mg

H2O .

Ti khoi hdi cua hdp chat Ad doi vd i khong khi la 2,69.

b ) Dot chay hoan toan 28,2mg hdp chat hffu cd Z vk cho cdc san pham sinh ra Ian

Iffdt

di qua cac binh dtfng CaClz khan va KO H dff thi thay binh

CaCl2

tSng them

19,4mg con binh K OH tSng them SOmg. Mat khac, kh i dot 18,6mg chat do sinh

ra 2,24ml khi nitd (dktc). Biet rkng phan tuf chat d6 chi chffa mpt nguyen tijf nitd.

J\ tii t la t ' 1

1

t .i

j f i 1

a)Kho-ilffdngmolphantuf:M = 2,6 9x2 9 = 78 ^^^ ^ , ,, ,

„ V M , H

. U

Khoi lifdng

cac nguyen to : ; W i .v

33,85 ' '

m c =

12.

mH = 2.

44

6,95

18

=

9,23 (mg)

=

0,77 (mg)

y • f

=> mo = 10,0©- 9,23 - 0,77 = 0=> khong

c6

oxi.

D 5 t

CTT Q cua hdp chat hu»

c eJ :

C^Hy. Ta c6 d I9 :

x : y =

^j^:

.2:21 = 0,77:0,7 7 = 1:1 => C T c d ' d a u ^ H ) „

T a

c6 : 13n = 78 => n = 6.

V Hy CTPT1 CfiHfi.

.

b ) Theo de ta c6 : ,. , ;„,,,, ;,„,

19 4

^H

= 19,4mg => mH = 2. = 2,15 (ja0

18

80

mco2 =80 mg => mc = 12. — = 21,81 mglr

^ 44

m 2,24 , iT¥

M • ) ' • ' ' ' ' ' :

niN (28,2mg) = 4,24 (mg)

=> mo = 28,2 - (2,15 + 21,81 + 4,24) = 0 ' h ,f ;;a .

J

V a y khong

c6

nguyen to oxi trong hdp chaft hffi ccfe l i l i ) i ; b 0 ',

D a t

CTTQ : C

^ H y N .

Ta c6 t i 1? : J

OS :

21,81 2,15 4,24 , „ ^ ^ , ^ „ ^ :iq u t i d i ^06:)

x : y : z = — • — : — — : — — = 1,82 : 2,15 : QB .

,:;;

;

- = 6 : 7 : 1 => CT cd dang (CeBilS^/ J r i ^ f)^?'

V I t rong phan tuf chi chffa mgt nguydn tuf nitd

=>-m=Û.

' "P

'^ "'^ i •

V a y

CTPT chaft

hffu

cd

m C 6 H 7 N . .

-"5' < « ^

B a i 5. Phan tich chat hffu cd X chffa

C , H ,

O ta c6 :

mc : mH

: mo = 2,24 : 0, 35 7^ ^' - '

' ' ' ' ' ' '

" '

a Lap cong thffc ddn gian nhat cua X.

t '

b ) Xdc dinh cong thffc phSn tff ciia X biet 1 gam J i ^ l a m bay hdi c6 th^tich

1,2108

l i t a 0"Cva 0,25 atm. '

a

D a t

CTTQ

c a a X : C x H A

. T a c 6 t i l : ' i : '

mc . mH . mp _ 2,24 . 0,357 . 2 '

>, r

j

, ~ 12 • 1 • 16 " 12 • 1 -16 .

= 0,187 :

0,375

: 0.125 = 3 :6 : Z:

,:, |*^^'

=>C6ng thffc

ddn giSn nhat cua X :

C 3 H 6 O 2 .

' •

^

^ u /

b ) S ^m ol X : nx = — =

^ ' ^ ^ ^ ^ ' ^ ^ ^

= 0,0135(mot) frfîwlspft.

R T ^ ^ 2 7 3 v i . ,

. , 273 , ,

Kh6i Iffdng mol cua X : Mx = — ^ - — = 74 , < ,

0,0135

(^. - ,

Ta c d : (3 x 12 + 6 + 2 x 16)n = 74 =i> n = 1

V § y cong thffc phan tff cua X la C 3 H 6 O 2 . r ' f f ) c >

19

7/23/2019 Tr1-30(3)


Phan d ng

va

phUdng phAp Q\i\a hgc

Hi?u

co - D5 XuSn

Hung

B a i

6 . TO

t i n h

d a u h o i , ng i ^d i ta t a ch d i f d c a ne t o l , m p t chaft t hd m d u dc d i i n g de san

x u a t keo cao s u. A n e t o l c6 k ho i l i f d ng mol phan tu f bang 148g/mo l . Pha n t i ch

n g u y e n

to

cho t ha y a n e t o l

c6 C

= 8 1 ,0 8 % ; % H =

8 , 1 % ;

c o n l a i

la

o x i . L l i p c o n g

thuTc ddn g ian nha t va cong th iJc phan tu f cua ane to l .

„ . „ : i d i

.. .

,., ,„

M = 148g/mo l .

T h a n h p h a n % o x i : 0 = 1 0 , 8 2 %

D a t C T T Q : Q H y O ,

"

^ . %C %H %0 8 1 , 0 8 % 8,1% 1 0 , 8 2 % ,

T a CO t i 1? : X : y : z = : : = : :

, 12 1 16 12 1 16

= 6,8 : 8 ,1 : 0 , 6 8 = 1 0

:

12 : 1 ,

V a y c o ng thuTc d d n g i a n n ha t cua a n e t o l : C i ( ) H | 20

=>

CT C O

d a n g

(CioHi20)„.

\

Taco: (10

x 12 + 12 +16)n =148 = > n = l » , ^ > r

V H ,

W , r r t » i

V | iy cong thiJcphan

t f

cua anetol

la

CioHuO.

( t

j

>l 3

Bai 7. D o t c h a y 2 0 0 m l h d i 1 c h a t Mu cd A chiJa C, H, O t r on g 9 0 0 m l O2 , the t ich

h 6n

h dp k hi thu di/dc 1^ 1,3 l i t . S a u k h i

cho

hd i n i f d c ng i f t i g t u , ch i

con

7 0 0 m l .

T i e p t heo cho q u a d u n g d i c h KOH dac c h i con 1 0 0 m l (cac the t i ch do d cUng

d i e u k i ^ n ) . T i m C T P T cu a A.

• - 'Kiisn'

n g t f n g t u . CO K O H dac

Sd

do

p h a n

t ich de

b a i :

m'kA +

O2

2 0 0 m l 9 0 0 m l

- C O 2

•

H 2 O

-O 2 d i O

- H2 O

1 3 0 0 m l

O 2 d i f

7 0 0 m l

D i f a

vao

sd

do

ta

t i n h

d i fdc :

Vo^ph int fng

= 9 0 0 - 1 0 0

=

8 0 0

m l

Vco^ = 7 0 0 - 1 0 0 = 6 0 0 m l

VHÔ = 1 3 0 0 - 7 0 0 = 6 0 0 ml

P h i f d ng t r i n h p ha n i Jng chd y :

O 2 c o n d i r

1 0 0 m l

1

Q H , 0 , + ( x 4

~ ) 0 2

1

4 2

^ 4

2^

2 0 0 800

= > X = 3;

y =

6 va z = 1

X C O 2 - H 2 O

2

J

Ml i

1

h nJÎ

C T P T cua A , l a C j H f i O ' ' W ^ ^ ^ i ^ ^

o n

Bai 8 . T r p n

4 0 0

c m ^ h o n h d p h d p ch a t Mu

cd

A va n i t d v d i

9 0 0

c m ' o x i d u f r o i d o t .

T h e

t i ch

ho n hd p sa u p ha n i Jng la 1 , 4 l i t . Sa u k h i

cho

hd i nufdc ngu ng tu th i con

8 0 0

c m ^ t i e p t u c c h o q u a d u n g d i c h K O H t h i c o n

4 0 0

cm \T cua A l a :

A . C 2 H 4 ' B . C H 4

S d

do

p h a n

t i ch de

b a i :

C . C 2 H 6

iai

C H y

(A)

N 2

+ O? dU

t

r c o 2 ^

H 2 0

N 2

^ 0 2 d i r

4 0 0 cm ^ 9 0 0 cm ^ 1 4 0 0 cm

Dt fa vao

sd do

ta t i n h

d i f d c : s, .

j

ngutig tu

-

H2O

C O 2

N 2

'J

ii'ii

^ qsof lq ,i 8 0 0 c m

D . C,u,

d d K O H

N2

0 2 d i r

4 0 0 c m

3

V c o , = 8 0 0 - 4 0 0 = 4 0 0

c m ^

s »»>

= 1 4 0 0 - 8 0 0 = 6 0 0 cm ^

V ( 0 2 c 6 n d U + N2 ) = 400 cm ' »'fH:;

^

Phi/dng

t f i n h

p h a n i Jng cha y : | i O; , f

• • CAM , -

Q H y O ,

+ (x + ^ - - ) 0 2

4 2

1

T a c d : V o j p h a n J n g =

Vco

+ 2 ^ H 2 0

=> condt/ = 900 - 700 = 200 cm '

= 4 0 0 - 2 0 0 = 2 0 0 c m ' =>

Ta cd t i 1? :

V r

» : V.

-> X C O 2

H2O

= 4 0 0 + — = 7 0 0 cm ' M'u , ,„•.•

^ C x H y = 400 - 200 = 200 cm'

C Hy •• V 0 2 phan tfng

: Vco

: VH^Q = 2 0 0 : 7 0 0 : 4 0 0 : 6 0 0 =

2 : 7

: 4 :

6

V i e t l a i p h i /d ng t n n h phan u-ng : 2 C x H y 7O2 -A- 4 C O 2 6 H 2 O

A p d u n g d i n h l ua t b a o t o a n ng u y en t o ' C va H ta cd :

2 x = 4

2 y

=

6 . 2 '

x = 2

y = 6

=>

C T P T c u a A l a : C 2 H 6

Bai 9. Cho 5 c m ' C,Hy d th e k h i vdi 3 0 c m ' O 2 (lay diT) vao k h i n h i e n k e . S a u k h i

bat t i a lu fa d i^n va l a m l a n h , t r o n g k h i n h i e n

ke

co n 2 0 c m ' m a

5

c m ' b i h a p t h u

b d i d u n g d i c h KOH. P ha n con la i bi hap t h u bdi p h o t p h o . Tim C T P T cua

h i d r o c a c b o n . , 1

T h e o de b a i : V c o ^ = 15 c m ' =^

= > phan ifng

= 30 - 5 = 25 cm '

i d i

. ,

V o d^ = 2 0 - 15 = 5 c m '

2 1

7/23/2019 Tr1-30(3)


ti tn âng^ phicnq phap giai Hfa^pc 11 Hi?u

co

-

P8 Xuan HiMg

•ûang trinhphan tfngchdy:

Q H y + (x +

^,)02

->-xC02 + ^HzO

5 25 15

=>x = 3vay = 8=> G T P T la C jH s .

B ^ O : Cho vao khi nhiSnske 10 cm ch it hffu cd A

(chiJa

C. H, N), 25 cm

H

va 40

tsd l^ z

Bat tia

liJa

di$ nn:ho hon hcfp no. Chuyen hSn

hcJp

khi nhan

diTdc

ve dieu

akiÎi'ban dau,

H2O

ngiteg tu he't, thu diTdc 20 cm hSn hdp khi, trong

66 c6

10

••Wf bi NaOH hap thu vi 'S cm hi photpho hap thu. C T P T cua A la:

i ^

ATCH s N

B. . C 2 H 7 N

C

. C 3 H 9 N

D.C4H,|N a ,

-Sdô phan tfng:

O 2 c6n dir

C O 2

N2

0 ^ ^ A ( C , H , N ) +

Photpho hafp thu O => V Q ^

j i

= 5 cm'

©oingdich NaOH

hap thuC O => VQO 10 cm' v>v

20 cm'

=> = 2 0 - ( 5 + 10) = 5 cm'

25

3

Wolchayhki ro

=

- = 12,5 Cm

^ .-iih b:) e l

(dotehayea A v i

hidm)

= 4 0 - 5 = 35 Cm'

%)j (dcfichiTy A = 35 - 4 2 , 5 = 2 2 , 5 cm'

-Ph«ang

trinhphin tfngxAdy: •

^gatf

n a i q ritvn^ g0biniq i fi j fS^V;

C^yN, + (x

+

^)02 X C O 2 + ^Hj O +

- N 2

•

feolriniasOTbqA

4 2 2

=>

VH,o<J

..*ftch4y A) = ^(V^, - Vco,) =

2 . ( 2 2 , 5

- 1 0 ) = 25 cm' ,

T a cd

tH?

the tich:

V A : - V e o ^ : V H ^ o : V N ^

= 1 0 : 1 0 : 2 5 : 5 =

2 : 2 : 5 : 1

fod

= > A : C : H : N

= 2 :2 :40 : 2 = 1 : 1 : 5 : 1

C T P T cua A la CH5N

,rii i\i .>..iO -;I ,»vrt

•,m ,ix^ .},> i i(6o;«iT

BAI TAP NANG CAO

Bai

1.

Mot hdp chat hilu cd A chuTa

5 4 , 8 % C ;

4 , 8 % H ; 9,3%N con lai la O. Cho biet

ph§n

tuf khoi cua nd la 15 3. Xac dinh c6ng thiJc phan tuf cua hdp chat. V i sao

phSn

tuf khoi cua c c hdp chat chtfa C, H, O la s6'

ch£n

ma phan tuf khoi ciia A

lai la so' le (khong ke phan thap phan) ? •

u

.• , |,

Gidi

% 0 = 1 0 0 % - 5 4 , 8 % - 4 , 8 % - 9 , 3 % =

3 1 , 1 % irf N

M A = 1 5 3 , / • •d i j oo^b j r i BtV

:r rs-r?

;

Dat C T T Q cua A : C H ^ N , . Ta c6 ti 1? : . i „ , : ^ ; , f c , M

, ^ ^ , g , « « > d vftrSo m

5 4 ^ , 4 ^ , 3 U . 9 ^

X : y : z : t =

12 1 16 14

= 4,57:4,8: 1,94 : 0,66 = 7 : 7 : 3 : 1

=> C T cua A cd dang C 7 H 7 0 3 N ) „.

Tac6 : 153n= 153 => n = 1

Vay C T P T cua A :

C7H7O3N.

* Phan tuf khoi cua A la so le vl nguyen to nitd c6 h6a tri 16 => nguyen tuT H la so

B^ i 2. Cho 400ml mOt hidro g6m

N C

^ H y

vô 900ml O dif roi dot. The tich hon

hdp khi thu dtfdc sau khi dot la 1400ml. Sau khi cho hdi niTdc ngifng tu thi con

800ml hon hdp ngi/di ta cho loi qua dung dich K O H tha'y con 400ml khi. Xac

dinh cong thiJc phan tuf cua hidrocacbon tren, biS't the tich

cic

khi deu do

d ciing

dilu ki^n nhi$t dp va ap sua't. ^^\,.

Giai

f

(,5fV(

tO.O

C x H y

y^

x + -

4

O2

>

X C O 2

+ ^H20;

• f . A S ;

Si

it) nvA A ;fe b

hy-

x + y

4

Dat the tich cua C ^ H y la a

=400-a

Theo de bai ta c6 :

ax

v j \

;

:Mit iirtJ

}flA

c ,,-

V H 2 0

= 1400-800 = 600ml =

« c

V c o 2

= 800 - 400 = 400ml = ax

^ j

|,

Vo2du

=900-

X ^ I

4

11

V o 2 d ^ + V N 2

=400 <* 90 0-

x + y

4;

a + 40 0- a = 400

23.

7/23/2019 Tr1-30(3)


Phan dgng va phuong ph^p giai H6a hpc

11 Qu eo

- B5 Xufln Hung

Vaytacd: ay

= 1200 (1)

• v ax = 400 (2)

G l a i ( l ) ,

(2), (3)

a = 9 00 - a (3) ' • 'i«i#iffc

• • - • , - i t i V j i'kih '•'pli ri/ ;y ,fy§^

a = 200ml fSsirj

'Îv;

Vjnv'i/i) bl u, ii lli

• X = 2 ,:ut}

y

= 6

•=

' ' r ; i '

.P

'fli

s.i'

•>]::

V a y CTPT cua hidrocacbon la C2H6.

B ai

3.

Dot chay hodn toan 3,6g chat huTu cd A (C, H, O) b^ng 4,48 l it O2 (dktc) thu

diTdc hon hdp khi

va

hdi trong do Vco , =3V o ;

^ ^ ^ ^ = 11.

Ti m cong thiJc

phan tuTcua A biet d the hdi; l,8g chat A chiem the t i c h bang the t i c h cua 0,8g oxi

cung dieu ki?n. , _ ^ , , •

Gidi

D a t CTT Q cua A : C^HyO

, 0,8

T a

CO : nA

\l f

'1

> J, , '

t 1 I . , .

nllt

' W

= 0,025 mol => M A =

1,8

= 72

32 ••

0,025

. , .

PhUdng tnnh dot chdy : , , ,„ , .. •„

O2 —

^ XCO2+

^H2

•« iTt'rf

a f t d U

^;^(m.

m i

jB'jof

{ :,ii

~.tciq

a t f r t i ;iin>; -> idofsi

y

x + - - -

4 2

0,05 mol

y

z

x +

4 2

0,05 0,05x

o

,05.^'-'^''^«^«''^'^^«^*^^

S

mol chat A dem dot: n = ^ = 0,05 (mo l)

So mol O2 da diing : n^ = = 0,2 (mol)

,

22,4

1,1 1

- • i . ; : t

,...,..H

,,D

H 5 n hdp khi thu difdc :

CO2

H2O

O2 dir

Theo d l

Vco2

= 3 d u « 0,05x = 3 0 ,2-

y z

x + - - -

4 2

0,05

^C02 11 44.0,05x 11

^

= — <=>

m

(1)

"29 18.0,05 i

3

'rrtU:

• ' • • - ' • 2

K; v': I ]• 1'X)9iaf;^(_'^^,,V

=> - = 1 ^ = - => y = - X . The va o( l) => z = 8 - 2 x

24

4

T a CO : 72 = 12x + y + 16z <=> 12x + - x + 16(8 - 2x) = 72

3

,r , • i ,,,/,. :

•;,(]

. => x = 3;y = 4;z = 2

a,. .,.,(1 <n;'.;:q

T ; • .

V a y

CTPT cua A

\k C3H4O2.

B ai 4. Dot chay hoan toan l ,37g chat

hufu

cd A thu diTdc 3,08g

CO2;

0,63g

H2O

va

0,126 lit N2 (30 C va 75 cmHg). Xac dinh cong thiJc phan tuT A . Biet l,37g A cho

bay hdi

d

100 C va 1 atm thi the t i c h thu diTdc la 306ml. •,.(;./

• « € ^ k i r >

m :

vi>d'riif .:îil.K\

n- :'

Sdmol A : H A = ^ = = 0,01 ( m o l ) '

^^^^^

R T 0,082.(100 + 273) ,

A.,, A i i : ' . b :

Y S H ^

K h o i l i f d n g mol A M A = = 137

0,01 . ' >: ,A f,Mb

^ ;

T T • •

Taco: mc= 12. — = 0,84(g);

m

= 2. — = 0,07 (g)

f ^ n ) -

44 18

S6 'molN2: n ^ = — ^ = 0,005 ,„ / ,,vi' , 1 ,

0,082.(30 + 273) .Û| , v * +

=> m N

= 28.0,005 = 0,14 (g) ; ;

mo = 1,37 - (0,84 + 0,07 + 0,14) = 0,32 (g) '

Dat CTT Q cua A : C,HyO,N, i /

0,84 0,07 0,32 0,14 + 7-f

T a cd

ti I 9 : X

:

y

: z :

t =

12 1 16 14

' • • ' = 0,07 : 0,07 : 0,02 : 0,01 = 7 : 7 : 2 : 1 ^ ^ .

V ay cong thu-c A cd dang

(C7H702N)„ .

T a c d :

(7 x 12 + 7 + 32 + 14)n = 137 n = 1 ^ > - -^

V ay C T P T cua A m C7H7O2N.

AX'kD

s^HO kJ ^

B a i

5.

Tri^dc kia, pham do dung de nhupm do cho^ng cho cac

Hong

y gido chu

diTdc tdch chie't tijf mpt l o ^ i oc bien. Do la mpt hdp cha't cd th^nh phan nguyen to

nhiTsau :

C : 45,7%>;

H

: 1,9%;

O

: 7,6%;

N

: 6,7%; Br : 3 8 , 1 % .

a)

Hay

xac dinh cong thi^c ddn gian nhat cua pham do .

b )

Phi/dng

phdp phd

khdi liTdng

cho biet trong dd phan tuT pham do

cd

chiJa hai

nguyen tur brom.

Hay

xac dinh cong thufc phan tiJf cua nd.

Gidi , „ _

a) D^ t C T T Q cua pham do : CxHyO ,N,Brk

45,7 1,9 7,6 6,7 38,1 = '

T a cd ti 1#

: x

:

y

: z :

t

: k =

12 • 1 • 16 14 80 •

= 3,808 : 1,9 : 0,475 : 0,478 : 0,476 = 8 : 4 : 1 : 1 : 1

7/23/2019 Tr1-30(3)


hSn dgng v k phtwng ph p giSi H a hpc 11 HOu CO D Xuan Hung

V a y cong IhiJc ddn gian nhat cua pham do la

C K H 4 0 N B r .

Vi phan tuf chiJa hai nguyen tuf Br nen A c6 dang ( C s H 4 0 N B r

2 .

V a y CTCT cua pham do la CfiHxOsNsBra.

B a i 6. Dot chdy hoan loan hon hdp X gom hai hidrocacbon A B, mach hd cilng

day dong dang. Cho toan bp san pham chay vao 4,5 lit dung

dich

Ca(OH 2

0,02M,

thu difdc ke't tua va

khoi liTdng

dung

dich

tang 3,78g. Cho dung

dich

B a ( O H 2

(in

vao dung

dich

thu

difdc,

ket tua lai tSng them, tong

khoi liTdng

ket

tua hai Ian la 18,85g. Ti khoi hdi cua hon hdp X so vdi hidro < 20. So mol cua A

b ^ n g 60% tong so mol cua A va B trong hon hdp X. Cdc phan iJng xay ra hoan

toan. Hay xac

dinh

cong thiJc phan tuT cua A va B. '

Gidi -J

D a t

CTTQ cua A : C Hy : 3a (mol); B : +„Hy 2 n 2a (mol)

V i n A

= 6 0

( n A

n B ) ^

• ^ = -

Phifdng trinh

dot chay : .

C x H y + x + —

4 j

O 2 > X C 0 2 + ^ H 2 0 , (1)

3ay

3a mol

i

3axmo l mol

1 *

M ,

,:S:

a :

2

C x + n H y +2n

+

A - 3„

r v +

2nV '

' - ' ' ' J ' T ^ ^ ^ ' i f i ^ ^ i

x + — + ^ O, >(x + n

C 0 2 +

H 2 O 2)

2 4 ;

2a mol :

i \ X 1 o.;); ' ; 1

(x + n)2a

2

y + 2n

.2a

•Sh V ;kv.(}:j

- { I V

n ca

( 0 H ) 2 =0,02. 4,5 = 0,09

( m o l )

, / ^ ^ , „^ „ ,

^, ^

C O 2 + Ca(OH 2

>

CaCOs + H 2 O ,/ . ,(:), 1-?":) l ; (3)

yJ:M

b mol b mol b m o L j i

ffiourU,.

-lib

gauf; "nf>

frartq"

jnJ;

•JSuil

2

}i:H

2 C O 2 + Ca(OH 2 )• Ca(HC03

2 l v : / M

& . n . > n

f

(4)

,1

-Îj

Â

2c mol c mol c mol

H M- in

Ca(HC03 2

+

B a ( O H 2

> CaCOj i +

BaCOji

+ 2 H 2 O (5)

c mol c mol c mol

, , ; s j y | i :,: ; }f: ^ iM, ( »

Theode: mcoj +mH20 =3,78 + mcaco3 3)

vî i^nf-JnM

(d

hay

m c o 2

+

m H 2 0

= 3,78 + 100b "

i^^H (6)

Theo (3) va (4 ) :

m c a f O H ) ,

= b + c = 0,09 (mol)

Mat kha c : 18,85 =

m c a c o a O )

+ m c a C 0 3

(4 )

+

m c a c o g r s )

„, , ,

hay 18,85 = 100b + 100c + 197c = 100b + 297c

T a c5 he phifdng

trinh

:

b + c = 0,09 fb = 0,04 mol)

100b + 297c = 18,85 [c = 0,05

mol)

= >

n c o 2

= b + 2c =

0,14 mol)=>

mcoj

(3 .4 )

=( b + 2c).44 = 0,14.44 = 6,16(g)

T tf

(6)=i> m H j O

= 3.78 +100.0 ,04 - 6,16 = 1,62 (g)

1,62

n i

- = 0,09 (mol)

Theo phifdng

trtnh

(1), (2) ta c6: nc oj = 3ax + 2a(x + n) = 0,14 1 r

1

3ay

-i-2a.

y + 2n

2

hay

n c o 2

= 5ax + 2an = 0,14

M a ta c6 :

d x

/H a < 20 Mx < 40 ,

(12x + y).3a + [12(x + n) + y + 2n].2a

hay 60x + 5y + 28n < 200

5ax-i-2an

= 0,14

T a c6 he phifdng trlnh : •

•

F

;/;,,;,)

r;/ ':' I ' i ' j i i , 'jLnJ. >•

(7)

(9)

Say = 4an = 0,18

Giai

ra ta c6 :

< 4 0

^'

60 x

+ 5y + 28 n<20 0 , „ , , ,..

a = 0,01

n

= 2

x = 2

V$y

C T P T

cua A m

C 2 H 2 ,

B la

C 4 H 6 .

, ^ , , g, W.

Bai 7. Dot chay ho^n toan m gam chat hiJu cd

X

chtfa ba nguyen to

C , H , O

thu

difdc a gam C O 2 va b gam H 2 O , biet a = va b = Hay xic

dinh

cong

thtfc phan

tvl

cua

X . Biet

3,6g hdi X cd the tich b^ng the tich cua l,76g

C O 2

ciJng

diluki?n. - ^ . j

• J : \

Gidi Li

Dat C T T Q cua X : C x H y O , .

2 2 m ^ 3m m , .

Tac6 : mc=

1 2 . ^ ^ = 0 ,4 m( g) ; mH = 2 . — = —(g)

15.44

J

-io

=>

mo = m—

0 , 4 m + —

1 5 j

8 m

I

(g)

27

7/23/2019 Tr1-30(3)


Ph an dgng phuo ng p h p

g \ k \ a hg c 1 1 H O u c O - D 8 X u Sn H a n g

T a c d t i l ? : X : y : z = : — : = 0 , 03 3 : 0 , 0 66 : 0 , 03 3 = 1 : 2 : 1

1 2 15.1 15.16

=> X c6 d a n g ( CH2 0 ) „ . ,

1 7 6 ' t .

M a t k h a c : n x =

n c o 2

= = 0 .0 4 ( m o l) ,

=> M x = = 90 30n = 90 n = 3

- ' 0,0 4 ' '

V $ y CTPT

c u a X m C j H f i Os .

v

* / r v , - , i .

v * - ' f

. ,.t •

B a i 8 . O x i h o a h o a n l o a n 18,6g c h a t h f f u c d A t h u d iTdc 52 ,8g C O 2 v a 12 ,6g H2 O .

M a t k h a c k h i

p h a n t i c h

l i T dng cha't h f f u c d

A

t h u d i f d c k h i N H 3 . D a n

t o a n

b p k h i

n a y v a o 1 2 5 m l

d u n g d i c h H2S O 4

2 M t h i

p h a n

a x i t du T d i f d c

t r u n g

h o a

vilfa

he ' t bd i

1 0 0 ml

d u n g d j c h

N a O H 3 M .

a ) X a c

d i n h

t h a n h

p h a n

% N c d

t r o n g A . u> ' , ^,

b ) L $ p c o n g

t h i J c p h a n

tuT

A

b i e t

d A/kk

< 3,25.

, r r , Gidi • :

T a c d : n H 2S04 = 0 ,125 .2 = 0 ,25 ( m o l ) ' ' ' ; ' ' ] • / ; ' ' ; : / / ' ' ' ^, f

N a O H

- 0 , 1 . 3 = 0 , 3 ( m o l ) ^ ^ ^

2 N H 3

+

H2S O 4 >

( NH4

) 2 S 0 4 ' ' (1)

0 ,2 m o l 0 , 1 m o l ;

H2S O 4 +

2 N a O H

>

Na2S04 +

2 H 2 O V * ( 2 )

0 ,1 5 m o l 0 ,3 m o l

: , -

n H 2 S O 4 ( ^ p u ( i ) = 0 , 2 5 - 0 , 1 5

= 0 , l ( m o l ) '''\

m N = 0,2 X 14 = 2,8 (g ) ' , - M n iM)

2,8

18,6

T h a n h

p h an % N t r on g A : % N = x 100% = 15 ,05% ... \

T

5 2 ,8 ^ 1 2,6 , , , ,

. s f o l T O x J l ^ -

T a c d : m c = 1 2 . = 1 4 , 4 ( g ) ; m n = 2 . — ^ = l ,4(g) ^

= > m o = 1 8 , 6 - ( 1 4 , 4 + 1 , 4 + 2 ,8 ) = 0 . . „

V S y t r o n g

A

k h o n g c d o x i . :

l D a t CTTQ cua A : C ^ H y N , rf:^='''^'M - ^ ''"^f"^'* - f*-^^' { •&;)•&• n . ^ i j ^ p t j r f l ^ •

Ta cd t i l e :

X

: y : z =

:

M : M = 1 ,2 :1 ,4 : 0 ,2 = 6 :7 : 1

^ ^

.

12 1 14 m^y . -^

^ A

c d d a n g

( C6H7N) „ . ... ... X

f i i o

pTTO

Theo

d e :

dA

/kk < 3,25 => M A < 3,25 x 29 = 94,25

Ta c d : (6 x 12 + 7 + 14)n < 94,25

V § y A cd CTPT : C 6 H 7 N .

7 S l

9 3 n < 94,25 => n < 1,01 => n = 1

B a i

9 . D o t c h d y 5 6 0 c m ^ h o n h d p k h i ( d k t c ) g o m 2 h id r o c a c b o n c d c i l n g s o n g u y e n

t i J ca cbon ta thu duTdc 4,4g C O 2 va 1,9125g hd i n iTdc. • '

a ) X d c d i n h

CTPT

c a c cha t hiJu cd ,.,, ,

....

j

^y

. . . . j : •

b ) T i n h % k h o i l i fd ng cac cha ' t

c ) N eu cho l i f dn g C O 2 t r e n v a o 1 00 m l d d K O H 1 , 3M ; T i n h

C M

m u o i t a o th a nh

G i a i

6

bk i

n ay , ta d i l n g p h i f dn g p hap so n g u yen tuf H t r un g bm h ke ' t hdp vd i p h i T dn g

p h a p b i ^ n lu a n d e g ia i. ij(|,;,,,|v) d d

a ) X d c

d i n h C T P T

c d c h id r o c a c b o n : , ||, ^ ,j ., ,,1

D a t C T P T

2 h yd r o c a c b o n t r e n :

B

: C

H y .

CTPT t r u n g b i n h 2 h y d r o c a c b o n t ren : C - H -

G ia su r y < y ' ^ y < y < y '

S o m o l h o n h d p k h i

n hh

= = 0 ,025 m o l

n c o 2

=4 ,4/ 44 = 0 .1 ( m o l ) _

n H 2 0

= 1,9125/18 = 0,10625 (mo l )

x + y O , - — ) .

XCO 2

+ ^ H 2 0

0,025 / • ' ^

0 0 2 = 0 , 0 2 5 = 0 , 1

n H 2 o = 0 , 0 2 5 ^ = 0 , 10 6 2 5

0 , 0 2 5 x - ^ 0 ,0 2 5 y / 2

x = 4 '

y = 8,5 '

CTPT

A ,

B cd d a n g :

A :

C 4 H ,

v a B :

C 4 H , .

-^^^ ^^ ' ^ ^ •

T a c d y < y

< y ' h a y y < 8 , 5 < y ' ( 1 )

^

8,5 < y ' chan^v^v ;

^ l

_ y ' ^ 2 x + 2 = 2 . 4 + 2 = 1 0 ' '

t < « « - ' '

• ' '

= ^ y ' = 1 0 r = > C T P T B : C 4 H , o tentO,0

T i W n g t i r b i ? n lu a n t i m CTPT A : y < 8,5, y chSn >|.ii ||''ai '

B i ^ n luHn

t i m CTPT B :

y

2 4

6 8

A C4 H2

C4H4

C 4 H 6 C H S

V a y c d 4 ca p n g h i e m :

A

: C 4 H

B :

C 4 H , )

v a

A : C H

B :

C 4 H

va

10

A

: C 4 H 6

B

: C H

va

10

A : C 4 H H

B :

C 4 H I ( )

2 9

W

7/23/2019 Tr1-30(3)


PhSn

d ing phUOng

phAp

gi^ i H6a hgc

11

HOu co

-

i Ju

X u a n

c ) T i n h

C M

c d c m u o l t a o t h a n h r

i

,

n K O H =

V . C M

= 0 ,1 .1 . 3 = 0 ,1 3 (m o l ) :

T a c 6 : ^ ^ = = 1,3 ^ T a o t h a n h 2 m u ^ . ^ X C*

,

, CO2 '- ''' •

CO 2 +2 K O H - >

K2CO3

H2O o l \ i r , ; (y r o: . . r y j f i i r s . • (••

a 2a a (mol)

CO 2 + K 0 H - > K H C 0 3

^y--

:r ;^:A:q

yuni

a

hyi()

b b b (mol)

••f *;;

a + b =

nco2

= 0,1

T a c6

2 a + b =nKOH = 0 - 1 3

0 , 0 3 0,07

a = 0 ,03 0.;

( m o l ) , : ,

b =0,07 ^• • •-••••-•••-••.•H:Tir VK;(r

C M ( K 2 C 0 3 )

=

-5J =

0 . 3 ( M ) ; C M ( K H C O 3 ) = - ^

=0 , 7 ( M )

, ,. „ .j

B a i 1 0 . M o t h i d r o c a c b o n A dt he kh i c6 the t i ch gap 4Ian the t ich cda lUU huynh

d i o x i t C O kho i l i / dng t i fdng d i fdng t rong cung d ieu k i?n . San pham chay cua A

dan qua b inh d iTng n i fdc vo i t rong diT th i c6 Ig ke t t ua don g thd i kho i l i / dng b inh

t a n g 0 , 8 g. T i m C T P T

A .

*

Tim MA :

^\

:^...«f«'n-i fOt^::r:r i ^

1V A = 4 . V S O

2

( d c u n g d i e u k i e n ) = > n A = 4.nso2 , ( y 1 .

=> = 4 ^^ ^ => i = (A SO 2 c6 khoi l iTdng t i^dng diTdng nhau)

M A

M S O 2

MA

M S O 2

: ^ M A = ^ = ^ = 16

. -<^V

> - .0

:>= j

D 3 t A

: Q H y v

'

, , - e r i ( .

, u

• : ::rG.. , ; v : , , n |

Binh dtfng C a ( O H 2 h a p th u C O va H O ^ ^ . ^ • .

C a ( 0 H 2 C O -> CaCOai + H O r '

i^ = m c a c o 3 =

Ig

nco2 = CaCOs = 1/100= 0 . 0 1 m o l ' h a / T T ^ m ) ) J ^ r i

n d i H

>nc = nco2 = 0 ,01 mo l mc = 12 .0 ,01 = 0 ,12g ^ j , , p;

r ^ , .

m c o j = 0 , 0 1 . 4 4 = 0 , 4 4 g ,, ,^.

^ , ^'[''i

rr.^'

n|:.)]

flfic ; j

«\arr

Ambi„ii= m c o 2 +m H 2 0 => m H 2 0 = 0.8 - 0 , 4 4 = 0 ,3 6g i t

m H = 2 ^ = 2 ^= 0 ' 0 4 g

" ^ M S "

nio f

5 . v i f V

D L B T k h o i l i f d n g A ) : m = mc + m H = 0 ,12 + 0 ,04 = 0 ,16 ' {

3 0

1 2 , y MA M A - m c 1 6 .0 ,1 2 ,

Xa c6 — = = —

—

= > X

=

—

—

—

= = I

m ^

m mA 12.mA 12

.0

»16

M A - m H _1 6 .0 ,0 4 ^^ /

m A 0,16 wV

V a y C T P T A : C H 4 | ,

< T c u m 4 .

j

£ fi

e m t Q ^

h i f £

p l i i k n

l i t

• • A

- . f i d

i

. . j ;

ri'fj;

î a

i x d a

p h i i i A i g phdf bifii l i i a t i

-

< T i f i i

oa»d f t i h . l i i a i b a o ^

o A t i

k h o i

• y :

BAr TAP MAU VA BAI TAP N NG CA O

Bai 1 . D ot chdy hoan to an ba h id rocacbo n A, B, C c6 c i ing so nguye n tuT cacbon t a th

diTdc ti le C O v a H O c6 gid tri tiTdng iJng la 0 ,8 : 1 : 2 . Xdc d inh con g th i Jc pha n

tu fcua A, B, C. , , , ,

^ ^^ Gidi

S,0

* ChS't A CO

:

— £ L =0,8 = > n c o , < " H I O = ^ A 1 ^ a n k a n : CnH2n+2 -

nH20

CnH2n+2

r3n n

O2 > nC02 + (n

+1

H20 • ^^ ]^p y: , , 5 . , ^ f f f ;

T a c 6 : - = 0 ,8 = > n = 4

.

/ , f >

i n , , i i \ f ,nr

n + 1 '

V a y C T P T A la C4H„.

* C h a t B c 6 : = 1 Ucoj = nH20

=>

B la

a n k e n h o S c x i c l o a n k e n :

C nH2n .

/

V i

A ,

B c i i n g s o n g u y e n tuT cacbon => B la C4H . ' f f *

I

* C h a t C c 6

:

= 2

=>

nco2

=

2nH20 hay

nH20

< " c o z

H 2 O ,T

D S t C T T Q c u a C :

C„H2^^2-2k (k

Ih s o l i e n k e t

71)

r 3 n + l - k \

CnH2n+2-2k

2

O2 n C 0 2 + (n + 1 -

k

H 2 0

n .••„'';|,,,„t I S - . . - ' I . . , ; U , > . . /

:/.,

, . 4 : • >••

T a c o : = 2 ... ... .

V i

A ,

B, C c i ing so cacbcn => n = 4 => k =

3

^ - ; ; if

; •

'

V§y C T P T c u a Cm C 4H 4 . v t

J -

7/23/2019 Tr1-30(3)


Phan dgng

vji

phuong pUip giSi

H6a

hi c

11 HOu ca - P5

Xuan Hang

B ai 2. Dot ch ay 3,7g chat hQ a

c d

X

( C ,

H , O ) d u n g

v Ua

du 6,72 h't o x i

( d k t c )

va thu

diTdc

0 , 2 5 mo l

H 2 O .

X a c d i n h c o n g thiJc phan tuT X b ie ' t 70 < Mx < 8 3 .

Gidi -i

T a c d :

= — =

0 , 3 ( m o l ) ^^

v^;-

X +

O 2 >• C O 2 + H 2 O

A p d u n g d i n h l u a t b ao t oan k h o i l iW n g : • îdvM •'iimb» '^/.i^,.

T a

CO

: mx + =

mco

^^lO '

.i TavAicit

fw o n # J ' -

=> mco = 3 J + 0 , 3 x 3 2 - 0 , 2 5 x 1 8 = 8 , 8 (g ) ,t ^ u V A tt U v W i' .* Kv ^ ^ -

8 8

'^ = 7 ^ :

= 2 . 4 ( g ) , ^ . , k \ ' i w n^ (

miiitm'ti^ m H = 2 X 0,25 = 0,5 (g ) . A ^

i ^ l i i i

i d 1 i i « n ? w H

' / a *

==> mo =

3 , 7 - ( 2 , 4

+ 0,5) =

0,8(g)

*

t' f ' j

J

)

M «. M ?' 1

. j ' i ,

D a t

C T T Q

cua X :

C ^ H y O , i ^ i i' ^ < '

T a c 6 t i l e : X : y : z = — : — : — - 0 .2 : 0 , 5 : 0 . 0 5 = 4 : 1 0 : 1

12 1 1 6

X c 6 d a n g

( C 4 H

, o O ) „ .

kU

„.-, •

O ^ H

Theode

7 0 < 7 4 n < 8 3 = > 0 ,9 5 < n < 1 ,1 2 = > n = 1

; j i f , f ^

V a y

C T P T c u a

X la

C 4 H , o O .

}/0'}.J i

t^r'^sf^'

' • •r

I *

B a i 3 .

Cho bon h dp cha 't hiJu c d A , B , C , D d e u b e n , ma c h

c a c b o n

l i e n tuc , kho ' i

l i f d n g

p h a n

t i j f

c u a c h i i n g l ap t h an h mot c ap s o c p n g . K h i do 't c h d y m o t l i f d n g b a t

k i

m o i c h a t d e u c h i t h u d i f d c

C O 2

va H 2 O ma kh oi l iTdng

C O 2

b a n g

1,8333

I an

khoi l iTdng

H 2 O .

X d c d i n h c o n g th iJc pha n tuf cua A , B , C , D.

Gidi „;>:;., ; e 3 M t ) *

D a t

C T T Q

c u a A , B ,

C ,

D la

C x H y O , .

y

J , I,

4 2.

T h e o d l : ^ = 1,8333 = . - ^ = 1,8333

- ' ' ^ - ^

'

n^HzO 1 8 . 1

x _ 3 • •

C6 ng thu-c cua A , B , C , D c6 d a n g

( C 3 H

« ) „ 0 ,

m ^ t a l u o n c d : 8 n < 2 x 3n + 2 n < 1 •• ^1^:^,

• f

V a y

c o n g

th iJc cua bo n chat c6 dan g C S H R O , , .

V i

kh oi l iTdng phan tuf cua ch i ing lap than h mp t cap so

c g n g

nen ta c6 z = 0 ; 1; 2; 3.

V a y C T P T c u a

A , B ,

C ,

D

Ia n

i d l

Ik

C a H g , C j H g O , C j H s O z . C J H H O J .

32

B ^ i 4 . Do t c h ay h oan t o^ n 3 ,2 4 g am h on h d p X g om h a i c h a t hiJu c d A v a B k h ac

day d ong d in g , t ro ng dd A hdn B mo t nguyen tuT cacbon , ngu 'd i ta ch i thu

diTdc

H 2 O

va

9 ,2 4 g

C O 2 . B i e 't t i k h o i h d i c u a X d o i v d i

H 2

b ^ n g 1 3 . 5. T i m

c o n g

thiJc

c i ia A , B va t i n h % k h o i l i f d n g c u a m 6 i c h a t t r on g X . 1 i '

„/sCt:, W ' li

^

h M f (Tr ich TS DHQ G TP HCM , d at 2 )

••ts , , . , , Gidi

(

,lin nj

' M

> r

in 4

n c o , = — = 0 , 2 1 ( m o l ) ' ' o> r, ,Mf . , /t.i *U ,

K h o i l i f d n g m o l p h a n

tuT

cua X 1 (Jxy t i j i fiX > r;> Mx = 13,5 x 2 = 27

n x = ^ = 0 . 1 2 ( m o l ) ^ ^ .

_ n c o , 0 2r^ -^^^ M W^^^^^^^

=> So ngu yen t i f C t run g b inh : n = ^ = — — = 1 ,75 '

X 0,12

-V

n,

= 1 < n = 1,75 < n 2 = 2 ma n = -

vo l a la so m ol cua A , b la so m ol cua B. , ; ;

^ . U 5 ^ | a = 0 . 0 9 . o i

, i ^ ^

a + b = 0 . 12 l l = » .03™ol - V .

Ta c6 h? phufdng

t r i n h

:

m x = 0 , 0 9 M A + 0 , 0 3 M B = 3,24 ^Mk^Â^^

--p 'i

^^m :>iî

itik Wl

= > 3 M A + M B = 1 0 8 M B = 1 0 8 - 3 M A > 0 ''.• 'iy-^^t

M A < 3 6 => A k h on g c 6 ox i v a t r on g A c 6 2 n g u y e n

tuT

cacbon . •

N e n A c6 the la

C 2 H 2 , C 2 H 4 h o a c C 2 H 6 .

+ A la

C 2 H 2

( M A = 26 ) =^ M B = 30

( C H 2 0 ) ^ * ' * ' ' '^ '^ '^ ' ' ' ' '

+ Ne\ A la

C 2 H 4

( M A = 28 ) ^ M B = 24 ( l o a i )

'^'' ^

+ A m

C 2 H 6

( M A

= 30) =^

M B

= 18

( l o a i )

^ ' ^ ' ' ^

« ' •

V $ y

CTPT

cu a A la

C 2 H 2 ,

B la

H C H O ( C H 2 O ) .

T h an h p h an % k h o i l i T d n g mo i c h a t t r on g X :

% C 2 H 2

=

— ^ . 1 0 0 %

= 7 2 , 2 2 %

, 3,24.

=>

% H C H O

= 2 7 , 7 8 % .

B a i 5 .

M Qt h i d r oc ac b on X c l t h e k h i c 6 t i k h o i h d i so v d i h i d r o l a 1 5 . L ap

c o n g

thiJc

phSn tuf c i5a h id rocac bon X .

i t M r a Gidi

) . f t

J , A U M ' * . r i .

1 ' '

D t C T T Q

c ua

X 1^

C^Hy.

\m\\x4 \J ^ \m I irA mh\ ' ^

M x = 1 5 x 2 = 3 0 ) . M f y j , i f j , . f ? ^ ( l J f t / . j *KS ) l ' '

T a c 6 :

12x + y = 30 => y = 30 - 12x ;

D i ^ u

k i? n y > 0 => 30 - 12x > 0 => X < 2,5 .it-sdatooifc-,? sA

:uwz

'^XnO m •.

7/23/2019 Tr1-30(3)


Phan d j n g vS phuong pMp giSi H6a hpc 11 HOu d - B5 Xuan Himg

;/i;,;'Neu

x = l => y = 18

(loai) ( K A q n o i p i ^ y n s n ^

x = 2. => y = 6 (nhan) ivs,t5|vvw; fa .Hn> n

.A ,

^ i M:

V $ y CTPT cua X : C 2 H 6 .

a i

6. Mp t hon hdp kh i A gom hai hidrocacbon thupc

ciing

day dong dSng. Dot

chay 2 li t kh i A can

diing

7,2 l i t k h i ox i thu diTdc 4.8 li t

C O 2

mgt

lUdng

hdi nifdc

(cdc khi do 3 C l ing dieu

ki#n).

sb©

a)

Tinh

the

tich

hdi nUdc thu dUdc. , , , , , ,^ ^

b )

Xdc

dinh

day dong

d ing

cua hai hidrocacbon trong A.

c) Lap cong thiJc phan tuf tinh thanh phan % theo t h i tich cac chat trong A.

Biet

h a i chat

dtfng

cdch nhau mOt chat trong day dong

ding.

G mi

'rr

a) DatCTTQtrungbinhciia A la C-H- . «

,„

••• n , daUi mmi D i n . t Z

x + y O 2 > X C O 2 + ^ H z O

2

li t 7,2 li t ^ 4,8 lit ^ ^

iaif:>

Sswt ia

41

d ,A afc

|r» i : r ro*

1 '^ •' 7 x - - '

,

^, . Ta c6 t i 1? : - = —f- = — => x = 2, 4; y = 4,8 .

j , . ^ a/ r f q

ho tsT,

•

T h I Uch hdi node :

V H 2 0

= f " "

^' ^ ^^'^^

'

•' _ 2 4 r • • < A M £ - . * ] i = ; : f l M : , ^

b ) Ta c6 t i l ? : - £ • = — = -

= >

H H

= 2nc thuOc day dong dang anken . ^

H H 4,8 2

c)

VI nc = 2,4 =

X

la anken nen n, = 2 n : = 4 (hai chat c^ch nhau rngt chlft).

V a y CTPT

ciia

hai hidrocacbon la C 2 H 4 C4Hg. ,,. , ,

^r

, .. ^

j ^

^

D a t

a la nc2H4 tron g

1

mol hon hdp A (1 - a) la sd'mol

C 4 H 8 .

^ ^ . ^ . j ^ ^

.^^^ Ta c6 : nc =

' ^ ^

=2 ,4 ^ a = 0.8 (mol )

' •^^^•^•^

^1A . v . TITO v£V

Thanh

phan % theo th i

tich

cdc

k h i :

^.

=> % V c , H 8 = 2 0 % . : : . . , > , ,

a i

7. Cho ba hidrocacbon A . B. C 5 thg khi trong

d i l u

k i ^ n

thiTdng.

A . B, C c6 phî

. ; la dong

d ing

cua nhau hay khong ? Bi^ t rJlng khi phan hOy deu tao ra cacbon va

j

hidro.

the

tich

hidro gap ba Ian

hi

tich

hidrocacbon ban d i u (5

cilng

d i l u

ki§n).

D $ t CTTQ cua ba hidrocacbon la CjHy.

x C +

a x ^

2

Theo de :

ay

= 3a

y =

6

va y

A , B, C Ian

liTdt

c6 CTPT

la : C 2 H 6 ,

C j Hf i ,

C 4 H 6 .

i

8. Khi phan

tich

chat

h i l u

cd A chi chtfa C,

H ,

O thi c6 mc + mw = 3,5.mo

a) Ti m cong thrfc ddn

g i i n

cua A. /

b) Lay hai

nfdu

ddn chi?c X, Y dem dun nong vdi H 2 S O 4 dam dac d nhipt dO

thich

hdp thl thu difdc A. Xac

dinh

con g thiJc cau tao mach hcl cua A, X, Y.

Biet

ring

Aia e t e .

m^ ^î.: m .uuici-a^;-*i

,m

<n: (jî,,,

DHQG,

dot

m<M i ..-i: K»

j ' b i

rof l . itjU^ dKm^m vi j f . i o f i •<-.s.i

Giai

D | t

CTTQ cua A : C.HA- ' ' ' ' ' '

„ ^ ¥ \r> J l i d s j Bhbj , i}rn^ ' r f j ; ui?

Ta O : m c + m 3,5.mo

H ay 12x + y = 3,5.16z => 12x + y = 56z

+

Nû z = 1=> 12x + y = 56 y = 56 - 12x

wfif,,

,'.f:ft

i j O H b

X

1

2

3

4

5

y

4 4

(loai)

32

(loai)

20

(loai)

8

nh$n

- 4

(loai)

=> A cd CTPT :

C 4 H 8 O .

' • ^ • ^ ^ w ^ , / - liw.^ *^;*^^^^--^^^ iCr ;

+

Nû z = n=> CT cua A c6 dang

( C 4 H g O ) „ . t •

V a y cong thtJc ddn gian cua A

la C 4 H 8 O .

cnîi '

Theo de : X la

C H 3 O H ,

Y la

C H 2

= C H

- C H 2 0 H .

A cd CTCT : C H 3 0- C H 2 - C H = C H 2 .

PhUdng trinhphan iJng :

T'

r'' '':^'

C H 3 O H +

C H 2

= C H

- C H 2 0 H

^l^ o*c> C H 3 0- C H 2 - C H = C H 2 +

H 2 O .

a i

9. Dot

c h i y h o a n t o an

1 hidrocacbon A

c i n dilng

28,8g ox i thu dUdc 13,44 li t

C O 2 (dktc).

Bi^ t t i khd i hdi cua A doi vdi khong khi la d vdi 2 < d < 2,5. Ti m

CTPTcia

A .

;fi •;,i-ffM •- -3

Gidi

V I A l a h i d r o c a c b o n . mi k h i 66t c h d y

1

h i d r o c a c b o n

ta

lu6n c6:

1

— 1

2

0 2

phJntfng - c o j +

7

H 2 0

" H 2 O = 2( no j

phiniTng

n c o 2 ) - 2

'28,8 13,44'

, 32 22,4 ,

=

0,6 mol

35

7/23/2019 Tr1-30(3)


Phan

djng

va

phii ng

ph^p g \ i \ a h p c 1 1 H O u C O - D 5

Xuan

ung

13 44

T a c d :

m c = 1 2

. n c o 2 =

1 2 . ^ =

7.2 g)

n

=

2

. n H 2 0 = 2.0,6=

1,2 (g)

D a t C T P T cua A la C,H y ta cd: x : y = — : — = 1 : 2

1 12 1

=> c o n g th iJc nguy f in cua A la

(CH2 )n.

M a 2 < d < 2 , 5 =^ 2 < — ^ < 2 , 5 ^ 58 < M A < 72 ,5.

9

H a y 5 8 < 1 4 n < 7 2 , 5 = ^ 4 , l < n < 5 , 2 ' ' ' '

.Ma n ngu yen =^ n = 5 =^

C T P T

cua A la

CsH.o.

.

/ ^ ^ ^

' ^

B a i

10 .

Ox i h o a h o a n l o a n mp t ch a t h u ^ cd Y ca n l , 2 8 g o x i , ch i t a o t h a n h 0 ,66g

^ ^H20

v a l , 7 6g

C O 2 .

Cho loan bp san ph am sau phdn iJng sue vao 25 0m l dung

d i c h B a ( 0 H ) 2

nong dp a mol/ l i t thu diTdc b gam ke ' t tua , Ipc ke ' t tua ro i dun nong

dung d ich loc du fdc c gam ke ' t tua n ffa . Cho b ie t b + c = 5,91 . T in h a , b , c va t im

c o n g thiJc ddn gia n nha't cua Y ? ' * ' '

; M

Giai . , ^

A p d ung d inh lua t bao loan kh o i liTdng : m y +

T Q

=

n i c Q ^

+ m ^ ^ Q , ^

my = 1 ,76 + 0 ,6 6 - 1 ,28 =

1,14 g)

\ t i

CO 2

+

B a ( O H )2

> B a C O j i +

H2O * ^ .

^ ^ X

i ,^4^ j

2CO2 +

B a ( O H )2

> B a

( H C 0 3)2

^ ^ H , , ,- ^

^

2y . y y ,. , ^

B a ( H C 0 3 )2

— ^ B a CO j l +

CO2 + H2O

1,76 0 66 0 22

^ ' I ' ^ r O c H a : A . ; ; >

^ ^ = 1 2 . — = 0 , 4 8 ( g ) ; mH = 2 . - i ^ = - : ^ g ) > ^ ^ I ^ ^ ^ ^ ^ ^ ^ ^

= ^ m o = 1 , 1 4-

o A S ^ ^ y h l l

^g)

,,r i^ 6 mo l k ^ t u a h a i l l n : . ^ ^ ^^

^5 ^3 ,

5 91 1 76

m B a C 0 3 = - j ^ = x + y = 0 ,03 ; n c o 2 = - ^ = x + 2y = 0,04

T a c o h e p h i T d n g t r i n h :

= 1^

= ° '^ .nfc«f AJV

[ x + 2 y = 0 , 04 [ x = 0 , 0 2 m o l v » y ' ^ ,

r p i

0 03 ' - ' H

" B a ( 0 H ) 2

= X + y = 0 ,0 3 ( m o l ) a = — — = 0 ,12

mol/lit)

,

' b = 0,02

X

197 = 3,94 (g )

« c =

5 , 9 1

- 3, 94 = 1,97 (g ) '

• •V^ ^ }:^ . :

K i i A i g f i i n i i r r

D a t C T T Q c u a

Y :

C^HyO, ,

0,48 0 ,22 1 ,76 . . . 0 ,22 0 ,11 ,^ , ,

T a rd t i l e : x : Y : z = : : = 0 ,0 4 : : = 12 :22 :11

T a c o u i e X y ^ ^ ^

C o n g

thiJc ddn gian nh at cua Y : Ci2H220n. ' i / ^ v ^ ;

B a i

11 .

B ot chdy h oan toan 0,1 m ol hdp chat hCu cd A ch i ch iJa C, H , O vd i ox i

t h e o t i 1? m ol 1 : 2 . T o a n bp san pham chay diTdc cho qua b inh 1 d i fng du ng d ich

pdCl2 d i i ro i qua b inh 2 d i/ng dung d ich C a ( O H )2 d i r . Sau th i ngh iem , b inh 1 tang

0,4g va xua ' t h i ^n 21 ,2g ke t tua , c6n b inh 2 cd 30g k e t tua . T im c o n g thiJc phan

tuTcua A.

B i n h

1 du-ng

P d C b

h u t CO v a

H2O,

g i a i p h dn g

C O 2 . j , ,

,

;

,^ „ ,^,,

CO + PdCl2 + H2O ^ Pd4 + C O j t + 2 H C1 ( 1 )

C02d phufdng t r i n (1 ) va CO 2 d phan iJng chay vao du ng d ich C a ( O H

) 2 .

CO 2

+ C a ( O H )2 - >

C a C O a l

+

H2O

30

100

Snc02 = n c a C 3 = — = 0 ,3 m o l i . I - m 0 <

2 1 2

T h e o

ptpi? (1 ) , ta cd: n c o j s i n h r a = co = n i p d = — ^ = 0 ,2 m ol •

= > S o mo l CO 2 do phan tfng chay la: 0 , 3 - 0 , 2 = 0 ,1 m ol . , ,

Kh o i l i f dn g b i n h P d C h tSng la : mco + m H 2 0 i " C 0 2 = ^,4 j

= > m H 2 0

= 0,4 +

m c o 2

-

m c o =

0,4 + 0,2.44 -

0 ,2 .28

= 3,6 (g ) i

>t

3,6 , <

=> n H , o = — = 0,2 mo l ' '

Ap dung d inh lua t bao toan kho i l i fdng ta cd:

S t

x i Y

n i A + m o 2

= m c o 2

+

' c o

+ m H 2 0

' • ^ ' S

=> m A = m c o 2 + m c o + m H 2 0 - m o 2 = 0,1.44 + 0 ,2 .28 + 3,6 - 0 . 2 . 32 = 7,2 g)

G p i c o n g thuTc ph an t i l c i l a A 1 ^ C^HyO^. , „ , , . „ , , » r i , A J S j . •

T a c d : m c =

1 2 . (U c o j + nco) = 12. (0,1 + 0.2) = 3,6 (g )

7{ioffl? b / ,

m H = 2. n H 2 0 = 2.0,2 = 0.4 (g) ;

m o =

7 , 2 - ( 3 . 6 +

0,4) = 3,2 (g )

12x _ y _ 16z _ M _ 72

m c

n i H

m o

m

7,2

12x y 16z ,^ A

o = ^ = = 10 => x = 3 ;y = 4 ; z = 2. •: ;

3,6 0,4 3,2 ^ V , , , . , R. -^ • ;

Va iy : C T P T cua A la C3H4O2 n * ' ' - ^ ^ / ' . • '

7/23/2019 Tr1-30(3)


Phan dgnp va phudng ph p

giai

H6a hgc

11

Hgu

co -

B8 Xufln Hung

B a i

1 2 . M p t h o n h d p g o m n i Q t so h i d r o c a c b o n h e n t i e' p t r o n g d a y d 6 n g d i n g c 6 k h o i

I t f d ng p ha n t i t t r u ng b i nh ( M )

=

6 4 .

6

1 0 0 C th i h o n h d p n a y

6

t h e k h i , l a m l a n h

d e n n h i f t dp p h o n g t h i m p t so c h a t bi ngift ig t u. C c c ha t k h i c 6 k ho i It f dng p h a n

tuf tr u ng b i nh b k ng 5 4 . Ci c c ha t l ong c 6 b Kng 7 4 . T o ng k ho i l if d ng c ^ c c ha t tr ong

h o n hdp dau la 2 5 2 . D i e t k h o i l i T dng p ha n ti3f c h a t n a n g n h a t g^p d 6 i c h a t n h e

n h ^ t . T i m C T P T ci c c h a t % t h e

t i ch

c d c c ha t t rong hon hd p

'

- - - - - - . . K ' ^ - ^

G i a i

G o i a j , 3 2 a n 1^ k h ^ i l if d n g p h a n

tijf

c u a c ^ c h i d r o c a c b o n t r d n .

* Ap d u n g t i l t h ch o t t o a n hg c

C d c

h i d r o c a c b o n l i e n t i e p t h u p c c i i n g m p t d a y d o n g d S n g se t a o n e n m p t c a p so

c O n g c6 c o n g s a i d

=

14

an

=

at

+

(n

-

l)d

S = — " " ' " ' * ^ *

V d i a n

=

2 a i

= > 2 a i =

ai + ( n - 1 ) . 1 4

,f. M) .

, i

=>ai

=

1 4 ( n -

1) •

=> S

=

l , 5 n a ,

=

2 5 2

' '

u

' n ,

„• 1 1 .

^,

( U ,

n . . i

( '

H a y 1 . 5 . 1 4 n ( n - 1)

=

2 52 = > 2 1 n , ^ - 2 1 n i - 2 5 2

= 0

=> n

=

4 ( n h a n ) va n = - 3 ( l o a i )

'

V a y : a ,

=

1 4 ( 4 - 0

=

4 2

" "

d a t h i d r o c a c b o n d a u la A | : Q H y (' I n - . , in

M , = 12 x + y =4 2

y c h i n

,,

, , ,, ' H

y < 2 x + 2

1

v , . ) K l

»A(

I

w/ ,

oi

i\i

i n ;/ ) , y n y b

q A

m ^

:

, My . -

X

1

2

> 4

y

30 18 6 < 0

V§y

A

la C3H6,

1^

h i d r oc a c b o n d a u t i d n t r ong c a p c p n g t r e n .

C d c

d o n g d S n g k e U e p c u a n 6 \ k C4H8, C5H,,,, C^Hu (M

=

8 4 ) ,<'

* T-inA

%

rfcA cd c c h i t t t r o n e h 3 n h<n> :

^ ^

^ ^ ^

G p i

a, b, c, d(m ol ) I an l i /dt 1^ so mo l c d c h i d r oc a c b on t i fd ng t f ng :

C3H6,

C 4 H S ,

— 4 2 a +5 6 b +7 0 c +8 4 d ' ^M - V r O j - .

M

= =

6 4

(1)

a

+

i

+

c

+

d

^ .

M « , . = ^ ^ ^ i ^

=

5 4 = * b

=

6 a

(2)

• •/:

.

^ C J ^ U / K € •

,

M l

=

7 0 c

+

8 4 d ^ ^ 4 ^ c

=

2 , 5 d

(3)

c +

d

T h a y ( 2 ) . ( 3 ) v a o ( l ) :

4 2 a

+

5 6 . 6 a + 7 0 . 2 , 5 d

+

8 4 d 3 7 8 a

+

2 5 9 d

M

=

=

0 4 = > — 5 4

a

+

6 a

+

2 , 5 d

+

d , , , 7 a

+

3 , 5 d

= >d

=

2 a

(4) . ,

c

=

2 , 5 . 2 a

=

5 a (3') i' ^

Hhh

= a

+

b

+

c

+

d

=

a

+

6 a

+

5 a

+

2 a

=

1 4 a

*'

d C l i n g d i e u k i# n , t i I9 ve so m o l b ^ n g ti 1§ v l t h e t i ch

% V C 3 H . = ^ x I 0 0

=

7 . 1 4 %

M

•

•

% V C 4 H K = x l 0 0 =4 2 , 8 5 %

« A ,

' , 1 , 1 - 1

1 , ,

.

% v c 3 H , o = - ^ x i o o % = 3 5 , 7 i % • •

% V C 6 H , 2 = — x l 0 0

=

1 4 , 2 8 %

,

14a

B a i

1 3 : D o t c h d y h o a n t o a n m

g )

h dp cha ' t

hiJu cd A

c h i I hu d i Td c

a(g)

C O 2

va b(g)

H 2 O .

B i e t 3a

=

1 lb va 7 m

=

3(a

+

b). T i m C T P T c ua A . ( b i e t l i k h o i h d i c u a

A

d 5 i

v d i k h o n g k h i n h o h d n 3 ) .

'

•

'X', rr:,,. J

•< • G i i ::;/V' ,.V;-;,YA'' 1

T a c 6 : dA/khongkhi

<

3

=>

M A

<

3 . 2 9

=

87

, , „ ,

,

, ,

ta c 6 : m c = 1 2 . n r o T

=

1 2 . —

=

—(g) l i ;

i

V I 3 a

=

1 lb => m f. =

=

b (g)

:

i 'J

> '

m H

=

2 . n H 2 0 = 2 . — = - ( g )

V 1 7 m

=

3 ( a

+ b) = 3 ( — + b ) =

14b m

=

2 b

=> m o =

m A

- ( m c +

m H )

=2 b -

D a t C T P T c u a

A

Id CxHyO, ; .

b

b 8b

T 2 ' 9 ' 9 . 1 6 1 2 ' 9

18

8 b

, ,

m^H ^ ' '^ ' '' '

_ A i ^

=

— i —

= 3 : 4 : 2

=> X

:

y

: z = . .

1 o •

o 18 * i

'

=>

Cong

thuTc

nguyen ci ia

A

la

C3H402

)n

, , j

M a t k h d c : M A < 8 7 7 2 n < 8 7

=>

n< 1 . 2 1

M a n nguyen =5. n = 1 = C T P T

cua

A

la : C3H4O2

7/23/2019 Tr1-30(3)


Phan dgng va phuong

phap

giSi H6a hpc 11 HOu co 5 Xuan

Htfng

Bai 14. Dot

c h a y h d p c h a t h i J U c d

A

( c h u T a

C, H, O)

p h a i d u n g

1

l i / d n g o x i b a n g

8 Ian

l U O n g

o x i

c6

t r o n g

A

va thu d U d c l u -d n g C O 2 v a H 2 O t h e o d

le

k h o i lU O n g

m c o 2 : mHjo = 22 : 9 . Biet t i k h o i h d i c u a X s o v d i H 2 la 2 9. Tun C T P T c u a X .

G ia s u f chat hffu

c d

A chiy sinh ra 22g

C O 2

v a 9g HzO: - -*

, =>mc=12.nco2

= 12.—=

6( g )

44

m H = 2

. n H 2 0

= 2

. - j ^ = l ( g )

Theo djnh luat hio tohn nguyen to la cO:

modrongA) + mophintfng =

mO

(trongC02)

' O l t r o n g

H 2 O )

,

:„ „,

,,,,

,. v;,:

= (22 - 6) + (9 - 1) = 24 (g) J 'i' ::-j-f,->V; S5

Ma

mopha

„.„g = 8 .mo

,..„g

A, hay = i mo(.,„gA, = = (g)

• O phaniJng' ' 'Ov. . -

•

' . ' C . i . , .... '4^&.:J

9

D t CTPT cua Al a QHyO,tac6:

r

12 • 1 "16 12

T

9.16

2

' r 6

\

. Cong thiJc nguyen c,5a A la (C 3 H.O)„

;,,eUV.r4v-.iM « / . ( M f

M a M , = 29.2 = 58 => 58n = 58 :=> n = 1 ' , , . . ^^ „ . ^

VayiCTPTcuaAlaCjHfiO , . ^ ..

B A I T A P T R A C N G H I | ; M

c a u

1. Mpt chat heru cd X khi d trang thai hdi c6 ti khoi hdi so vdi chat hdî cd Y la

2. Biet 2,2 gam chat Y c6 the tich b^ng the tich ciia 1,6 gam oxi trong ciang dieu

ki?n nhi?t dp, ap suat. Vay ti khoi cua X doi vdi C H la:

A . 11 B. 10 . C.6 D.5,5

C a u 2. Dot chay hoan toan 4,64 gam mpt hidrocacbon X (chdt khi a d'lhu kien

thuong) r6i dem toan bp san pham chay hip thy h^t vao binh dung dung djch

Ba(0H ) 2 .

Sau cac phan irng thu dugc 39,4 gam k t tiia va

khoi

lugng phin dung

djch giam bat 19,912 gam. Cong thirc phan tCr cua X la

A

. C 3 H 4 .

B. C H 4 . ' ^:

^

C . C ^ f ^m r t

Q^^^. nm'i..

' Trich di thi tuyin sinh Dai hgc khoi A nam 2012

c a u 3. Nicotin cd trong thuoc la la mot hdp chat rat doc, c6 the gay ung thiT phoi.

D ot

chdy 16,2 gam nicotin b^ng oxi vâ dii thu difdc 44 gam C O 2 , 12,6 gam

nifdc va 2,24 lit N 2 (dktc). Cho 85 < M

„i,.,i i„

< 230. Cong thiJc phan tuT dung cua

nicotin 1^:

A.C5H7NO B . C 5 H 7 N O 2 C . C , O H M N 2 D . C H , H , 4 N 3

C f i u 4.

Do't chdy hoan toan 5,2 gam hdp chat huTu cd X roi cho san pham chdy Ian

l ifdt

qua binh (1) di/ng H 2 S O 4 dam dac, binh (2) chu"a nifdc voi trong diT, thay

blnh (1) tang ISn 1,8 gam, binh (2) thu diTdc 15 gam ket tua. Khi hoa hdi 10,4

gam A thu di/dc mot the tich dung bang the tich cua 3,2 gam oxi trong cung dieu

kien nhiet dp dp suat. V$yc6ng thu'c phan tuTcua A 1 ^:

'i 'i/i • i •*

A. C5H12O2 B. C 3 H 4 O 4 C. C 7 H 4 O 6 D. C ; H S 0 3

c a u

5. Khi d^t chdy hoan toan 100ml hdi cha't B can 250ml oxi , tao ra 200ml C O

va 200ml hdi niTdc (cac the

tich

do d cung dieu ki?n). Vay cong thu'c phan li j

cua B la:

A.

C 2 H 4 O 2

B.

C 3 H 4 O 4

' • C.

C 2 H 4 O

D.

C 3 H 6 O

c a u 6. Do't chay hoan toSn 3,5 gam mpt hidrocacbon thu difdc 10,68 gam khi

cacbonic va 5,25 gam niTdc. Khoi

liTdng

oxi can

diing

de dot chdy la:

A . 6,21gam,; M B . 11,04 gam C. 12,43 gam D. 12,73 gam

c a u

7. Dot chdy hoan toan 0,366 gam chat hiJu cd X, thu du-pc 0,792 gam C O 2 v^

0,234 gam H 2 O . Mat khdc phan buy 0,549 gam X do thu difdc 37,42cm nitd (do

d 27°C va 750 mmHg), biet trong phan tuf cua X chi chiJa 1 nguyen tijf nitd. Vay

cong thu'c phan tuf cua X la:

A

. C 9 H 1 3 O 3 N B.C9H15O2N C.CyHi303N2 D. C K H 1 2 O 3 N

c a u 8 . Dot chay 400 ml mpt hon hdp gom nitd va mpt hidrocacbon

Y

3 the khi

bing 900 ml khi oxi (dif). The tich hon hdp thu dUdc sau khi dot la 1,4 l i t . Sau

k hi ngu-ng tu hdi niTdc thi the tich hon hdp con 800 ml, tiep tuc dan qua dung

dich KOH thay con 400 ml k h i . Biet rkng cac khi do d cilng dieu ki?n nhi?t dp,

dp suat. Vay cong thu'c cua Y Id:

A.

C 2 H 6

B.

C 3 H 4 r i i c i

C.

C 2 H 4

D.

C 4 H X

Cau9

.Sod6ngphancuaC5H

,2la:

, . •

A . 3 B.4 C.5 D.7

c a u

10 . Do't chdy hodn toan mpt

liTdng

hidrocacbon X. Hap thu toan bp san

pham chay vao dung dich Ba(0H

)2

(dU) tao ra 29,55 gam ket tua, dung djch

sau phan iJng c6 khoi

liTdng

giam 19,35 gam so vdi dung dich Ba(0H

)2

ban

dau. Cong thiJc phan tur cua X Id '

A

. C 3 H 4 .

B

. C 2 H 6 .

^^- ^^^

C . C 3 H 6 : ^ - ^ ^ ^ ^ ^ ^

^

- ^ iOsl l i * „ y^^^ ^^ ^^ ^.^ f.^.^^ 2070"

c a u

11.

Dot chdy hodn todn 0,9 gam hdp cha't hiJu cd A chuTa C,H,0 thu diTdc 1,32

gam

C O 2

vd 0,54 gam

H 2 O .

Ti khoi hdi cua A so vdi hidro la 90. Vay A cd cong

thu'c phan tuf Id:

A.CeHnOft B. C,oH,203 C. C

» H 2 o 0 4

j i n • , D. CTHifiO,

7/23/2019 Tr1-30(3)


P h an d g n g va p h u o n g p h ^ p

giai

H 6 a h p c

11

H Q u c d

-

D8 X u f l n H u n g

C f i u 12. Kh i dot cMy

1

lit khi X can

5

lit khi o x i , sau phan

tfng

thu dUdc

3

lit C O 2 va

4 l it hdi

nifdc. Biet

cdc khi diTdc do d cdng dieu

ki#n

ve

nhi^t

dp va dp

suS't.

V§y

cong thiJc phan

tuT

cua X la:

A .

C 3 H 6

B.

CjHx

' C .

C J H K O

D . C 3 H 6 O 2

CSu

13.

Trong mot

binh

kin chiJa hdi chat h O u cd X (c6 dang

C „ H 2 n 0 2 )

mach hcl

O 2 s 6

mol

O 2

gap doi so mol can cho phan uTng chdy) 6 139,9"C, dp

suat

trong

binh

la 0,8 atm. Dot chdy hoan toan X sau do diTa ve nhi?t dp ban dau, ap suat

trong binh luc nay la 0,95 atm. X cd cong thuTc phan

tijf

la:

A . C 2 H 4 O 2 . B. C H 2 O 2 . C. C 4 H 8 O 2 . D . C 3 H 6 O 2 .

r i c h

d e t h i t u y e n s i n k d a i h o c k h o i B )

CSu

14.

Dot chay hoan toan 100 ml hdi chat A, can

diing

250 ml ox i, chi tao ra 200

m l

C O 2

va 200 ml hdi nifdc (cac the

tich

khi do d cOng dieu

ki?n).

Xic

dinh

cong thtfc phan tur

ciia

A. ,^ , , ^

A . C 2 H 4 B. C 2 H 6 O C . C 2 H 4 O

D

. C j H f i O

CSu 15. Dot chay hoan toan 1 l it khi X can 5 h't khi o x i , sau phan uTng thu diTdc 3 li t C O 2

v ^ 4 li t hdi nufdc.

Biet

cac kh i dUOc do d

ciing

dieu ki?n ve

nhi^t

dp va ap suat.

Cong

thifc

phan tiy cua X la: f .

A

. C 3 H 6 B

C H 8

,

C . C 3 H S O

D

. C j H f i O z ^ ' '

Cau 16.

Co 3 chat hOU cd A , B ,

C

ma phan t&

ciia chijng

I§ p t h 1 cap so cpng. Bat c i? chat

nô khi chay cung chi tao

C O 2

v^

H 2 O ,

trong do nô^: n ^ ^ ^ o = 2 : 3 .

CTPT ciia

A, B,

ClanlUdtla:

, , . . . .

A C 2 H 4 , C 2 H 4 O , C 2 H 4 O 2 . I

i/ :> B.

C 2 H 4 . CzHfiO.

C j H f i O j . \„J

c.

C 3 H S , C 3 H S 0 , C 3 H X O 2 . ? r, D . C Z H ^ . C Z H S O C 2 H 6 O 2 . 'I

'^f .,

H i ' S U M i

HUdNGDlNGlArxa

cau 1.

Ta c6: Mx = 2My ^,.

„ .

A ,

:Ji i: j if i u v,

rMq -^nhb o^ uk )

M Y = — , 3 2 = 44

=>Mx=

2.44 = 88

u

' . . ^ ' ' ' ' ' v '

1.6 , ' ' -

=> Chp nD . • :<

Cfiu 2

: So mol BaCOs = 0,2 mol so mol CO j = 0,2 mol ->

khdi liTdng

= 8,8 gam

Khoi liA;Jng

dung dich giam =

khoi

l i M n g kel tua -

( C O 2 + H 2 O ) , ; •

Tdngkhoi I t f d n gC O 2 + H 2 O

= 19,488

= > 4 4. n c o 2

+

1 8 .

n H 2 0 =

19,488 (1)

D o t

X thu diidc C O 2 va H 2 O .

sfjri îr f;,, rp ^

. j s i L i J * ^

.OO nik^

A p dung D L B T K L ta c6 : mx + moj = mc oj +

^ ^ 2 ;4i j r n a ^ q

= > m o 2

=

n i c o 2

^ » 2 ~ ' " x = 14,848 g => IXQ =0,464 mol V f l i j ' ) A

Bao toSn nguyen to o x i : 2.

n c o 2 +

H 2 O

2.

n o 2

= 0,464 'f'

TO

( 1 ) . (2)

n c o 2

= 0,348mol; n H 2 0 = 0'232 mol . ' "

=>nc: n H = 0,348 :0,4 64 = 3

:4=>XmankinC3H4

r -< , .

Cflu3,ChonC.

,„,„„.••.

„: 1 . ; T •

Nicotin

c6 cong thtfc cau tao

nhiT

sau:

Cfiu 4, Ta

c6

phiMng trinh phan iJng:

' : = =

.

c > '

C O 2 + Ca(OH 2 -> CaC03i +

H2O , ( , ; 0 0 f

_mCaC03^

15 Oud

100 100

mH=7^-2

=

0,2(g)

mo =5 .2 -( 0, 2+ 1,8) =

3.2(g)

M , = i M : ^ 1 4

A

3 2

12x y 16z 104

1,8 0,2 3,2 5,2

V a y

C T P T

la C 3 H 4 O 4 => Chon B.

Cfiu 5. Goi

C T P T

cua B I k

C x H y O ,

x = 3;y = 4;z = 4,

X

CxHy O,+

1

y z

y z

O j - J - x C O s + l H j O

2

2 0 0 200

00 250

x = 2; y = 4

Taco: 2,5 = x + | - - | = > z = l =>ChonC.

C S u 6.

A p dung

dinh

luat bao toSn

khdi lifdng:

T O H C + m o 2 = m c o 2 + 1 8 2 0

= ^ ' 0 2 = (10,68 + 5,25) -3, 5 = 12,43

=> Ch on C • - ' • "

Cfiu

7. Goi CT cua

X I h : C x H y O , N

•

''i

750 37.42 „v

, „ , „ * , v f c / ; l

X; / . ; M f i^ ^ ^ ; : H V ' t • ^ . ^ v V ^ J . o ; . ; «

<

nx

= — = 7 6 0 1000

=oooi inol 7,fHr; î;

s'ro fcf:?r'i f-,.;. ,•>

^ RT 0,082.300 ' '

=> V i

khi phan

tich

0,366g

X

thi

n N ,

=

' ^ ^ ' ^ ^ ^

=

O.OOlmol

^ 0,549

(2 )

7/23/2019 Tr1-30(3)


Phan d j ing phuang p h p giSi H6a hp c 11 HOu co - D Xuan iflig

^ m„ = 0,028(g) ; mc = 0,216(g) ;

ITIH

=

0,026(g); = 0,096(g)

0,216

0,026 0,096 0,028

,^ ^ ,

x : y : z : t = - ^ :— :— :— =9 13 : 3 • 1

12 1 16 1

=>

Cong thiJc nguyen (C9H 303N)n.

D o

XC O mot nguyen

tu

N n6n C T P T la

C9H13O3N =>

Chon A. -

C a u 8. Ta c6: V^^ +

Vco + ^ H ^ O + V ÛU

= 1-4

•

Khingufngtu:

VH2o=600ml

/

•

Di n qua

K O H :

Vcoj = 800 - 400 = 4()0ml . , , .

•

,

+Vo2du=400ml

. U

^

/« .(,

,1 , A<,

V O , P / , = 1 V H , O + V C O , ~ +400 = 700ml j ^ , ,

= > V o 2 d u =

200ml „ . H ^ • ) . , ^ . ^

=

200ml

'

The tich cua Y = 400 - 200 = 200ml

y

24 ,

O 2

->xC02+-H2O

200 400 600 ' a' î( m / W F ^ )

^ , . )

??.<»ir

= > x

= 2;y = 6=> Chon A.

u

9.

CH3-CH2-CH2-CH2-CH3

C H 3 - C H

- C H 2

- C H 3

. C H 3 - C H - C H 3

C H 3

, V

.

CH3

C H 3

I

<....

=>Chpn A.

i

, : ; , , :v

/;- r i; , : : ; ; ; :y:

, '>

>

29 55 *

' i I J I f

C a u

10.

Taco: ng coj = = 0.15mol => nco2 = 0.15mol

1 - , . . , , .

,

, / b um ' i l l ' 4 1

\At

I i'* , u l

Matkhac: m d j g i i „ = mBaco3 - ( m c o 2 +

I H 2 0 )

= > m c o 2 + mH20= mBaC03 ~ Jg ^

=

29,55

- 19,35 = 10,2 g

^^^^^

^

M a

m c o 2 =

0,15.44

= 6,6 g=i> m H 2 0 = 3.6 g=> nH20 = 0.2 mol - „

Ta thay: n^jo >

002

hidrocacbon X la ankan va:

nx = n^ jo -

n c 0 2

= 0,2 - 0,15 = 0,05 mol

t i l I

ii

so nguyen tuf C trong X = = — = 3 X la C3H8

^ Dap an

D . v.

12x

CSu

11.

Ta C O dp dung cong thiJc:

niyf mQ

=i>

x = 6; y = 12; z = 6 ^ CTPT

:

CftHijOg

Chon A.

> . 1.

•• 1 . J - .

C a u

12.

Goi CTPT cua X la CxHyO,.

16z _ M

a

Ta c6:

C . H y O ,

+

1

x

+

4 2.

5

x

= 3;y = 8.

O 2

^x C0 2 +IH2O

X y

2

3

4 •

, , ^ ' , - - . , • • 1 '

Matkhac : x + 4 - - = 5=>3 + 2 - - = 5=>z = 0

• ( 5

V l y

X

Id C3H8 = >

Chon B.

/wlfe

1 i » / i i f i r| v' '

if J

fit/HQ

Caul3.

C„H2„02 + ^ ^ 0 2

1

mol

2

3n-2

nC02 + nH20

m o l n

mol n mol

G id

suf c6

Imol

X tham gia phan iJng. Theo phifdng trinh phan tfng ta c6:

no2 bandlu

= ^^Y~

-2 = ;(3n - 2) mol

3n-2 ,

n02

phan iJng -

1

mOl

=>

Tong so mol cua X va O 2 luc dau la: 3n - 2 +

1

= (3n - 1) mol

Sau phan iJng chdy va d 139,9°C thi H 2 O

dang

d the hdi nen tong so mol khi va

h d i

sau phan iJng la:

Skhi =

n c o 2

+

nH20

+ 0 2

J

= n + n + ^ ^ ^ = (3,5n - 1) mol

Nhiet

dO

bmh tnfdc vd sau khong

d o i ,

gia suf thi tich binh khong

d o i ,

ta c6:

s a u

0,8

3 n - l

0,95 3,5n-l

=>

n = 3

V | i y

CTPT cua X Id

C 3 H 6 O 2

=> Dap an D.

Caul4. Tac6sdd6sau:

C,HyO,

+

2,502

->

2CO2

+ 2H2O

Cdn ciJ vdo h0 so phan iJng vd dp dung dinh luat bao todn nguyen to, de

dang

c6

AldC2H40

Dap an

C .

45

Phan d j n g \ phudng p h i p g i i l H6a h p 11 HiTu o Pg Xuan Hung

7/23/2019 Tr1-30(3)


CSu 15 : Goi cong

thiJc

p h a n

ttif

c ua

X \ k :

C , H „ 0 , ( z cd t h e b k n g

0)

QHyO,

+ (x +

^ - ^ ) 0 2

- >

X C O 2

+ ^HzO

K - f

^ f

= > x =

3 v a y

= 8

y

4

2

M a t k h a c :

x + = 5

=:>3

+ 2 - - =

5 = > z

= 0

V y

: X m C 3 H 8 = >

D a p a n B ,

n • ' ' v: •

C a u l ( » . T a c 6 : - ^

= - n H 2 0 > C O ,

Y ;^

.

= >

A ,

B, C

d eu no , d eu

c6

c i l n g

so

n g u y e n

tuT C

va H .

NhiT

v a y

de

p h a n

tuT khoi

cua chiing l|ip

t h ^ n h 1 cap so

cgng th l chi ing khd c nhau

\i so

n g u y e n

tiJ

oxi

t r o n g

phan tuf .

.' ^^

1

- f J*

D S t c o n g t h i J c t o n g q u d t c u a

A , B C i a C „ H 2 „ + 2 0 x

(x> 0)

C„H2„ + 2 0 x —'^^2_>

nCOz

+

( n + 1 ) H 2 0 , U j o i

c o i

n 2

n T ; : T = T ^ = 2= > ^^ ^ '^^ g C 2 H 6 O X . ; , ; ; m fn m>

H 2 0 n +

1

j

V i

C 2 H 6 O X

la

hdp chaft

no

n e n d i ^ u k i ^ n

de

t 6 n t a i

Ik so

n g u y d n

tiJ

o x i

< so

n g u y e n

t u r C

=> x £ 2 => x = 0;

1 ;

2.

, , ,r

V | y :

CTPT cu a

3

cM t

A, B. C

l l n IU*?t

^

C 2 H 6 .

C 2 H 6 O .

CjHf iOa . „ f,,,,,

da

H I D U O C A C B O N N O

A

.T6MTATLfTHUY

§ T

I . A N K A N

(parafin): C „ H 2 „ + 2

(n ^ 1)

* C H 4 . C 2 H 6 , C j H g .

C

„ H 2 „ + 2 1 $ P

thanh d ay d ong d ^ng c i ia ankan.

* TCf C 4 tr 3 d i cd d ong phan cau tao . j ^

* D a n h p h d p :

^ ^ ^ j . .

, . 1^ : . : : . ^ } ^ : .

<

i g i i

+

A n k a n k h d n g p h a n n h d n h

: ,

T e n a n k a n

=

t e n m a c h c h i n h

+

an

. . ,,

+

A n k a n p h a n n h d n h

:

T 6 n

a n k a n

= s6

c h i v i t r i n h d n h

+

t e n n h d n h

+


+

a n

*

T i n h

cha't hda hoc

:

p h a n tfng the ' , td ch, 0 x 1 h d a .

' .

,1

+

Phan

i t ng

t h e ' : C H 4 + C I 2 C H 3 C I

+

H C i

- >

p h a n

tfng

halogen hd a . j . ^ . j

+

P h a n i J n g t i c h :

, \ d

1 i h

i l ' ^ r

C a H 2 a + 2

+

C h H 2 b

. ,

H f

iS'.J)".' - :,.)

: M.:i s y ?

ankan thap anken

p h a n

iJng cr ack inh . . ..

^ . r ^ , j y ^ . ^ ,q ^.m M iH ^ \ V

I A O J

i-

r3n+n

C n H 2 n + 2

r t

t

C n H 2 n + 2

x t

t

( n > 3 )

+ Ph an i?ng ox i hda:

C„H2n+2

+

O2

>

n C 0 2

+

( n

+

1 ) H 2 0

2

K h i

c d x d c t d c t h i c h h d p :

C H 4

+

O 2

H C H O

+H 2 O

* P h 5 n i J n g d i l u c h e an k a n:

; i j A f i

R - C O O N a

+

N a O H R - H

+

N a j C O j

(R

Ik C„ H 2 „ + , - )

h ^ C n H 2 „

+

H 2

C n H z n r f . ^

a X I C L O A N K A N : C o H t o (n ^ 3)

* X i c l o a n k a n

la

nhOhg hid rocacbon

no

m a c h

v 6 n g

( g 6 m m o n o x i c l o a n k a n ( d d n

v 6 n g ) p o l l

x i c l o a n k a n ( d a

v 6 n g ) ) . ; . 1

* D a n h p h d p :

T 6 n

x i c l o a n k a n

= s6

c h i v i

t r i

n h d n h

+

t e n n h d n h

+

x i c l o

+


+

an:

Ch i i y :

- M a c h c h i n h

la

m a c h

v 6 n g .

- D d n h

s^

sao c ho t d n g c c s6 c h i v i t r i c d c m a c h n h d n h Ik nh6 nhSt .

*

T i n h

ch^t hd a hoc

: ^

41

K l f A M f i V l E T

7/23/2019 Tr1-30(3)


Phan dgng

vS

phuong ph^p gl5i

H6a

hgc

11

HO u c o - B6 XuSn Hung

+

Phan u-ng the : , , » , ^

Xiclopentan brom xiclopentan

+

Phan u-ng cpng md vong : THIYUHF

i i

(M

f4C: .fi

/ \

Br2

(dd)

C H 2 - C H 2 - C H 2 i ; y > : f « » B t « t J

1,3-dibrompropan

A

+ HBr ' - ^

C H

- C H 2 - C H 2

- B r

'

•MJ ^q.::«fla

^

,:

1-brompropan

I ' ' into

a n i n f j i i t

n&T

+ H2 C H 3 - C H 2 - C H 2 - C H 3

Xiclobutan

...rfix fU

1 f butan i { iin Au

M i

^ i i / i . - 0 i^n fl'Vi

+

Phan u-ng tdch : /

r

u r

, 4 *

- f i »'{.'ii*

i ^ f i

i S i h i \

, JIMI t

Metylxiclohexan metylbenzen (toluen) ri < > i^ i r i ri\ +

+

Phan u-ng oxi h6a : C„ H 2 „ + —Og nCO^ n H 2 0

* V '

*

Dieuche: CH - ( C H 2 4- C H 3 f j +H2. H . : ;

B. PHA N LOA I VA P H O O N G

P H A P

G I A I C A C D A N G BAI TAP

<ViA eAng.

thi te emi tâ im gnl llii atihati^ adehmnkati , ,«

BA I

TAP

MAU VA BAI

TAP

N A N G C A O " ^ ^

I S ' ' * ^

B a i 1.

Viet cong thiJc phan tuf cua ankan va goc hidrocacbon tifcfng i?ng •

.

^ )Ch^^ lOH b) Ch,fa8C

, . , H . . , & : > ' : « ^ ^

•

c)

Chifa n nguyen tur C ' d) ChiJa (x + 1) nguydn

t

) ^C.

' 2^

C 4 H H

=>

C 4 H 9 -

b)

Q H

, 8 => Q H n - * '

B^ inf io i i i a .

*

'

:

C ) C

„ H 2„ . 2 ^

C

„ H 2 „ . , - : . d)

Q

„ H 2 x ^ = ^

Q

„ H 2 x

. 3 - . '

^ 1

a)

Viet

cong tMc phan

tijf

cua hidrocacbon tifdng iJng vdi cic goc ankyl sau f ' ^ ^

- C H 3 - C 3 H 7 - C 6 H 1 3 .

b)

Viet

c6ng

thuTc

cau tao cua c^c ankan sau : pentan, 2-metylbutan, isobutan. Cic chat

tren c6 ten goi nao khdc khong ?

a) - C H 3

:

hidrocacbon :

CFL,

•

5^;^,; ^ i : / : ? ^ ^

y , ^ ; . ;

; ^ e i V i : l : i

v /Ths,

-CsHv

: hidrocacbon : C 3 H S v j j ^ ; ; ; ) /

- C 6 H 1 3

: hidrocacbon :

C 6 H 1 4 .

v , / ; , ,

• , , g j s

# , : S i l

k :

vûkn

i t d s .• a, .,5

b)

:.il;:> , f i H f . : ; > , . , A H ; : > ,.a

^Ten goi khdc .^^.J;,.

C H 3

- C H 2 - C H 2 - C H 2

- C H 3 pentan n-pentan

H J . M

C H 3

- C H

- C H 2 - C H 3 2-metylbutan isopentan , : ;

2-metylpropan.

C H 3

C H 3

- C H

- C H 3

, isobutan

C H 3

Ba i

3.

a) 0ng vdi propan c6 hai nh6m

ankyl

la

propyl

va isopropyl. Hay

viet

cong thiJc cau

tao cua chiing va cho biet b| c cua nguyen tuf cacbon mang hoa tr i tif do.

b) Hay vie't c6ng thiJc cau tao thu gon va thu gon nhat cua cac chat sau : ,

* isopentan, neopentan, hexan.

* 2,3-dimetylbutan, 3-etyl-2-metylheptan, 3,3-dietylpentan.

idi

Propan :C3Hs

a) Propyl: C H 3 - C H 2 - C H 2 - (bac 1) '

Isopropyl: C H 3

- C H -

(bac 2)

, 011,11, .K rfl 'j ( '

ii

b)

Isopentan :

Neopentan :

C H 3

Cong

thtfc cau tao

C H 3 - C H - C H 2 - C H 3

CTCT

thu gon nh^t

C H 3

C H 3

- C

- C H 3

C H ,

Hexan:

2,3-dimetylbutan

:

C H 3 - C H 2 - C H 2 - C H 2 - C H 2 - C H s i . . , ;

C H 3 - C H - C H - C H 3 ,

C H ,

C H 3

3-etyl-2-metylheptan: C H 3 - C H - C H - C H 2 - C H 2 - C H 2 - C H 3

C H 3 C 2 H 5 • • •;v -

3,3-dietylpentan :

C 2 H 5

C H 3 - C H 2

- C

- C H 2 - C H 3

C 2 H 5

4 9

7/23/2019 Tr1-30(3)


Phan

dgng phuong

phAp g i i i

H6a hpc 11 HOu cO -

D8

Xufln Hung

B a i 4.

D i f a

vao

k h d i

ni§m

d o ng d i n g h o a c d i f a t r en

so

e l e c t r o n

h6a

t r i

cua

cacbon

v a h i d r o .

Hay

chuTng

m i n h

c o ng t h t fc t d ng qu^ t cua ank a n

Ik C„H2„+2.

id i

* D i/a vô

k h a i

n i $ m d o n g d ^ n g :

D o n g d i n g cua m e t a n : CH* , CaHft, CsHx, C4H10

>CH

CH 2

) x

« Q + , H 2 , ^ ; i f

) . . . . ,^r l^ v V , ;

D a t n = x + x = n - 1 ' J

- i : - . ; f : ^M- , / J ; - .

N e n s o n g u y e n t u f H : 2x + 4 = 2 ( n - l ) + 4 = 2n + 2

V a y C T T Q cua a n k a n C„H2„+2. •;,;vyi

>i r

/; i l J-.H

* D i f a vô so e l e ct r o n h 6a t r i : < • ,

D a t so n g u y e n

tuT

C 1^ n => So e l ect ron h6a

tr i

c i i a nC 1^ 4n

So l i e n k e t d d n g i f f a

C - C

(n - 1)

D o do so e h6a

tr i

da

suT

d u n g Ik 2(n - 1) (vi moi l i 6 n k e 't d d n d i i n g 2e h6a tri)

=> So e l e c t r o n h6a tri c6n la i la 4n - 2(n - 1) = 2n + 2 dCing g6p c h u n g vdi

e l e c t r o n h6a tr i cua H tao t h a n h l i e n ke t c o ng h 6a

tr i

C-H .

j> So n g u y e n tu f H la 2 n + 2. „. ^:, , , ,

V $ y C T T Q

cua

a n k a n

m C„H2„+2. ?v .

B a i 5. Hay v i e t co ng t h i Jc cau tao va go i ten t h eo l UPA C c c a n k a n c6 cong thiJc

a

C H

,o b)C5H , 2 c ) Q H , 4 . ,,,,,,,^^^^^^^^^^

id i

a)

C 4 H

) o : CH3-CH2-CH2-CH3 : b u t a n - '

^•/••••x:- - -, ,- :*.:

CH 3-C H-CH3 ' :

2 - m c t y lp r o p a n

b) C 5 H i 2 : CH3-CH2-CH2-CH2-CH3 : p e n t a n ' • '

CH 3-C H-CH2-CH3 , . : 2 - m e t y lb u t a n

'

CH3 .

CH 3-C -CH3

' '^J->: ' -^' '^--

": 2 ,2 -d i m e t y l p r o pan

CH3

c) C f i H u : CH3-CH2-CH2-CH2-CH2-CH3 : h e x a n

. C H 3 - C H - C H 2 - C H 2 - C H 3 : 2 - m e t y l p e n t a n

CH3 '^

C H 3 - < : H 2 - C H - C H 2 - C H 3 : 3 - m e t y l p e n t a n

~ CH3

CH 3-CH-CH -CH3 : 2 , 3 - d i m e t y l b u t a n

CH3CH3

CH •

•

•

CH 3-C-CH 2-CH 3 : 2 , 2 - d i m e t y l b u t a n .

CH3 -<.m.;:i?M/ vy:J,: /H',

B i i

6. V i e t co n g t h i Jc c au tao cdc c h a t c6 t d n s au :

a) X i c l o p e n t a n \ ^---^ yi^S ti b) 3 ,3 -d i b r o m -5 ,5 ,6 - t r i m e t y l o c t an

c)

3 -m e t y l -3 - e t y l p ent an

d)

, 1 ,4 -d i m e t y l x i c l o hex an

e) l - c l o - 3 - x i c l o p r o p y l p e n t a n f) e t y l x i c l o h e x a n

g) l - c l o -3 - e t y l -2 -n i t r o x i c l o pent an .

a) X i c l o p e n t a n : ' '

I ^

b) 3 , 3 - d i b r o m - 5 , 5 ,6 - t r i m e t y lo c t a n : C H 3

- C H 2 - C - C H 2

- ( p — J

: H - C H 2 - C H 3

*

':X, . \: »

' , ' r - r j : r - . . ; r .

' Br

CH3CH3

CH3

'^-^ •

c) 3 - m e t y l - 3 - e t y l p e n t a n : C H 3 - C H 2 - C - C H 2 - C H 3

. ^^Hs .

d) 1 ,4 -d i n i e t y l x i c l o hex an : f | a , i^ ^1^ *

CH • ^ •

e) l - c I o - 3 - x i c l o p r o p y l p e n t a n :

C H 2 - C H 2

- C H

- C H 2 - C H 3

C I

^ .

f) e t y l x i c l o h e x a n : [ ] 1 j, ;

...,... mm ^ ij^:t c i • 'iM •

g) l - c l o -3 - e t y l -2 -n i t r o x i c l o pent an :

B & i

7, Hay v i^t c6ng thuTc cau tao \k go i ten c c h i d r o c a c b on no uTng vd i cong th iJc

phSn

tuf C5H 10. r,v t.i li

> ir - n < m^ -

thlK\ ,;*>'•;;•-'.'

\ . : x i c l o pent an ^

CH3 : 1 ,2 -d i m e t y I x i c l o p ro pan

51

PhSn djng vS phUOng ph^p giSi H6a hpc 11 HOu co

D

Xufln Hung

7/23/2019 Tr1-30(3)


: 1,1-dimetylxiclopropan

^^ ^^ : metylxiclobutan ; j . | . f

^ ^ - C 2H 5

: etylxiclopropan. ,,H,>

Bai 8. Doc ten cac chat iJng vdi cac cong thiJc cau tao sau : ; ,

J J-JJU

,|,,,,,..„ , j^ ; y j

a) CH3-CH2- CH -CH2-CH3 , , j b) C H 3- C H - C H 2 -C H- CH 3 ,

^^ ^^ , ,_^^^^H3GdeH

CH3

, CjHs-„;,,,.,

CH3

•

CH2-CH3

4 0 : CH3 ^

c) C H 3

C H 2

C H C H C H 3 d) C H 3- C H - C H 2 - C - C H 2 - C H 3 i i j

C H

2 < :

H

2 <

H 3

CH 3 CI

e) C H 3 C H

C H 2

C H C H

C H 2

C H

C H 3

C 2 H 5

C H- € H 3

C H C H 3

- (V iet ehi tdi f t l ian ihig

-

Jii ih

ehut hdu hê eua ai tk t ut , aeSdtui t ikat t

BAI TAP M A U V A BAI TAP N A NG C A O

B^ i

1 Viet phiTdng trinh hda hoc cua cdc phan rfng sau :

a) Propan t ac dung vdi clo (theo ti le mol

1 : 1 )

khi chie'u sang.

b) Tdeh mpt phan tuf hidro tilf phan tuf propan.

c) Dd'tchiy hexan.

d) Khi syc khi xiclopropan vao dung dich brom. ;

e) Dan hon hdp xiclopropan, xiclopentan va hidro di vSo trong

6 ng

c 6 bpt niken,

nung

n6ng .

f )

Dun ndng xiclohexan vdi brom theo ti

1

m o l l :

1.

KHAJgGVDBT

Gidi

CH3-CH2-CH2CI + HCl

1

-clopropan

'1 t¥>

<.H''>'4I''>

^<l ?

a) C H 3 C H 2 C H 3 + C I2 ^ CH3-C H-CH3 + HCl

b) CH3-CH2-CH3 ' ' °>

C H 3 H= C H 2

+ Ha;:: ifv rs is

liz

G ) C

6

H 4

—-O2 6CO2 + 7H2O •• • .

d)

+ Br2(dd) CH2-CH2-CH2 1,3-dibrompropan) .

X i c l o p r o pa n ;, Br

c) Khi dan hSn hdp xiclopropan, xiclopentan va H di vao trong ong

c6

bgt Ni,

nung nong thi chi c6 xiclopropan phan iJng cpng vdi hidro,j:>.. H. ) -f J1 ,

/ \ H2

CH3-CH2-CH3

,H : , )

f) +

Br2

^

^ +

HBr. «

•^^; f i :>~l i : |^ f

Xiclohexan , .

Bî 2.

Hai chat A, B

c 6

cung cong thi?c phan

tu C5H12

tdc dung vdi clo theo ti le mol

1 : 1

thi tao ra mot dan xuaft duy nhat con B thi cho bon dan xuat. Yie't cong ti i i fc

cS'u tao cua A, B, cac dan xuat clo cua chiing va goi ten. .

j

Gidi , .

C5H12 CO cac cong thiJc cau tao sau: ; , . . . ; ^

CH3 , ,

C H 3 C H 2 C H 2 C H 2 C H 3 ; C H 3 C H 3 ;

C H 3

H

C H 2 C H 3

„ ,

C H

4

(*,

Qli^>,H

/ t r i

Theo de ta

O :

' i CH3 l > t . 4 H l l .

I -

* A cdC T C T :

C H 3 C H 3

2,2-dimetylpropan (neopcntan)

I 'i • • • '•

=> DSn xug't duy nhat mot Ian the clo :

CH3-C-CH2CI . , ; , ,:.

B c6

CTCT : CH3-CH-C H2-€H3

2

-metylbutan (isopentan)

mt wa

7/23/2019 Tr1-30(3)


P n

dgng

Aphijgng pMp qai

H a

hoc

11

HOu co

Dfi

Xuan

Himg

=> Cic dan xuat mpt Ian the

d o

:

CI

CH3-CH -CH2-CH2CI ;

CH3-C-CH2-CH3

• c j .JH

.

,,i-r>- :

CH3 - r

^+ . H : :

^

.-

CH3-CH-CH-CH3 ; C H 2 - C H - C H 2 - C H 3 .

1 1

H I

CI CH3

B a i

3 Vi^t phUtfng trlnh gpi ten phan iJng cQa isobutan trong cdc tnfdng hdp sau:

a

L a y

1

mol isobutan cho tac dung vdi

1

mol clo c6 chieu sang. ,, i^ ,

b)

L a y 1

mol isobutan dun nong vdi

1

mol brom.

c) Nupg nong isobutan vdi xQc tac Cr203 de tao thanh

C4HX

(isobutilen).

d) Dot isobutan trong khon g khi.

Gid i - v.

J

K

. i.y.U

ofe-f Cl jix

,rufq'/;qokux Qdt Aii lUh JiDs : :

a) CH3

-C H

-CH3

+CI2

CH3-C-CH3 +HC 1 => phan tfng the

. . C H 3 ^ CH3 ' v : . -

h)

CH3

-C H

-CH3

+ Br: C H 3 - C - C H 3 + HBr => pha n iJng thg (1

c) CH3

-C H

-CH3 CH3-C

=CH2

+ H2 iniiv => ph5n i?ng tach

.

s ; - . . . . . jCH I- - • CH3 iM> • » ^ ^

• v

^ i ' A s i fi ;* ;

* • .

13

n ob imr.

.I it

'jk> <H;J mh

d)

C 4 H , „ + y 0 2

).4C02

+

5H20.

^

,

Bai4. w-.,

a) Trinh

b^y

phi/dng phap

hda

hoc

phan bipt

hai khi

khong

mû

propan

va

xiclopropan di/ng trong cdc blnh rieng bi?t.

'^--i O -iU^ JÔ -rl f^

b) Nhan biet cdc khi sau

:

C O 2 .

C H 4 ,

H 2 .

a) Dan

hai

khi

di

qua dung dich brom

c6

mû v^ng da cam, khi nô

l^m

mat

mû dung dich brom

1^

xiclopropan. Con lai la propan. ^

PhiTdng tr\nh phan iJng :

/ \

Br2(dd) > C H 2 - < : H 2 -C H 2

: VBr

Br

b) Dan ba khi Ian li/dt di qua dung dich niTdc v6i trong (diT), khi nao lam due nUdc

voi trong

l£k

C O 2 , con lai C H 4 v a H 2 . , . .

C O 2 +

Ca(OH

)2

> CaCOj

+ H 2O sf ? •

H

Bot chiy hai khi con lai sau dd dan san pham Ian liTdt di qua nifdc v6i trong diT,

khi nao lam due nUdc vol trong la

C O 2 ,

khi ban dau la CR, con lai la H 2 .

CH4 + 2O2

C O 2 + 2H2O

2H 2 + O2 > 2H2O .

B ^ i 5. Vi^t c c phiTdng trlnh phan iJng thifc hi#n chu5i phan tfng sau :

CHjCOONa

— V ^ ^

C H C I 3

C - ^ - / _ _ ( 6 ) _ ^

^

C H 4

H o

HCHO

AI4C3 C2H2.

•

Gid i

' \

(1) CHjCOO Na + NaOH C H 4 + NajCOj ' ' *

>

(2 ) C+ 2 H 2 - ^ C H 4

11

•

(3) C 4 H , 0 CH 4 + C3H6 . • ;

(4)

AUC3 +

I 2 H 2 O 4Al(OH)3 + 3CH4t h

.^: i

^Mh¥ t 5;

(5) C H4 + 3 C I 2 - 2 ^ C H C I 3

+

3HC1 Jii

(6)

CH 4 +

O 2

C + 2H2O „ .ryr -,^ -v-. ^.-m: .. nHih

^r-

itjiJ - -

,,liy' 'r^^..

7) C H 4 O 2 HCHO + H 2 O

(8) 2 C H 4

- ^ J g ^ C 2 H 2 - H 3 H 2 .

Ba i 6 6ng nghiem A chiJa dung dich KOH, ong nghiem B chiJa dung dich H2SO4,

ong nghiem

C

chi?a dung djch KMn04, ong nghiem D chiJa nifdc brom, cho vao

moi ong nghi em dd 1ml octan l^c deu roi de yen. Dif doan

hi§n

tiTdng xay

ra d

moi ong nghiem va giai thich

?

Gid i '

Khi cho 1ml octan

Ian

lifdt vao moi ong nghiem

thi

khong thay hi^n tUdng gi

xay ra trong tCfng d'ng nghi§mvi : ? ,

-

d

ong A

:

octan khong cd phan iJng vdi dung dich KOH.

ong B

:

octan khong phan itng vdi axit

H2SO4.

ong C

:

octan khong cd phan iJng oxi hda vdi dung dich KMn04.

- d6'ng D

:

octan khong cd phan tJng cgng vdi nifdc brom.

B ^ i

7. Viet ciic phUOng trlnh phan

uTng

hda hoc dudi dang cong thiJc cau tao rut gon cua

c5c chat theo cdc day chuyen hda sau (cho biet

A 2

va

A 3

la cac san pham chinh).

nu xt t° / >

Cao

su

buna

^

>

C 4 H 7 B r

{A2, A3

55

Phan d j i n g p h u O n g p h ^ p

g i i i

H 6 a

hgc 11

HO u

co - P8

X u a n H u n g

7/23/2019 Tr1-30(3)


xt , t "

T i x t , t ° , . / C a o s u

' ^ > C.H

•> Cao su isopren

; ..,;in,-',,ui«fi

a;"; ;;.;

, , u . . ; . / - . n u i j . i i h i (Trich TSDH , khoi A)

Gidi

Theo

sd do ta c6 cdc phifOng t r inh phan iJng sau : '' J ^ - ^ •'

* CnH2n+2 : C 4 H 1 0 1'A-. N.i' • . ; > • • /SI

r.. C H 3 - C H 2 - C H 2 - C H 3 ^ C H 2

= C H - C H

= C H 2 + 2 H 2 :1 , i V H . ,

C H 2

= C H - C H

= C H 2 - ( C H 2

- C H = C H

- C H 2 } „

^ ^ ' 5 > C H 3 - C H < H < H 2 ^

1,2 I

C H 2

= C H - C H

= C H 2 +

H B r — (

Br A2 )

C y n g

^ C H 3- C H = C H - C H 2 B r

.'3 f f t )

; / . ^ ^ ; : r n : i

I /

A3) ,

- m n 2 m + 2 • * - 5 M | 2

'

. "1

C H 3 - C H 2 - C H - C H 3 C H 2 = C H - C = C H 2 2 H 2 ^ ^ f

,

..,„..,. ,

M^. ,

. u ,

A 4

< ^ 3

>

. r i >

C H 2

= C H - C

= C H 2

- r C H 2

- C H = C

- C H 2

4 (cao su isopre n) ' '

I

4 111,1

C i i v

o r f i i

, : v i f i ( i

vîiii r f t j i i

I

J O ^ K I

C H 2 = C H - C = C H 2 B r 2

C H 3

: , , : i i >

A 5 )

:')io i l }5« |nfe

- ^ J ^ C H 2 - C H = C - C H 2 B r

^ f i r f J

' . . . B r CH3 (A ,) . '

| ^ # a ' • • ^ v l y u ' i Vi ^ ^ ^ ^ 3 j - * C H 2 - C H - C = C H 2 j ^ , } ,5,,^

j ^ ; ^

.. . Br Br C H 3 (Av);,^-,,,,,,,

a) C ong th i?c c i ia mpt h idrocacbon

A c6

d ang

(CnHzn+i . Vay A

t hup c

day

d ong

d i n g n^ o ?

b ) V i e t cdc phi fdng

tr inh

phan ufng : C„H2„+2 tac d y ng vdi C I 2 (ask t ) , c rack inh b ^ng

n h i e t .

Cho v i d u

?

. v

x j

;-:,,;•.•• ,

^••••r'

• dd

^\iiv : -^krk

a)

(C n H 2„ + i ) x

hay

C n x H j n x + x .

• , X : : , . . . ' L : , * : L , . „

" . '

V i

so

n g u y e n tuT

H

t rong h idrocacbon luo n 1^ s6 ' chS'n nen

* K h i X = 1 ==> C n H 2„ i : la goc h id r oc ac b on c6 hd a t r i 1 ( l o a i ) v i A la m p t hd p c ha t .

6

* K h i

X

= 2 =>

C 2 n H 4 n

+ 2 ha y C2nH2.2nf2 la m O t ank an (nhan ) .

* K h i

X =

3 => C3nH6„+3

( l o a i )

v i H c6 so le.

* K h i

X = 4 =>

C4nHsn

( l o a i ) vi

k hong p hu hd p

vdi hda

tr i

cua

c ac b on .

Vi

d u

:

C 4 H i 2 ( n = l

) l o a i .

, •/

V $ y A t hup c d ay d ong d in g c ua ank an . ^ :

b)

C „ H 2n . 2

+

x C l 2 C „ H 2 „ . 2 - x C l x +

x H C l

V i

d u : C 2 H 6 + C I 2 C 2 H 5 C I + H C l

C „ H 2 „ . 2

^ y ' ' >

C aH 2a. 2

+

C h H 2b

( v d i

a + b = n)

V i d u : C 3 H « C H 4 . C 2 H 4 . '

i 9 . Xac d i n h

co ng thiJc cafu

tao

c i i a C f iH u , b i e t r ang k h i

tac

d u n g v d i

e ld

theo t

1

m o l 1 : 1 c h i t hu d iT d c ha i d ong p han . G p i t e n ha i d ong p han do.

Con g th iJc cCu tao Cf iHM : f ^ ^ ^ ^ ' ' * '

; C H 3

- C H - C H

- C H 2 +HC1 '

V ^ . i

C H 3 C H 3 C I J •

: vMMtQM:>M

/

l - c l o -2 ,3 -d im e t y lb u t an

* : i U ' ' i - ' '

C H 3 - C H - C H - C H 3 + C I 2

••0..'V-.-.-.'-^.'';'^

C H 3 C H 3 C H j - C H - C - C H j +HC1

2 -c l o -2 ,3 -d im e t y lb u t an

,1

Bai

1 0 . Cho b i e t CTCT c u a p e n t a n t r o n g cac

t r i ^ n g

hd p s au :

a) Tac d u n g v d i C I 2 ( a s k t ) t i le 1:1 ch o 4 sa n p h a m

b ) K h i c r a c k i n g c h o

2

sa n p h a m .

Gidi

L f n g v d i p e n t a n

C 5 H 1 2

c 6

cac

da n g k h u n g

C

s a u t i

- i i >

1

Si •

•• ' lii

iV ,

v . ^ i i -tVV:/' '^

i.^^ i r ' i i . V •

/'''i^''

I )

C (2) ^ C

(3 ) '•••-•',e',:.

a ) K h i t h iT c h i e n p ha n t fng the^:

(1 )

C O

3 v i t r i t h e ( C 1 , C 2 , C3) t a o 3 sa n p h a m ( l o a i ) I t ' '

( 2 ) C O 4 v i t r i t h e ( C I , C 2 , C 3 , C4) ta o 4 s a n p h a m ( n h a n )

(3 )

C O

1 v i t r i t h e (CI h o S c C3) -> t a o m p t s a n p h a m ( I p a i )

V a y CTCT c u a p e n t a n la (2) : 2 - m e t y l b u t a n ( i s o p e n t a n ) '

P t p i ? : ^ ^

Phfln d?ng va phuong ph p gi i H6a hpc 11 HQu cd - Oi XuSn Htmg

JgWAlgft VWOT

7/23/2019 Tr1-30(3)


C H 3 - C H - C H j - C H 3 C I 2

C H 3

*)

C H 2 C I — H - C H 2 - C H 3

C H 3

C H 3 —

C C I

- C H 2 — C H 3

C H 3

C H 3 — C H - C H C I - C H 3

C H 3

C H 3 —

C H

- C H 2 — C H 2 C I

b)

TMngM--

CTCT cua pentan 1 (3): 2,2-dimetylpropan (neopentan), ichi cracking chi cho

2 san pham : ^

•.>.;•,;.•• •V':;?i

? ai

;V

)-.v ruvb

J U X r-

C H 3 Srlvj ;lf f ?i:5--ri,f<iit ;.;;;, /vyfV ;, ( i / . =,» ;> 4,?L ;;r;;, ta;,. j • ; U V

o •

C H 3

- C H 3

cracking,t _

> - C H 2 - C - C H 3 + CH4 ,„,^^,Mr;,n5:>

. .,H fv:>...,}; C H 3 "•'"^

a) Viet c6ng thil-c cau tao va goi t6n cdc d6ng phan cua CjHsBrj.

b)

Nh$n

b i^ t

cdc cha sau :

C 2 H 6

S O 2

N H 3 ,

O^.

,

^, j ,.,M:>. fs :> ,

H 5 >

Gidi

a) CaHsBrs cd cac dong phan :

C H 3 - C H 2 - C - B r

B r

CH 3-CH-CH-Br

I

B r

Br

C H 2 - C H 2 - C H - B r

B r

Br

C H 2 - C H - C H 2

M I ^

B r Br Br

B r

C H 3

- C

- C H 2

- B r

B r

1.1.1- tribrompropan

1.1.2- tribrompropan

• I

1,1,3-tribrompropan

1,2,3-tribrompropan

1,2,2-tribrompropan

)) 5 (

1

1 '

b) C 2 H 6 , S O 2 NH 3 O 2

- Dan bon khi l^n IiTdt di vao dung

djch

brom, khi nô lam mat mau dung

dich

brom

S O 2 , con lai la C 2 H 6 , O 2 N H 3 .

S02 + B r 2 + 2H20

>2HBr

+ H2S04

Cho quy tim am Ian

I tfdt

vao ba kh i c6n

l a i ,

khi nô lam quy tim am h6a xanh la

N H 3 , c6n lai la CzHg,

O 2 .

Cho que ddm c6n t^n 66 \ko hai khi c6n lai , khi l^m cho que d6m c6n tSn do

bilng chdy m O 2 , c6n lai

m

C 2H6 .

, ; ,1 ; , ^ ; .

( p a t u i 3. :

.x ;: ' . ' , : : , ' '

,'.

~ ^ d i o e a n k a i t f x i e lo t u i k a n

grinh

H t d n h

fihdit

-

£ g f t

e u g . I l t i t e ( u t k a i t f x i e ie m t i u u i

B A I

T A P M A U

B a i

1.

CT ddn gian nhat cua 1 ankan \k ( C 2 H 5 ) „ . Hay bi^n luan de t im CTPT cua

chat tr6n.

Cdch 1: CT ddn gian tren la 1 goc ankan h6a tri 1

ttfc

c6 kha nSng ket hdp them

v di 1 goc

nhir

vay nO-a = > n = 2 CTPT ankan

C 4H , , ,

- •

GicA

2: CTPT cua ankan tren: ( C 2 H 5 ) „ =

CxH2x+2

^:f^€--: ^my^\i' ' H

=> 2n =

X

5n = 2x + 2 jj^Jeicfs&iiV>

=> 5n = 2.2n + 2 => n = 2 => CTPT ankan :

C4H10

u-i.« , > .

C / f c A i ; Ankan tren phdithoadieukien S H<2 .so C+ 2 ,

=>5n<2.2n + 2

=:>n^2

«' >n,' '; V c-. vsîJ,11, : $ ' a O . - ' ' : ^ - •

n = 1 thi s6'H 16 => loai ,, , . , ,,,

n = 2 => CTPT ankan la

C4H o

(nhan)

j

<

• j „

V § y

CTPT ankan

1^

C4H10

, i "

n

B a i 1. Xicloankan ddn v6ng X c6 ti khoi so vdi nitd bJlng 2. L§p cong thiJc phan tuT

cua

X .

Vie't phiTdng trinh h6a hoc (d dang cong thtfc cau tao),

minh

hoa tinh chat

h6a hoc cua X, bie't rJlng X tic dung vdi

H 2

(xuc tdc Ni) chi tao ra mpt san

pham.

•

.

:

S l i t ftiifkninnT -

D | t CTTQ cua xicloankan X :

CnH2n

,; • . v

ij;

,3

TheodI :

M x

= 2x28 = 56

r:[{-H-l'l ' '

T •['-'•''^-

Ta c6: 14n = 56 o n = 4 .

fntV;.^.

V

V |y CTPT cua X la

C 4H8 .

V i X +

H 2

( N i , t ) -> mpt sin pham duy nha't nen X c6 CTCT - . [ ^

* PhiTdng t f inh hoa hoc

minh

hpa tinh chat hda hpc ciia X :

Tr1-30(3)

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