Nhung Bt Co Nhieu Pp Giai

8/9/2019 Nhung Bt Co Nhieu Pp Giai

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PHN MTMT S PHNG PHP GII TON HO

1. Phng php p dng s bo ton khi lng, s mol nguyn tC s Trong cc qu trnh ho hc th :Tng khi lng ca cc cht trc phn ng lun bng tng khi lngca cc cht sau phn ng :

(tr- cphnng) (sauphn ng)m m= Tng s mol nguyn t ca nguyn t A trc phn ng lun bng tngs mol nguyn t ca nguyn t A sau phn ng.

= A(tr- c phn ng) A(sauphn ng)n nCch p dngKhi gii bi tp trc nghim ta nn lp s tm tt cc phn ng, rdng nhng s bo ton trn tm ra cc i lng khc nh : s mkhi lng cc cht trong s phn ng th bi ton s c gii nhhn. Bi tp minh ha

Bi 1.Ngi ta cho t t lung kh CO i qua mt ng s ng 5,44 g hn hpgm FeO, Fe3O4, Fe2O3, CuO nung nng, kt thc phn ng thu c hnhp cht rn B v hn hp kh C. Sc hn hp kh C vo dung dch nvi trong d thy c 9 g kt ta v kh D bay ra. Khi lng cht rn B tc lA. 3g B. 4g

C. 5g D. 3,4g

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Li giiS phn ng:

FeO

CO + Fe2O3 ot A + CO2

Fe3O4CuO

CO2 + Ca(OH)2 CaCO3 + H2O

0,09

9 0,09(mol)100

=

Theo nh lut BTKL th 2CO A B COm m m m+ = +

0,09.28 + 5,44 = mB + 0,09.44m = 4g

Bi 2. Cho mg hn hp A gm ba mui XCO3, YCO3 v M2CO3 tc dng vidung dch H2SO4 long, phn ng xy ra hon ton thu c 4,48 lt CO2(ktc), dung dch B v cht rn C. C cn dung dch B thu c 20 g mukhan. Nung cht rn C n khi lng khng i thy c 11,2 lt kh CO2(ktc) bay ra v cht rn D c khi lng 145,2 g. m c gi tr lA. 170g B. 180gC. 190g D. 200gLi gii

XCO3

YCO3 + H2SO4 mui B + CO2 + H2O + C

M2CO3 Nhit phn B

C ot D + CO2

2C D CO

1,12m m m 145,2 .44 167,2(g)22,4= + = + =

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Phng trnh ion rt gn khi cho A tc dng vi H2SO423 2 2CO 2H CO H O

4,480,4 0,2 0,222,4

++ +

=

m + 0,2.98 = 20 + 0,2.44 + 0,2.18 + 167,2 m = 180 gBi 3. Ha tan hon ton 3,22g hn hp X gm Fe, Mg, Zn bng mt lng v

dung dch H2SO4 long, thu c V lt H2 (ktc) v dung dch cha m ghn hp mui Y. Cho ton b lng H2 trn i t t qua ng s ng 4 ghn hp gm Fe2O3, CuO nung nng, thu c 3,04g hn hp kim loi. mc gi tr lA. 8,98g B. 8,89gC. 7,89g D. 6,98gLi gii

S (1) phn ng ca X tc dng vi H2SO4 long:

2 4 2

FeMg H SO Hnh p mui Y + HZn

+

S (2) phn ng kh Fe2O3, CuO bi kh H2 :

2 3Fe OCuO + H2

FeCu + H2O

Bn cht cc phn ng xy ra theo s (2) lH2 + O(oxit) H2O

2H O4 3,04n n 0,06(mol)

16= = =

Theo s (1) th mmui= 24X SOm m 3,22 0,06.96 8,98g+ = + =

Bi 4.Nung nng m g hn hp X gm ACO3 v BCO3 thu c m g hn hp rnY v 4,48 lt kh CO2. Nung nng Y n khi lng khng i thu thm

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c kh CO2 v hn hp rn Z. Cho ton b kh CO2 thu c khi nung Yqua dung dch NaOH d, sau cho dung dch BaCl2 d vo dung dch trnth thu c 19,7 g kt ta. Mt khc cho CO d qua hn hp Z nung n

thu c 18,4 g hn hp Q v 4,48 lt kh CO2 (ktc) . m c gi tr lA. 34,8 g B. 25,7gC. 44,1g D. 19,8gLi giiS phn ng nhit phn :

3

3

ACO

BCO Y + CO2 (1)

Y 0t Z + CO2 (2)

2BaClNaOH 22 3 3CO CO BaCO

19,70,1 0,1(mol)197

=

2CO Z Q CO+ + (3)

Bn cht ca s (3) l :

CO + O(trong Z) CO2

m(trong Z)=4,48 0,2(mol)22,4 =

Z Q Om m m 18,4 0,2.16 21,6(gam) = + = + =

2 Y Z COm m m 21,6 0,1.44 26(gam) = + = + =

2X Y COm m m 26 0,2.44 34,8(gam) = + = + =

Bi 5. Ho tan hon ton hn hp gm 0,2 mol FeO, 0,3 mol Fe2O3, 0,4 mol

Fe3O4 vo dung dch HNO3 2M va , thu c dung dch mui v 5,6 lt

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kh hn hp kh NO v N2O4 (ktc) c t khi so vi H2 l 33,6. Th tchdung dch HNO3 tham gia phn ng l

A. 3,6 lt B. 2,4 ltC. 3,2 lt D. 4,8 ltLi giiS phn ng :

ot2 3 3 3 3

3 4

FeO

Fe O HNO Fe(NO )Fe O

+ +2 4

NON O + H2O

t 2 4NO N On x(mol) ; n y(mol)= =

5,6x+y =0,25x 0,1mol22,4 Ta c h

30x 92y y 0,15mol33,62(x y)

= = + ==+

p dng s bo ton nguyn t Fe tnh s mol Fe(NO3)3 :

3 3 2 3 3 4

3 3 2 3 3 4

Fe(Fe(NO ) Fe(FeO,Fe O ,Fe O )

Fe(NO ) FeO Fe O Fe O

n nn n 2n 3n 0,2 2.0,3 3.0,4 2mol

= = + + = + + =

p dng s bo ton nguyn t N :

3 3 3 2 4 3N(HNO ) N(Fe(NO ) NO N O ) HNOn n n 3.2 0,1 2.0,15 6,4mol+ += = + + =

Vy 3HNO 6,4V 3,2lt2= =

Bi 6. Ha tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo dungdch HNO3 (va ), thu c dung dch X (ch cha 2 mui sunfat) v mkh duy nht l NO. Gi tr ca a lA. 0,12 mol B. 0,04 mol

C. 0,075 mol D. 0,06 molLi gii

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S phn ng :

2

2

FeSCu S + HNO3

2 4 3

4

Fe (SO )CuSO + NO + H2O

p dng nh lut bo ton nguyn t Fe, Cu, S

FeS2 Fe2(SO4)30,12 0,06

Cu2S CuSO4a 2a

2 2 2 4 3 4S(FeS ) S(Cu S) S(Fe (SO ) ) S(CuSO )n n n n+ = +2 2 2 4 3 4FeS Cu S Fe (SO ) CuSO2n n 3n n+ = +

2.0,12 + a = 3.0,06 + 2a a = 0,06 mol

Bi 7.Thi t t hn hp kh X gm CO v H2 i qua ng ng 16,8 g hn hp Ygm 3 oxit gm CuO, Fe3O4, Al2O3 nung nng. Sau khi phn ng hon tonthu c m g cht rn Z v mt hn hp kh T, hn hp T nng hn hn hX l 0,32 g. Gi tr ca m lA. 14,28g B. 16,46gC. 16,48g D. 17,12gLi giiS phn ng

2 3

2 3

CuOFe OAl O

+2

COH Z + T

Ta thy X + O(oxit) T

T X O(oxit)m m m 0,32g = = m Y Z Om m m= +

Z Y Om m m 16,8 0,32 16,48g = = =

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Bi 8. Kh hon ton m g hn hp CuO, Fe3O4 bng kh CO nhit cao, thuc hn hp kim loi v kh CO2. Sc kh CO2 vo dung dch Ca(OH)2 thuc 20 g kt ta v dung dch A, lc b kt ta, cho Ba(OH)2 d vo dungdch A thu c 89,1 g kt ta na. Nu dng H2 kh hon ton m g hnhp trn th cn bao nhiu lt kh H2 (ktc) ?

A. 16,46 lt B. 19,72 ltC. 17,92 lt D. 16,45 ltLi giiS phn ng :

3 4

CuOFe O + CO

ot CuFe + CO2 (1)

Cho CO2 vo dung dch Ca(OH)2 th

2

2

3

Ca(OH)

2 Ba(OH)3 2 3 3

20CaCO 0,2(mol)100

CO (2)

Ca(HCO ) CaCO BaCOx x

=

+

Z

]

100x 197x 89,1 x 0,3(mol)+ = =p dng s bo ton nguyn t C

2 3 3

2 3 3

C(CO) C(CO ) C(CaCO ) C(BaCO )

CO CaCO BaCO

n n n n

n n n (0,2 0,3) 0,3 0,8(mol)

= = +

= + = + + =

Bn cht cc phn ng xy ra trong (1) l :

CO + O(oxit) CO2 0,8 0,8 0,8

Nu dng H2 kh m g hn hp CuO, Fe3O4 th bn cht cc phn ng

l

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H2 + O(oxit) H2OTng s mol nguyn t oxi trong hai qu trnh ny bng nhau nn

2 2H O H (ktc)n n 0,8(mol) V 0,8.22,4 17,92 (lit)= = = =

Bi tp vn dng

Bi 1. kh hon ton 27,2 g hn hp gm Fe, FeO, Fe3O4 v Fe2O3 cn va 6,72 lt CO (ktc). Khi lng Fe thu c lA. 18,9 g B. 22,4 g

C. 19,8 g D. 16,8 g Hng dnS phn ng

3 4

2 3

FeFeOFe O

Fe O

+ CO Fe + CO2

Bn cht cc phn ng xy ra trong s trn ch l :

CO + O(oxit) CO2

nCO= nO =6,7222,4 = 0,1 (mol)

mFe = mhh mO (oxit) = 27,2 16.0,3 = 22,4 g.Bi 2.Cho t t mt lung kh CO i qua ng s ng m g hn hp gm Fe, Fe

Fe3O4, Fe2O3 nung nng, kt thc phn ng thu c 64g st, kh i ra gmCO v CO2 cho sc qua dung dch Ca(OH)2 d c 40g kt ta. Vy m cgi tr lA. 70,4g B. 74gC. 47g D. 104g

Hng dn

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Kh i ra sau phn ng gm CO2 v CO d cho i qua dung dch Ca(OH)2 d :

CO2 + Ca(OH)2 CaCO3+ H2O

0,4 (mol) 4,010040

= (mol)S phn ng:

FeO

CO + Fe2O3 Fe + CO2

Fe3O4

28.0,4 + m = 64 + 44.0,4m = 70,4gBi 3. Kh hon ton 24 g hn hp CuO v FexOy bng H2 d nhit cao thu

c 17,6 g hn hp hai kim loi. Khi lng nc to thnh lA. 3,6 g B. 7,2 g C. 1,8 g D. 5,4 g

Bi 4. tc dng ht 5,44 g hn hp CuO, FeO, Fe2O3 v Fe3O4 cn dng va

90ml dung dch HCl 1M. Mt khc, nu kh hon ton 5,44 g hn htrn bng kh CO nhit cao th khi lng st thu c lA. 3,20g B. 4,72 g C. 2,11 g D. 3,08 g

Bi 5.Thi rt chm 2,24 lt (ktc) hn hp X gm CO v H2 (ly d) qua ng sng 24 g hn hp Al2O3, CuO, Fe2O3 v Fe3O4 un nng. Sau khi kt thc

phn ng, khi lng cht rn cn li trong ng s lA. 12,4 g B. 14,2 gC. 22,8 g D. 22,4 g

Hng dn

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2 3

2 3

3 4

Al OCuOFe O

Fe O

+2

COH Cht rn +

2

2

COH O

Ta thy 1 mol CO hoc 1 mol H2 u phn ng vi 1mol O :

2

COH + O(oxit)

2

2

COH O

0,1 0,1Khi lng cht rn cn li l 24 0,1.16 = 22,4 g

Bi 6. Cho hn hp gm : FeO (0,01 mol), Fe2O3 (0,02 mol), Fe3O4 (0,03 mol)tan va ht trong dung dch HNO3 thu c dung dch cha mt mui v0,448 lt kh N2O4 (ktc). Khi lng mui v s mol HNO3 tham gia phnng lA. 32,8 g ; 0,4 mol B. 33,88 g ; 0,46 molC. 33,88 g ; 0,06 mol D. 33,28 g ; 0,46 mol

Hng dnS phn ng :

FeO

Fe2O3 + HNO3 Fe(NO3)3 + N2O4 + H2O

Fe3O4

p dng s bo ton nguyn t Fe :3 3 2 3 3 4Fe[Fe(NO ) ] Fe[FeO,Fe O ,Fe O ]n n=

3 3Fe[Fe(NO ) ]n = 2 3 3 4FeO Fe O Fe On 2n 3n+ +

= 0,01 2.0,02 3.0,03 0,14(mol)+ + =

3 3Fe(NO )m 0,14.242 33,88(g)= =

p dng s bo ton nguyn t N :

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3 3 3 2 4

3 3 3 2 4

N [HNO ] N [Fe(NO ) N O ]

HNO Fe(NO ) N O

n nn 3n 2n 3.0,14 2.0,02 0,46(mol)

+= = + = + =

Bi 7. Cho 1,1 g hn hp Fe, Al phn ng vi dung dch HCl thu c dungdch X, cht rn Y v kh Z, ho tan ht Y cn s mol H2SO4 (long) bng 2 ln s mol HCl trn, thu c dung dch T v kh Z. Tng thtch kh Z (ktc) sinh ra trong c hai phn ng trn l 0,896 lt. Tng khlng mui sinh ra trong hai trng hp trn lA. 2,92 g B. 2,67 gC. 3,36 g D. 1,06 g

Hng dnS phn ng :

FeAl + 2 4

HClH SO hn hp mui (X+T) + H2

t nHCl= x mol ; 2 4H SOn = y mol

p dng s bo ton nguyn t H :

2 4 2

2 4 2

H (HCl H SO ) H(H )

HCl H SO H

n nn 2n 2n 0,04mol

+ =+ = =

x + 2y = 0,04y =2x x = 0,008 ; y = 0,016

mmui= m(Al,Fe)+ 2Cl SO4m m + = 1,1 + 0,008.35,5 + 0,016.96 = 2,92 (g)

Bi 8. Cho 2,48 g hn hp 3 kim loi Fe, Al, Zn phn ng va ht vi dung dH2SO4 long thu c 0,784 lt kh H2 (ktc). C cn dung dch, khi lngmui khan thu c lA. 4,84 g B. 5,84 gC. 5,48 g D. 4,56 g

Hng dnS phn ng :

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Fe FeSO4

Al + H2SO4 Al2(SO4)3 + H2

Zn ZnSO4p dng s bo ton nguyn t H:

n 2 4H SO = n 2H =0,78422,4 = 0,035 (mol)

p dng s bo ton nguyn t S:

2 24 2 4 4SO (H SO ) SO (mui)n n = = 0,035 (mol)

mmui= m(Fe, Al, Zn)+ 24SOm = 2,48 + 0,035.96 = 5,84 (g)

Bi 9. Ho tan 2,57g hp kim Cu, Mg, Al bng mt lng va dung dcH2SO4 long thu c 1,456 lt kh X (ktc), 1,28g cht rn Y v dung dchZ. C cn dung dch Z thu c m g mui khan, m c gi tr lA. 7,53g B. 3,25gC. 5,79g D. 5,58g

Hng dnCch gii tng t bi 8S phn ng :

CuMgAl

+ H2SO4 42 4 3

MgSOAl (SO ) + Cu + H2

Cu khng tc dng vi HNO3 long nn 1,28 gam cht rn Y l Cu.+= + = + =2

4(Al Mg) SOm m m (2,57 1,28) 0,065.96 7,53(g)

Bi 10.Cho 17,5 g hn hp gm 3 kim loi Fe, Al, Zn tan hon ton trong dungdch H2SO4 long thu c 5,6 lt kh H2 ( 0oC, 2 atm). C cn dung dch,khi lng mui khan thu c l

A. 65,5 g B. 55,5 g

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C. 56,5 g D. 55,6g Hng dn

2H

2.5,6n 0,5(mol)0,082.273= =

ZnFeAl

+ H2SO4 dung dch 3 mui + H2

24

(Al,Zn,Fe) SOm m m = + = 17,5 + 0,5.98 = 65,5 (g)

Bi 11. Cho 35g hn hp Na2CO3, K 2CO3 tc dng va ht vi dung dch BaCl2.Sau phn ng thu c 59,1g kt ta. Lc tch kt ta, c cn dung dch c m g mui clorua. Vy m c gi tr lA. 38,3g B. 22,6gC. 26,6g D. 6,26g

Hng dn

S phn ng: 2 3

2 3

Na COK CO + BaCl2 BaCO3 +

NaClKCl

= =2 3BaCl BaCOn n 0,3(mol)

p dng nh lut bo ton khi lng:+ = +2hh BaClm m m m

m = 35 + 0,3.208 59,1 = 38,3 (g)

Bi 12. Cho 4,48g hn hp cht rn Na2SO4, K 2SO4, (NH4)2SO4 tan vo ncc dung dch A. Cho A tc dng va vi 300 ml dung dch Ba(NO3)20,1M. Kt thc phn ng thu c kt ta B v dung dch C. Lc tch kta, c cn dung dch C thu c m(g) mui nitrat. Vy m c gi tr lA. 5,32g B. 5,23gC. 5,26g D. 6,25g

Hng dn

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S phn ng:

Na2SO4 NaNO3

K 2SO4 + Ba(NO3)2 BaSO4 + KNO3(NH4)2SO4 NH4 NO3

= =3 2 4Ba(NO ) BaSOn n 0,03(mol)

C C4,48 7,83 6,99 m m 5,32(g)+ = + =

Bi 13.Ho tan hon ton 3,72g hn hp 2 kim loi A, B trong dung dch HCl dthy to ra 1,344 lt kh H2 (ktc). C cn dung dch sau phn ng thu c

mui khan c khi lng lA. 7,12g B. 7,98gC. 3,42g D. 6,12g

Hng dnS phn ng :

A

B + HCl n

m

ACl

BCl + H2

+ = = =H Cl1,344n n 2. 0,12(mol)22,4

mmui= mKL+ Clm = 3,72 + 0,12.35,5 = 7,98 (g)

Bi 14.Nung m g hn hp A gm 2 mui MgCO3 v CaCO3 cho n khi khngcn kh thot ra thu c 3,52g cht rn B v kh C. Cho ton b kh C hth ht bi 2 lt dung dch Ba(OH)2 thu c 7,88g kt ta. un nng dungdch li thy to thnh thm 3,94g kt ta na. Cc phn ng xy ra hoton, m c gi tr lA. 7,44g B. 7,40gC. 7,04g D. 4,74g

Hng dn

m = mB + 2COm

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CO2 + Ba(OH)2 BaCO3 + H2O2CO2 + Ba(OH)2 Ba(HCO3)2

Ba(HCO3)2 BaCO3 + CO2 + H2Om = 3,52 + (7,88 3,942. ).44 7,04197 197

+ = (g)

Bi 15.Cho hn hp A gm 0,1 mol Cu, 0,2 mol Ag phn ng ht vi V lt dundch HNO3 1M thu c dung dch X v hn hp Y gm 2 kh NO, NO2 (

2NO NOn n 0,1mol= = ). V c gi tr l

A. 1 lt B. 0,6 ltC. 1,5 lt D. 2 lt

Hng dnS phn ng:CuAg + HNO3

3 2

3

Cu(NO )AgNO + 2

NONO + H2O

p dng s bo ton nguyn t Cu, Ag ta c :nCu = 3 2Cu(NO )n = 0,1 mol v nAg = n 3AgNO = 0,2 mol

p dng cho nguyn t N :

3N (HNO )n = 3 2 3 2N(Cu(NO ) AgNO NO NO )n + + +

3(HNO )n = 3 2 3 2Cu(NO ) AgNO NO NO2n n n n+ + +

3(HNO )n = 2.0,1 + 0,2 + 0,1 + 0,1 = 0,6 mol

3HNO0,6V 0,6(lit)1

= =

Bi 16.A, B l 2 kim loi thuc nhm IIA. Ha tan hon ton 10,94 g hn hp Xgm 2 mui clorua ca A v B vo nc c 100 g dung dch Y. kta ht ion Cl- c trong 50 g dung dch Y phi dng dung dch c cha 10,2

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g AgNO3. Nu cho Y tc dng vi dung dch H2SO4 d (gi thit ASO4 vBSO4 u kt ta), th thu khi lng kt ta thu c lA: 12,44 B: 13,44

C: 14,33 D: 13,23Hng dn

ACl2

+ AgNO3 AgCl + ...

BCl2

Trong 50 g dung dch Y : AgClCl Agn n n 0,06(mol) += = =Trong 100 g dung dch Y :

(A,B)Cln 0,12(mol) m 10,94 0,12.35,5 6,68 g = = =

S mol in tch 24Cl SO2n n = m = 6,68 + 0,06.96 = 12,44 g

Bi 17.t chy m g mt hirocacbon A vi 11,76 lt O2 (ktc) va . Phn ng

to ra 8,1 g nc v mt lng CO2. Cng thc phn t ca A lA. C2H6 B. C2H4

C. C3H6 D. C2H8 Hng dn

x y 2 2 2C H O CO H O+ +

p dng s bo ton nguyn t O : = + = + =2 2 2 2 2 2 2O(O ) O(CO ) O(H O) O CO H O COn n n 2n 2n n n 0,3(mol)

= =C Hn 0,3(mol) ; n 0,9(mol) CTGN l CH3, CTPT l (CH3)n

+ 2 63n 2n 2 n 2chn n=2 CTPTC H

Bi 18.t chy m g hp cht A (CnHn1ONa) vi mt lng va l 6,272 ltO2 (ktc) thu c 2,12 g Na2CO3 v hn hp X cha CO2, H2O. Nu cho

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hn hp X qua bnh ng H2SO4 c th khi lng bnh tng 1,8 g. Vy mc gi tr lA. 6,46 B. 4,64

C. 4,46 D. 6,44 Hng dn

CnHn1ONa 2O+ Na2CO3 + CO2 + H2O

2 3 2 3Na(A) Na(Na CO ) Na COn n 2n 2.0,02 0,04(mol)= = = =

2 2 3 2 2

2 2

O(A) (O ) Na CO CO H O

CO CO

n 2n 3n 2n n

2n 2.0,28 0,04 3.0,02 0,1 0,44(mol) n 0,22(mol)

+ = + +

= + = =Am 12(0,02 0,22) 0,1.2 0,04.(16 23) 4,64(g)= + + + + =

Bi 19 . Thu phn hon ton 1 este n chc A cn va 100ml NaOH 1M thc ancol etylic v mui ca axit hu c B. Phn hu hon ton B thc 5,6 lt kh CO2 (ktc), 4,5 g H2O v m g Na2CO3. Cng thc cu toca A l

A. C2H5COOC2H5 B. CH3COOC3H7C. C3H7COOC2H5 D. C3H7COOCH3

Hng dnS phn ng :

2 5

2 2 2 3

A NaOH C H OH B

B CO H O Na CO

+ +

+ +2 5A NaOH C H OH Bn n n n= = = = 0,1 (mol).

2 5 2 2 3

2 5 2 2 3

C(A) C(C H OH) C(CO ) C(Na CO )

C H OH CO Na CO

n n n n2n n n 2.0,1 0,25 0,05 0,5(mol)

= + += + + = + + =

S nguyn t C trong A l0,5 5

0,1

= (nguyn t).

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2 5 2

2 5 2

H(A) H(C H OH) H(H O) H(NaOH)

H(A) C H OH H O NaOH

n n n nn 6n 2n n 6.0,1 2.0,25 0,1 1(mol)

= + = + = + =

S nguyn t H trong A l1

100,1 = (nguyn t)

CTPT A. C5H10O2 CTCT A C2H5COOC2H5

Bi 20 . t chy hon ton m(g) hn hp X gm C2H2, C3H6 v C4H10 thu c2,688 lt kh CO2 (ktc) v 2,16 g H2O. Vy m c gi tr lA. 1,48g B. 2,86 g

C. 14,8g D. 1,68g Hng dn

X C H2,688 2,16m m m .12 .2 1,68(g)22,4 18

= + = + =

Bi 21. Cho 13,8g hn hp gm ancol etylic v glixerol tc dng va vi Nthu c 4,48 lt H2 (ktc) v dung dch mui. C cn dung dch mui, khi

lng cht rn thu c lA. 22,6 g B. 22,4 gC. 34,2 g D. 25,0 g Hng dn

p n A (mmui = 22,6 g)

Bi 23. t chy hon ton hp cht A c CTPTn 2n 1C H COONa vi oxi thu

c 21,2g Na2CO3, 10,8g H2O v mt lng CO2. Lng CO2 ny cho tcdng vi dung dch Ca(OH)2 d thu c 100 g kt ta. Cng thc phn tA l

2 5 3

2 3 3 7

A. C H COONa B. CH COONaC. C H COONa D. C H COONa

Hng dn

n 2n 1 2 2 3 2 2C H COONa + O Na CO + CO + H O

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3 2A Na CO H H O210,8n 2n 2.0,2 0,4(mol) ; n 2.n 2. 1,2(mol)18= = = = = =

2 3

3 2

CO CaCO

C(A) C(CaCO ) C(CO )

n n 1 (mol)

n n n 0,2 1 1,2(mol)

= =

= + = + =

2 2 3 2CO Ca(OH) CaCO H O+ +

O(A) A Na(A) An 2n 0,8(mol) ; n n 0,4(mol)= = = =

= + + + = + + + =

= = + = =

A C H O Na

A 2 3

m m m m m 1,2.12 1,2 0,8.16 0,4.23 37,6 (g)37,6M 94 14n 66 94 n 2 CTPT A: C H COONa0,4

Bi 24.un 13,8 g hn hp 3 ancol no, n chc vi H2SO4 c 1400C thuc 11,1g hn hp cc ete c s mol bng nhau. Tnh s mol mi ete.A. 0,025 mol B. 0,1 molC. 0,15 mol D. 0,2 mol

Hng dn

un hn hp 3 ancol c3.(3 1) 62+ = ete.

Theo nh lut bo ton khi lng: mancol= mete + 2H Om

2H Om = mancol mete = 13,8 11,1 = 2,7 (g)

Tng s mol cc ete = s mol H2O =2,7

18= 0,15 (mol)

S mol mi ete = =0,15 0,0256 (mol)

Bi 25. t chy hon ton mt cht hu c A cha 1 nguyn t Oxi thu hn hp sn phm B. Cho B i qua dung dch Ca(OH)2 d thy c 15 g ktta v khi lng dung dch gim 4,8 g. CTPT ca A l

A : CH4O B : C2H6O

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C : C3H8O D : C4H10O

Hng dnTheo L bo ton khi lng th :

mdung dch u+ 2 2CO H O ddsaum m m m+ = +

Nu khi lng dung dch tng th :

mdung dch tng= mdung dch sau mdung dch u= 2 2CO H O(m m ) m+

Nu khi lng dung dch gim th

mdung dch gim= mdung dch u mdung dch sau=2 2

CO H Om (m m ) +


4,8 = 0,15.100 0,15.44 2 2H O H Om m 3,6(gam) =

CnH2n+2O nCO2 + (n+1)H2O0,15 0,2

0,2n = 0,15(n+1) n =3CTPT C3H8OBi 26.Hn hp X gm 0,1 mol2 4 2C H O v 0,2 mol hirocacbon A. t chy ht

X cn 21,28 lt O2 (ktc) v ch thu c 35,2 g CO2 v 19,8 g H2O. Cngthc phn t khi ca A l

A. 7 8C H B. 8 8C H

C. 6 6C H D. 8 6C H

Hng dn

2 2 2X O CO H O+ +

2 4 2 2 2 2A C H O O CO H Om m m m m+ + +

A21,28m (35,2 19,8) ( .32 0,1.62) 18,4g22,4 = + + =

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A18,4M 920,2= = g/mol 12x y 92 + =

Vy gi tr ph hp x = 7 ; y = 8 7 8CTPT C H

Bi 28. Nhit phn 8,8 g C3H8 thu c hn hp kh A theo phng trnh phnng:

C3H8 CH4 + C2H4C3H8 C3H6 + H2

t hon ton A khi lng CO2 v H2O to thnh l

A. 24,6 g; 14,4 g B. 26,4 g; 16,4 gC. 23,5 g ; 15,5 g D. 32,5 g ; 14,8 g

Hng dnTheo nh lut bo ton khi lng th :

3 8C H Am m= t chy cht A cng chnh l t chy3 8C H hoc tchy C v H :

C + O2 CO20,6 0,6 0,6

H2 +12 O2 H2O

0,8 0,4 0,8

== == =

22

2

O

CO

H O

V 22,4ltm 0,6.44 26,4(g)m 0,8.18 14,4(g)

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2. Phng php tng gim khi lngC s

Khi mt nguyn t hay nhm nguyn t X trong cht tham gia phn n(gi l cht u) c thay th bng mt nguyn t hay nhm nguyn t Yto ra cht mi (cht cui), th s chnh lch khi lng gia cht u v cui chnh bng hiu khi lng ca hai nhm nguyn t X v Y (|XY|).

Th d :CaCO3 CaSO4

Ta thy th s chnh lch khi lng gia hai mui CaCO3 v CaSO4 :

M (40 96) (40 60) 36g/ mol = + + = ng bng s chnh lch khilng ca hai anion 23CO (60g) v 24SO (96 g): M 96 60 36g/ mol = =.Cch p dngKhi mt cht thay anion c bng anion mi sinh ra cht mi th s chlch khi lng gia cht c v cht mi chnh l s chnh lch khi l

ca anion c v anion mi.Khi mt cht thay cation c bng cation mi sinh ra cht mi th schnh lch khi lng gia cht c v cht mi chnh l s chnh lch klng ca cation c v cation mi.

Bi tp minh ho

Bi 1.Cho 41,2 g hn hp X gm Na2CO3, K 2CO3 v mui cacbonat ca kim loiho tr 2 tc dng vi dung dch H2SO4 d. Kt thc phn ng thu c hnhp Y gm ba mui sunfat v 8,96 lt kh CO2 (ktc). Khi lng ca Y l

A. 58,6 g B. 55,6 gC. 45,0 g D. 48,5 gLi gii

S phn ng :

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2 3

2 3

3

Na COK COMCO

+ H2SO4 2 4

2 4

4

Na SOK SOMSO

+ H2O + CO2

1 mol X chuyn thnh 1 mol Y th tng khi lng lM 96 60 36(g/ mol) = =

Theo nh lut bo ton nguyn t C :2 23 COCOn n 0,4(mol) = = khi

lng Y ln hn khi lng ca X l 0,4.36 = 14,4 (g)

Vy mY = 41,2 + 14,4 =55,6 (g)

Bi 2. Cho 84,6 g hn hp A gm BaCl2 v CaCl2 vo 1 lt hn hp Na2CO30,3M v (NH4)2CO3 0,8 M. Sau khi cc phn ng kt thc ta thu c79,1 g kt ta A v dung dch B. Phn trm khi lng BaCl2 v CaCl2trong A ln lt lA. 70,15% ; 29,25% B. 60,25% ; 39,75%C. 73,75%; 26,25% D. 75,50% ; 24,50%

Li giit 2 2BaCl CaCln x(mol); n y(mol)= =

2

2

BaClCaCl +

2 3

4 2 3

Na CO(NH ) CO

3

3

BaCOCaCO + 4

NaClNH Cl

C 2 mol Cl mt i (71 g) c 1 mol mui 23CO thm vo (60 g)

chnh lch (gim) khi lng ca 1 mol mui lM = 71 60 =11(g) gim khi lng mui :m = 84,6 79,1 = 5,5 (g)Vy s mol mui clorua bng s mol mui cacbonat phn ng

=5,5 0,5(mol)11

M s mol CO32 (theo gi thit) = 0,3 + 0,8 = 1,1 (mol) > 0,5 mol (phnng). Vy mui cacbonat phn ng d.

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x +y = 0,5 (1)208x + 111y = 84,6 (2)

x 0,3moly 0,2mol

==

= == =

2

2

BaCl

CaCl

0,3.208%m .100% 73,75%84.6%m 100 73,75 26,25%

Bi 3. Nhng mt thanh nhm nng 50g vo 200ml dung dch CuSO4 0,5M. Saumt thi gian ly thanh nhm ra cn nng 51,38g. Gi s kim loi thot u bm c vo thanh nhm. Khi lng Cu thot ra lA. 0,64g B. 1,28g

C. 1,92g D. 2,56gLi gii

2Al + 3Cu2+ 2Al3+ + 3Cu2 mol (tan ra) 3 mol Cu (bm vo)

Th khi lng thanh kim loi tng l 3.64 2.27 = 138 (g)

ng vi khi lng tng 51,38 50 = 1,38gs mol Cu =3.1,38138 = 0,03

(mol)

Theo gi thit s mol Cu2+ = 0,1 mol > 0,03 molmCu= 0,03.64 = 1,92 (g)

Bi 4. Ly mt inh st nng 20g nhng vo dung dch CuSO4 bo ha. Sau mtthi gian ly inh st ra sy kh, cn nng 20,4g. Khi lng Cu bm trinh st l

A. 0,4884 g B. 3,4188 gC. 3,9072 g D. 0,9768 gLi gii

Fe + CuSO4 FeSO4 + Cu tng khi lng khi chuyn 1 mol Fe thnh 1 mol Cu l 64 56 = 8g

tng khi lng thc lm = 20,4 20 = 0,4 g Cu0,4

n 0,05mol8 = =

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Khi lng Cu = 0,05.64 = 3,2 g

Bi 5.Ho tan 10g hn hp 2 mui ACO3 v B2(CO3)3 bng dung dch HCl ta thuc dung dch A v 0,672 lt kh bay ra ktc. C cn dung dch A th thu

c m(g) mui khan. Vy m c gi tr lA. 1,033g B. 10,33gC. 9,265g D. 92,65gLi gii

3

2 3

ACOB CO + HCl

2

3

AClBCl + CO2 + H2O

C 1 mol mui 23CO i ra (mt i 60g) c 2 mol Cl kt hp (thm 71g)

chnh lch (tng) khi lng ca 1 mol mui lM = 71 60 =11 (g)

m: 2 23 COCO0,672n n 0,03(mol)22,4 = = =

Vy khi lng mui tng :m = 11.0,03 = 0,33 (g)Tng khi lng mui clorua = 10 + 0,33 = 10,33 (g)

Bi 6. Nung m g hn hp X gm hai mui cacbonat ca hai kim loi nhm IIASau mt thi gian thu c 2,24 lt kh v cht rn Y. Ha tan Y vo dungdch HCl d thu c thm 4,48 lt kh v dung dch Z. C cn dung dch Zthu c 33 g mui khan (cc th tch kh o ktc). Gi tr ca m lA. 35,3 g B. 29,7 g

C. 23,6 g D. 37,9 gLi gii

Gi cng thc chung ca X l 3MCOot

3 2MCO MO CO +

Ho tan Y ( 3MCO v MO) vo dung dch HCl

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o

o

2

t3 2 2 2

t2 2

CO

MCO 2HCl MCl H O CO

MO 2HCl MCl H O

n 0,2 0,1 0,3(mol)

+ + +

+ +

= + =Khi 1 mol mui 23CO chuyn thnh muiCl th M = 2.35,5 60 = 11g

Vi 0,3 mol mui 23CO th khi lng mui clorua nng hn khi lng

mui 23CO l m 11.0,3 0,33(gam) = =

Khi lng 3MCOM 33 0,33 29,7(gam)= =

Bi 7 . Hn hp A gm 10 g MgCO3, CaCO3 v BaCO3 c ho tan bng HCl dthu c dung dch B v kh C. C cn dung dch B c 14,4 g mui khanSc kh C vo dung dch c cha 0,3 mol Ca(OH)2 thu c s g kt ta lA. 10g B. 20gC. 30g D. 40gLi gii

CO32 + 2H+ CO2 + H2O

p dng phng php tng gim khi lng khi chuyn23CO thnhCl ta

tnh c s mol A = = = =2 23 COCO14,4 10n n 0,4(mol)

11


0,4 0,3 0,3CO2 + H2O + CaCO3 Ca(HCO3)2 0,1 0,1

3CaCOm 0,2.100 20g= =

Bi 8. Cho 68g hn hp 2 mui CuSO4 v MgSO4 tc dng vi 1lt dung dchcha KOH 1M v NaOH 0,4M. Sau phn ng thu c 37g kt ta v dun

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dch B. Vy phn trm khi lng CuSO4 v MgSO4 trong hn hp ban uln lt lA, 47,50% ; 52,95%. B. 47,05 % ; 52,95%.

C. 47,05% ; 53,59%. D. 47,50% ; 53,59%. Li gii

t 4 4CuSO MgSOn x(mol);n y(mol)= =

4

4

CuSOMgSO +

NaOHKOH

2

2

Cu(OH)Mg(OH) +

2 4

2 4

Na SOK SO

T chnh lch khi lng ca mui sunfat v khi lng kt ta trntnh c tng s mol hai mui sunfat l

68 37 = 0,596 34 (mol)

x + y = 0,5 (1)160x + 120y = 68(2)

x 0,2y 0,3

= =

= =

= =4

4

CuSO

MgSO

0,2.160%m .100% 47,05%68

%m 100 47,05 52,95%

Bi 9.Nhng mt thanh kim loi A (ho tr II) vo dung dch CuSO4. Sau phnng khi lng thanh kim loi A gim 0,12g. Mt khc cng thanh kim loA c nhng vo dung dch AgNO3 d th kt thc phn ng khilng thanh tng 0,26g. Bit s mol A tham gia hai phn ng bng nhaKim loi A l

A. Zn B. MgC. Cd D. Fe Li gii Phng trnh phn ng :

A + Cu2+ d A2+ + Cua a

A + 2Ag+ d A2+ + 2Ag

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a 2a

Khi lng thanh kim loi tng = mA mCu = 0,12g

a.MA 64a = 0,12 MA.a = 64a + 0,12 (1)

Mt khc khi lng thanh kim loi gim = mAg + mA = 0,26 g

2a.108 MA.a = 0,26 MA.a = 2a.108 0,26 (2)

x = 2,5.10 3mol MA =3

364.2,5.10 0,12 112(g/mol)

2,5.10

+ =

Cht X l Cd.

Bi 10.C 2 dung dch FeCl2 v CuSO4 c cng nng mol.

Nhng thanh kim loi vo M (nhm IIA) vo V lt dung dch FeCl2, ktthc phn ng khi lng thanh kim loi tng 16g.

Nhng cng thanh kim loi y vo V lt dung dch CuSO4 kt thc phnng khi lng thanh kim tng 20g. Gi thit cc phn ng xy ra hon tov kim loi thot ra bm ht vo M. Kim loi M l

A. Zn B. MgC. Cd D. Fe Li gii Cc phng trnh phn ng xy ra :

M + Fe2+ M2+ + Fex x x

M + Cu2+ M2+ + Cux x x

Theo gi thit cc phn ng u xy ra hon ton nn cc ion2 2Fe vCu+ +

phn ng ht 2 2Fe Cun n x(mol)+ += =

Khi lng thanh kim loi tng (1) l : m = mFe mM = 16g

56x MM.x = 16M.x = 56x 16

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Khi lng thanh kim loi tng (2) l : m = mCu mM = 20 g

64x M.x = 20M.x = 64x 20M = 24. Vy kim loi M l Mg.

Bi tp vn dngBi 1.Cho 20 g hn hp X gm hai axit cacboxylic no, n chc tc dng va

vi dung dch Na2CO3 thu c V lt kh CO2 (ktc) v dung dch mui. Ccn dung dch thu c 28,8 g mui. Gi tr ca V l

A. 3,36 lt B. 4,48 ltC. 2,24 lt D. 6,72 lt Hng dn

t cng thc chung ca hai axit cacboxylic ln 2n 1C H COOH+

2 3 2 22RCOOH Na CO 2RCOONa H O CO+ + +

C 1 mol hn hp axit hn hp mui th (tng)M

= 22 g/molx mol (tng)m = 28,8 20 = 8,8 g

x = 8,822 = 0,4 (mol) 2COn 0,2 (mol) V 4,48lt = =

Bi 2. Cho 4,16 g hn hp 2 axit cacboxylic no, n chc k tip nhau trong dng ng tc dng vi mt lng d kim loi Ca thu c 5,3 g hn h

mui v gii phng kh H2. CTPT ca 2 axit trn lA. CH3COOH ; C2H5COOH. B. C3H7COOH ; C2H5COOH.

C. HCOOH ; CH3COOH. D. C3H7COOH ; C4H9COOH. Hng dn

t cng thc chung ca hai axit cacboxylic lRCOOH

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+ += = = =

+ + = =

2 2

muimui

2RCOOH Ca (RCOO) Ca H5,3 4,16 5,3n 0,03(mol) M 176,7(g/mol)

40 2 0,03

2(14n 44) 40 176,6 n 1,73CTPT ca 2 axit l 3 2 5CH COOH ; C H COOH.

Bi 3. Cho 5,5 g hn hp 2 ancol n chc l ng ng k tip nhau tc dva vi Na kim loi to ra 8,8 g cht rn v V lt kh H2(ktc). Cng thcca 2 ancol l

A. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.

C. C3H7OH v C4H9OH. D. C2H3OH v C3H5OH. Hng dn

t cng thc chung ca hai ancol n chc lROH

+ + 21ROH Na RONa H2

= =hh

8,8 5,5n 0,15(mol)23 1

hh5,5M 36,67 14n 18 36,67 n 1,330,15= = + = =

CTPT ca hai ancol l CH3OH v C2H5OHBi 4. Khi thy phn hon ton 5,9 g este hai chc to t axit n chc v anco

hai chc th tiu tn ht 5,6 g KOH v thu c 8,4 g mui. Cng thc ceste lA. (HCOO)2C2H4 B. (CH3COO)2C2H4

C. (CH3COO)2CH2CH2CH3 D. CH2(COOC2H5)2 Hng dn

KOHn 0,1(mol)=

t cng thc chung ca hai este l 2(RCOO) R'

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2 2(RCOO) R' 2KOH 2RCOOK R'(OH)+ +

C 1 mol este mui th (tng)M = 78 R g/mol

x mol (tng)m = 8,4 5,9 = 2,5 g

R' 2 4R'

2,5 1x 0,1 M 28 R':C H78 M 2= = =

este R R5,9M 118(g/mol) 2(M 44) 28 118 M 10,05= = + + = =

Vy cng thc ca este l (HCOO)2C2H4.

Bi 5. Thy phn 0,01mol este ca 1 ancol a chc vi 1 axit n chc tiu tht 1,2g NaOH. Mt khc khi thy phn 4,36g este th tiu tn ht 2,4 NaOH v thu c 4,92g mui. Cng thc ca este l

A. (CH3COO)3C3H5 B. (C2H3COO)3C3H5

C. C3H5(COOCH3)3 D. C3H5(COOC2H3)3 Hng dn

NaOH1,2n 0,03(mol)40= =

V n NaOH= 3nesteeste 3 chc (c to t ancol 3 chc + axit n chc)

t cng thc este (RCOO)3R'.

(RCOO)3R' + 3NaOH (RCOONa)3 + R'(OH)3

1 mol 3 mol 1 molKhi lng tng : 23.3 R' = 69 R' (g)

0,02 mol 0,06 mol 0,06 mol

Khi lng tng : 4,92 4,36 = 0,56 (g)este0,56n 0,02

0,9 R' = =

0,56 = 0,02 (69R') R = 41R' l C3H5.

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Meste= =4,36 2180,02 (g/mol)

mR

= =218 41 44.3 153

R: CH3

Vy cng thc ca este l (CH3COO)3C3H5Bi 6 . Thc hin phn ng este ho gia axit axetic (d) v hn hp gm 7,52 g

ancol k tip nhau trong cng mt dy ng ng. Sau phn ng thu 15,92 g 3 este. Gi s hiu sut phn ng l 100%.a) CTPT ca ba ancol l

A. 3 2 5 3 5CH OH;C H OH;C H OH B. 2 5 3 5 4 7C H OH;C H OH;C H OHC. 3 5 4 7 5 9C H OH;C H OH;C H OH D. 3 7 4 9 5 11C H OH;C H OH;C H OH

b) Ly sn phm ca phn ng este ho trn thc hin phn ng x phho va vi NaOH th thu c s g mui thu c lA. 14,5 g B. 16,4 gC. 16,5 g D. 17,8 g

Bi 7. Cho 16,15 g hn hp NaX v NaY (X, Y l hai halogen hai chu k ktip) vo dung dch AgNO3 d thu c 33,15 g kt ta. Hn hp hai mui ban u lA. NaF v NaCl B. NaCl v NaBr C. NaBr v NaI D. Khng xc nh

Bi 8. Ha tan hon ton 20,85 g hn hp X gm NaCl v NaI vo nc

dung dch A. Sc kh Cl2 d vo dung dch A. Kt thc th nghim, c cndung dch thu c 11,7 g mui khan. Khi lng NaCl c trong X lA. 5,85 g B. 7,55 gC. 2,95 g D. 5,10 g

Bi 9. Cho kh CO qua ng s cha 15,2g hn hp cht rn CuO v FeO nunnng. Sau mt thi gian thu c hn hp kh B v 13,6g cht rn C. Ch

hn hp kh B hp th hon ton vo dung dch Ca(OH)2 d thy c kt ta.Lc ly kt ta v sy kh ri cn th khi lng kt ta thu c l

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A. 12g B. 11gC. 10g D. 9 g

Bi 10. Nhng thanh kim loi M (ha tr II) vo dung dch CuSO4, sau mt thigian ly thanh kim loi ra thy khi lng gim 0,1%. Mt khc cng nhthanh kim loi trn vo dung dch AgNO3. Sau mt thi gian thy khilng tng 7,55%. Bit s mol CuSO4 v AgNO3 tham gia phn ng haitrng hp nh nhau. Kim loi M lA. Zn B. MgC. Ni D. Ca

Bi 11. Ha tan 3,23 g hn hp mui CuCl2 v Cu(NO3)2 vo nc thu cdung dch X. Nhng thanh Mg vo dung dch X cho n khi mt mu xanhca dung dch, ly thanh Mg ra cn li, thy tng thm 0,8 g. C cn dundch sau phn ng thu c m g mui khan. Gi tr m lA. 3,08 B. 4,03 C. 2,48 D. 2,84

Bi 12.Cho 16,2 g hn hp este ca ancol metylic v hai axit cacboxlic no, nchc tc dng va vi dung dch NaOH 1M thu c dung dch A. Ccn dung dch A thu c 17,8 g hn hp hai mui khan, th tch dung dc NaOH 1 M dng lA. 0,2 lt B. 0,3 ltC. 0,4 lt D. 0,5 lt

Bi 13. un nng 3,188 g este ca glixerol vi ba axit cacboxylic no, n chc

mch h X, Y, Z (X, Y l ng phn ca nhau v k tip vi Z) vi dundch NaOH d, phn ng kt thc thu c 3,468 g hn hp mui. Cnthc phn t ca cc axit l

A. C2H4O2 v C3H6O2 B. C3H6O2 v C4H8O2

C. C4H8O2 v C5H10O2 D. C3H4O2 v C4H8O2

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3. Phng php s dng nh lut bo ton in tchC s

Trong phn ng oxi ha kh th tng s mol e m cc cht kh cho bng tng s mol e m cc cht oxi ho thu vo :

= echo enhnn n Nu bi ton c nhiu cht oxi ho v nhiu cht kh tham gia trong s phn ng, hoc qu trnh phn ng phi i qua nhiu giai on th p d phng php ny gii s rt nhanh v kt qu thu c chnh xc.

Cc bc p dng phng php bo ton electron nh sau : Phi xc nh c t cc cht ban u tham gia phn ng n cc csn phm c bao nhiu cht cho electron v s mol tng cht, c bao nhicht nhn electron v s mol tng cht (c th phi t n s). Vit cc qu trnh cho electron tnh tng s mol e m cc cht kh chi ( e chon ). Vit cc qu trnh nhn electron tnh tng s mol e m cc oxi ho nhvo ( enhnn ). p dng nh lut bo ton electron :

= echo enhnn ni vi nhng h trung ho in Nu trong h tn ti ng thi cc ht mang in th ta lun c tng s min tch dngn t(+) bng tng s mol in tch mn t() : n t(+) = n t()Vi nt = s mol ion s n v in tch ca ion .

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Bi tp minh ho

Bi 1.Ho tan ht 7,5 g hn hp Al v Mg trong HNO3 long thu c dung dchA gm 2 mui v 3,36 lt ( ktc) hn hp 2 kh NO v N2O, khi lng

ca hn hp kh l 5,2 g. Khi lng ca Al, Mg trong hn hp ln lt lA. 3,5g v 4,0g. B. 2,1g v 5,4g.C. 2,7g v 4,8g. D. 4g v 3,5g.Li gii

t = =2NO N On a(mol); n b(mol)

3,36

a b 0,15 a 0,1mol22,4b 0,2mol30a 44b 5,2

+ = = = =+ =S phn ng :

AlMg + HNO3

3 3

3 2

Al(NO )Mg(NO ) + 2

NON O + H2O

Cc cht cho electron : Al : x (mol) ; Mg : y (mol)

Al0 Al3+ + 3e

x 3x = + echon 3x 2y (mol)Mg0 Mg2+ + 2ey 2y

Cht nhn electron l HNO3 c hai qu trnh nhn e : N+5 + 3e N+2 (NO)

0,3 0,1 0,1 enhnn 0,7 (mol)= N+5 + 4e N+ (N2O)

0,4 0,10,05

p dng s bo ton electron ta c : 3x + 2y = 0,7 (1)

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Phng trnh khi lng : 27x + 24y = 7,5 (2)

Gii h (1, 2) Al

Mg

m 2,7gx 0,1(mol)m 4,8gy 0,2(mol)

== ==

Bi 2. Ha tan hon ton hn hp gm 0,06 mol FeS2 v x mol Cu2S vo axitHNO3 (va ) thu c dung dch A (ch cha hai mui sunfat) v kh duynht NO. Gi tr ca x lA. 0,04 B. 0,06C. 0,12 D. 0,03Li gii

3 22 4

2 22 4

FeS Fe 2SO0,06 0,06 0,12Cu S 2Cu SOa 2a a

+

+

+

+

Theo nh lut bo ton in tch n t(+) = n t()3.0,06 2.2a 2.0,12 2.a a 0,03(mol) + = + =

Bi 3. Ho tan 3,81 g hn hp Zn v Fe vo dung dch HNO3 thu c dung dchA v 1,12 lt hn hp kh D ( ktc) gm NO v NO2 (gi thuyt NO2 tn ti ktc). T khi ca hn hp D so vi hiro l 16,75. Khi lng kim loFe v Zn phn ng ln lt lA. 0,6g v 3,21g. B. 0,56g v 3,25g.

C. 1,15g v 2,76g. D. 1,68g v 2,13g.Li gii

= =2NO NOn a(mol); n b(mol)

+ = = = =+ =

1,12a b 0,05 a 0,04(mol)22,4

b 0,01(mol)30a 46b 1,675

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FeZn + HNO3

3 3

3 2

Fe(NO )Zn(NO ) + 2

NONO + H2O

Cc cht cho electron Al : x (mol) ; Zn : y (mol)

Fe Fe3+ + 3e

x 3x = + echon 3x 2y Zn Zn2+ + 2ey 2y

Cht nhn electron : HNO3

N+5 + 3e N+2 (NO)

0,12 0,040,04 = + = enhnn 0,12 0,01 0,13N+5 + e N+4 (NO2)

0,01 0,01 0,01p dng s bo ton electron ta c : 3x + 2y = 0,13 (1)

Phng trnh khi lng : 56x + 65y = 6,5 2,69 = 3,81 (2)

Gii h (1, 2)==

Al

Mg

m 0,56gm 3,25g

Bi 4. Hn hp X gm Fe v kim loi M ha tr khng i. Ha tan ht 2,51 g Xtrong dung dch HCl thy c 0,896 lt H2 (ktc) bay ra. Nu ha tan cnglng hn hp X trn vo dung dch HNO3 thu c 0,672 lt NO duy nht(ktc). Kim loi M lA. Zn B. AlC. Cu D. Cr Li gii

Gi Fe Mn x(mol) ;n y(mol)= =Ta c : 56x + My = 2,51 (I)Cc cht cho e l Fe ; M

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2

n

Fe Fe 2ex 2xM M ney ny

+

+

+

+e(cho)n 2x ny= +

Cht nhn e :

22H 2e H0,08 0,04

+ + e(nhn)n 0,08=

Ta c 2x + ny = 0,08 (II)

Tng t ta c phng trnh : 3x + ny = 0,09 (III)Gii v bin lun h (I), (II) v (III) ta c MM = 65 M l Zn

Bi 5 . Ho tan hon ton 24,3 g Al vo dung dch HNO3 long d thu c hnhp kh NO v N2O c t khi hi so vi H2 l 20,25 v dung dch B chcha mt mui. Th tch kh thot ra ktc lA. 8,96 lt B. 4,48 lt

C. 11,2 lt D. 2,24 ltLi giiS phn ng

2

3 3 3 2

N O

Al + HNO Al(NO ) + + H ONO

Cht cho electron l Al c s mol =24,3 0,9(mol)27

3+Al Al + 3e0,9 2,7

= echon 2,7mol .Cht nhn electron l HNO3

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+5 +2

+5 +2

N + 3e N NO3x x x

2N + 8e 2N N O8y 2y y

Ta c+ =+ = +

3x 8y 2,730x 44y 40,5.(x y)

x 0,1 V 0,4.22,4 8,96y 0,3

= = == (lt)

Bi 6. Khi cho m g Fe tc dng va vi V lt dung dch HNO3 thu c dungdch mui, 0,1 mol kh NO v 0,3 mol kh NO2. m c gi tr l

A. 12,5g B. 11,5gC. 11,2g D. 15,2gLi giiCht cho e Cht nhn e

3 5 2

5 42

Fe Fe 3e N 3e N NOm m3 0,3 0,1 0,156 56

N e N NO0,3 0,3 0,3

+ + +

+ +

+ +

+

p dng nh lut bo ton e3m56 = 0,6m 11,2g. =

Bi 7. Ho tan 8,45g Zn vo 3 lt dung dch HNO3, thu c dung dch A v

2,688 lt hn hp NO v N2O4 (ktc). Khi lng ca 1 lt hn hp kh ktclA. 2,689g B. 7,252gC. 7,068g D. 3,646gLi gii

+ + +3 3 2 22 4

NOZn HNO Zn(NO ) H O

N O

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t 2 4NO N On x mol ; n y mol= = .

Cht cho electron : Zn (0,13 mol) Cht nhn electron : HNO3

2 5 2

5 42 4

Zn Zn 2e N 3e N (NO)0,13 0,26 3x x x

N e N (N O )2y 2y y

+ + +

+ +

+ +

+

e(cho) enhnn 0,26 3x 2y n (1)= = + = Tng s mol khx y 0,12+ = (2)t (1) v (2) x = 0,02 mol ; y = 0,2 mol.

mhn hp= 0,02.30 + 0,2.92 = 19 (g) ng vi 2,688 lt

Vy 1 lt hn hp c khi lng l19.12,688

= 7,068g.

Bi 8. Cho 15,5g hn hp 2 kim loi Al, Cu tc dng vi HNO3 va thu c

6,72 lt hn hp hai kh NO, NO2 ( =2NO NOn : n 2:1 ) v hn hp hai mui.Thnh phn % khi lng Cu v Al trong hn hp ln lt lA. 82,25% ; 17,75%. B. 82,58% ; 17,42%.C. 63,84% ; 36,16%. D. 36,16% ; 63,84%.Li gii

Gi Al Cun x(mol) ; n y(mol)= = . Ta c 27x + 64y = 15,5 (I)

Theo gi thit ta tm c 2NO NOn 0,2(mol) ; n 0,1(mol)= =

Cht cho e Cht nhn e3 5 2

2 5 42

Al Al 3e N 3e N (NO)x 3x 0,6 0,2Cu Cu 2e N e N (NO )

y 2y 0,1 0,1

+ + +

+ + +

+ +

+ +

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p dng nh lut bo ton e : 3x + 2y = 0,7 (II)Gii h (I) v (II) thu c x = 0,1 (mol) ; y = 0,2 (mol)

%mAl = 17,42% ; %mCu = 82,58%

Bi 9.Ho tan 11,2 g Fe bng dung dch H2SO4 (long, d) thu c dung dch Xv 5,6 lt kh bay ra ktc. Dung dch X phn ng va vi V ml dundch KMnO4 0,5M. V c gi tr lA. 500ml B. 100mlC. 150ml D. 200mlLi giiCht cho e Cht nhn e

32

7 2

Fe Fe 3e 2H 2e H0,2 0,6 0,5 0,25

Mn 5e Mn0,1 0,6 0,5

+ +

+ +

+ +

+

4KMnO 0,1V 0,2(lit)0,5 = =

Bi 10. m g Fe ngoi khng kh mt thi gian nn b g (gi s g st ch tol oxit st) cn nng 10 g. Lng g st trn lm mt mu hon ton 200 mdung dch KMnO4 0,5M trong dung dch H2SO4 d. m c gi tr l

A. 17,2g B. 9,8gC. 9,0g D. 15,0gLi giiS phn ng :

2 a 4 2 4 2 4 3 4 2Fe +O FeO KMnO H SO Fe (SO ) MnSO H + + + +

Cht cho electron : Fe :m (mol)56

Fe Fe3+ + 3e

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m56 3

m56 = echo

mn 3.56

Cht nhn electron O :10 m

16

(mol) v KMnO4 : 0,1 (mol)

O + 2e O2

10 m16 210 m16

Mn+7 + 5e Mn+2 0,1 0,5

e nhn10 mn 2. 0,5

16 = +

p dng s bo ton electron :10 m m2. 0,5 3. m 9,8(g)

16 56 + = =

Bi 11. Thi lung khng kh i qua m(g) bt st nung nng sau mt thi gia

bin thnh hn hp A c khi lng 30g gm Fe, FeO, Fe2O3, Fe3O4. Cho A phn ng hon ton vi dung dch HNO3 thy gii phng ra 5,6 lt kh NOduy nht (ktc). Khi lng ca m lA. 27,5g B. 22,5gC. 26,2g D. 25,2g Li gii

S :

+ + +2 2 3 3 3 3 23 4

Fe, FeOFe + O Fe O HNO Fe(NO ) NO H O

Fe O

Cht cho electron Fe, s mol :m .56

Fe Fe3+

+ 3e

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m56 3

m56 = echo

mn 3.56

Cht nhn electron O2, s mol :30 m

32v HNO3.

O + 2e O2

30 m16 230 m16

N+5 + 3e N+2 (NO)0,75 0,25 0,25

= + e nhn 30 mn 2. 0,7516p dng s bo ton electron :

+ = =30 m m2. 0,75 3. m 25,2(g)16 56Bi 12 . Nung nng 5,6 g bt st trong bnh ng O2 thu c 7,36 g hn hp X

gm Fe, Fe2O3 v Fe3O4. Cho X tan hon ton trong dung dch HNO3 thuc V lt (ktc) hn hp kh Y gm NO v N2O4, t khi hi ca Y so viH2 l 25,33. V c gi tr lA. 22,4 lt B. 0,672 ltC. 0,372 lt D. 1,12 ltLi gii

Gi 2 4NO N On x(mol); n y(mol)= =

2 Y/H30x 92yd 25,33(x y).2

+= =+ (1)

+ + + 2 2 3 3 3 3 22 4

3 4

FeNO

Fe + O Fe O HNO Fe(NO ) H ON O

Fe O

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Cht cho electron Fe : 0,1 mol+ +3Fe Fe 3e

0,1 0,3 = echon 0,3mol e

Cht nhn electron O : =7,36 5,6 0,11(mol)16 v HNO3.

2

5 2

5 4 2 4

O 2e O0,11 0,22N 3e N (NO)

3x x x

N e N (N O )2y 2y y

+ +

+ +

+

+

+

= + + enhnn 3x 2y 0,22 (mole)

0,3 3x 2y 0,22 = + + (2)T (1), (2) x = 0,02 mol ; y = 0,01 mol.Vy V = (0,02 + 0,01)22,4 = 0,672 lt.

Bi 13 . Cho 6,64 g hn hp A gm Fe, FeO, Fe2O

3, Fe

3O

4vo dung dch HNO

3long, d thu c V lt hn hp kh B ( 30oC, 1 atm) gm NO, NO2 (vi

=2NO NOn :n 2 ). Mt khc khi cho lung kh H2 d i qua hn hp A nungnng, sau khi phn ng hon ton thu c 5,04 g Fe. Th tch hn hp kB lA. 0,464 lt B. 0,672 ltC. 0,242 lt D. 0,738 lt Li gii

Fe

FeO + H2 Fe + H2O

Fe2O3

Fe3O4

O(A) Om 6,64 5,04 1,6(gam) n 0,1(mol)= = =

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FeFeO NO

Fe + O2 Fe2O3 + HNO3 Fe(NO3)3 + + H2OFe3O4 NO2

Cht cho electron l Fe :5,04 0,09(mol)56 =

Fe Fe3+ + 3e0,09 0,27

Cht nhn electron l :

O + 2e O2

0,1 0,2

N+5 + 3e N+2 (NO)6x 2x

N+5 + e N+4 (NO2)x x

0,2 + 6x + x = 0,27 x 0,01 = tng s mol 2 kh = 3x = 0,03 mol.

Gii h tm c 0,03.0,082.300V 0,7381= = lt.

Bi 14. Cho lung kh H2 i qua ng s ng m g oxit Fe2O3 nhit caomt thi gian, ngi ta thu c 6,72 g hn hp A gm 4 cht rn kh

nhau. em ho tan hon ton hn hp ny vo dung dch HNO3 d thyto thnh 0,448 lt kh B ktc (duy nht) c t khi so vi hiro l 15th m c gi tr lA. 7,5 g B. 7,2 gC. 8,0 g D. 8,4 g Li gii

S :

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2 3 2 2Fe O H H O A+ + 3HNO+ Fe(NO3)3 + NO + H2O

Xt c qu trnh th : Fe+3 Fe+3 hnh nh khng c s cho v nhn e.

Cht cho electron : H2 m 6,72 (mol)16 ( do 2 (oxit) 2H O H O+ )

m 6,72 m 6,7216 16

H2 2H+ + 2e

m 6,72

16

2 m 6,72

16

= echon Cht nhn electron : HNO3, kh B l NO.

N+5 + 3e N+2 (NO )

0,06 0,02 0,02 = enhnn 0,06 = =m 6,72 0,06 m 7,2g8

Bi 15. Trn 60g bt Fe vi 30g bt lu hunh ri un nng (trong iu kikhng c khng kh) thu c cht rn X. Ho tan X bng dung dch axH2SO4 long, d c dung dch B v kh C. t chy C cn V lt O2 (ktc).Cc phn ng xy ra hon ton th V c gi tr lA. 39,2 lt B. 32,928 ltC. 32,29 lt D. 38,292 lt

Li giiS :

o2 4 2H SO + Ot 2 2

42 2

H H OFe Fe FeSO +S FeS H S SO

Xt c qu trnh phn ng th Fe v S cho electron, cn O2 nhn electron.

Cht cho electron : Fe :60

(mol)56 ; S :30

(mol)32

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Fe Fe2+ + 2e

6056 26056

S S+4 (SO2) + 4e

3032 43032

Cht nhn electron : gi s mol O2 l x mol.

O2 + 4e 2O 2 x 4x

p dng s bo ton electron : 4.3230

2.5660

4 += x

Gii ra x =330224 mol 2O330V 22,4 33224

= = (lt)

Bi 16. Hn hp X gm FeS2 v MS c s mol nh nhau (M l kim loi ha trII). Cho 6,51g X tc dng hon ton vi lng d dung dch HNO3 unnng, thu c dung dch Y (ch cha mui sunfat) v 13,216 lt (ktc) hhp kh Z gm NO v NO2 c khi lng 26,34g. Kim loi M l

A. Mg B. ZnC. Mn D. CuLi giiS phn ng :

2FeSMS + HNO3 Y + 2

NONO + H2O

T gi thit ta tnh c = =2NO NOn 0,05(mol) ; n 0,54(mol)

Cht cho e : 2FeS , MS

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+ +

+ +

+ +

+ +

3 62

2 6

FeS Fe 2S 15ex 15xMS M 2S 8ex 8x

= e(cho)n 23x(mole)

Cht nhn e : HNO3+ +

+ +

+

+

5 2

5 42

N 3e N (NO)0,15 0,05

N e N (NO )

0,54 0,54

= e(nhn)n 0,69(mole)

23x = 0,69x = 0,03 (mol)120x + (M+32)x = 6,51 M = 65 : Cu

Bi 17 . Ho tan hon ton 12,8g kim loi Cu vo dung dch HNO3 long, tt ckh NO thu c em oxi ho thnh NO2 ri sc vo nc c dng oxi chuyn ht thnh HNO3. Th tch kh oxi ktc tham gia vo qu trnh

trn lA. 2,28 lt B. 4,48 ltC. 2,24 lt D. 6,72 ltLi giiS :

HNO3 + Cu Cu(NO3)2 + H2O + NO 2O+ NO2 2O+ HNO3Xt c qu trnh th coi nh Cu cho e v O2 nhn e :Cht cho electron : Cu ; 0,2 mol.

Cu Cu2+ + 2e0,2 0,4

Cht nhn electron : O2

O2 + 4e 2O2

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x 4x

4x = 0,4x = 0,1

2OV = 0,1 . 22,4 = 2,24 (lt)

Bi 18. Chia 5,56 g hn hp gm Fe v kim loi A (c ho tr n) lm hai phn bng nhau : Phn 1 ho tan ht trong dung dch HCl c 1,568 lt hiro.

Phn 2 ho tan ht trong dung dch HNO3 long thu c 1,344 lt kh NOduy nht v khng to ra NH4 NO3.Cc kh c o ktc. Tn kim loi A lA. Mg B. AlC. Zn D. Cr Li giiTrong mi phn th Fe : x(mol) ; A : y(mol) l cc cht cho electron :

H+ v N+5 l cc cht nhn electron.

p dng nh lut bo ton e cho tng phn ta c h:56x My 2,78 x 0,042x ny 0,14 ny 0,063x ny 0,18 A 9n

+ = = + = = + = =

gi tr ph hp n = 3, A = 27 :Al

Bi 19. Cho 1,15 g hn hp gm Cu, Mg, Al tc dng ht vi dung dch HNO3thu c hn hp kh gm 0,1 mol NO vo 0,4 mol NO2. Khi lng mui

nitrat to ra trong dung dch lA. 45,69g B. 64,59gC. 44,55g D. 34,69gLi giit x, y, z ln lt l s mol Cu, Mg, Al.

Cu Cu(NO3)2 NO

Mg + HNO3 Mg(NO3)2 + + H2O

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Al Al(NO3)3 NO2

mmui= m3KL+ 3NOm

Ta c: 2x + 2y + 3z = 0,3 + 0,4 = 0,7 (1) Nhng 0,07 cng chnh l s mol3NO to mui vi ion kim loi.

Khi lng mui nitrat l : 1,15 + 62.0,7 = 44,55g.Bi 20 . Cho m g hn hp ba kim loi Al, Fe, Cu tan hon ton trong dung dch

HNO3 thu c V lt hn hp kh D (ktc) gm NO2 (gi thit tn ti NO2 ktc) v NO (khng sinh mui NH4 NO3). T khi hi ca D so vi hiro

bng 18,2. Tng s g mui khan to thnh theo m v V l

A. m +V

22,4 B. m V

22,4

C. 2m +V

22,4 D. 2m V

22,4

Li gii

NOM + HNO3 M(NO3)n + + H2O

NO2

T gi thit ta tnh c 2NO NO4V 2Vn (mol);n (mol)112 112= =

Cht nhn electron : HNO3

N+5 + 3e N+2 (NO)

12V112 4V112

N+5 + e N+4 (NO2) = enhn 14Vn 112

2V112

2V112

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Cht cho electron :

M cho i14V112 mol e = s mol in tch (+) = s mol in tch () ca NO3

to mui = 3NOn :

mmui= mKL + 3NOm = m + 62.14V112 = m +

V (gam)7,75

Bi 21. Ha tan hon ton hn hp hai kim loi gm 0,02 mol A (ha tr II) v0,03 mol B (ha tr III) cn V ml dung dch HNO3 2M. Sau phn ng thuc V1 lt hn hp hai kh NO v N2O c t khi so vi H2 l 15,35. V cgi tr lA. 0,076 B. 0,086C. 0,069 D. 0,179Li giiS phn ng

AB + 3HNO

3 2

3 3

A(NO )B(NO ) + 2

NONO + 2H O

p dng phng php bo ton electron

o 2 5 2

o 3 5

2

Qu trnh cho e Qu trnh nhn e

A A 2e N 3e N (NO)0,02 0,04 3x x x

B B 3e N 4e N (N O)0,03 0,09 8y 2y y

+ + +

+ + +

+ +

+ +

Ta c h2

NO

N O

3x 8y 0,13 x n 0,038mol30x 44y y n 0,002mol15,352(x y)

+ = = = + = ==+

p dng nh lut bo ton nguyn t N :

3 3 2 3 3 2HNO A(NO ) B(NO ) NO N On 2n 3n n n 0,172(mol)= + + + = V = 0,086 (lt)

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54/119

A. 0,4M; 0,2M B. 0,2M; 0,4MC. 0,4M; 0,6M D. 0,2M; 0,3M Hng dn

S :Al Ag+ Ag

+ Cu2+ Fe(d) + CuFe H+ H2

Cht cho electron : Al (0,02 mol) v Fe (0,06 mol).

Al Al3+

+ 3e0,02 0,06 = echon 0,14molFe Fe2+ + 2e0,04 0,08

Cht nhn electron : Ag+ (x mol) ; Cu2+ (y mol) ; H+ (0,04 mol)

Ag+

+ e Agx x x

Cu2+ + 2e Cu = + + enhnn x 2y 0,04y 2y y

2H+ + 2e H2 0,04 0,02

(Fed+ 2H+ Fe2+ + H2)

=+ + = = = + + = = =

3

3 2

M AgNO )

D M Cu(NO )

C 0,2Mx 2y 0,04 0,14 x 0,02m x.108 y.64 0,02.56 5,84 y 0,04 C 0,4M

Bi 4 . Ho tan hon ton mg hn hp 3 kim loi Al, Fe, Mg trong dung dch HCthy thot ra 13,44 lt kh. Cng lng hn hp trn tc dng vi dung dc

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CuSO4 d, lc ly ton b cht rn thu c sau phn ng tc dng vi dudch HNO3 nng d th thu c V lt kh NO ( ktc).Th tch kh NO thu c lA. 17,92 lt B. 13,76 ltC. 13,44 lt D. 44,8 lt Hng dn

Ta c m g (Al, Mg, Fe) cho electron ; Cu2+ nhn c s mol electron = smol e m H+ nhn c = s mol e m N+5 nhn c.

2H+ H2 + 2e0,6 1,2 mol e

Cu2+ Cu + 2e1,2 mol e

N+5 + 3e N+2 (NO)1,2 0,4 0,4

2 4N OV 0,4.22,4 8,96(lt)= = Bi 5. Ho tan m g hn hp A gm Fe v kim loi R (c ho tr khng i) tron

dung dch HCl d th thu c 1,008 lt kh (ktc) v dung dch cha 4,575gmui khan. Cng lng hn hp trn ha tan trong dung dch cha hn hHNO3 c v H2SO4 nhit thch hp th thu c 0,063 mol kh NO2v 0,021 mol kh SO2. Kim loi R l

A. Mg B. AlC. Ca D. Zn Hng dn

Trong phn ng vi HCl, Fe x(mol) ; R y (mol) cho e. H+ nhn e.

Fe Fe2+ + 2ex 2x

R R n+ + ne

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y ny

2H+ + 2e H20,09 0,09 0,045

Ta c : 2x + ny = 0,09 (1) Phng trnh khi lng 2 mui :

56x + My + 0,09.35,5 = 4,575 (2)

Trong phn ng vi HNO3, H2SO4 c cc qu trnh :

Fe Fe3+ + 3e

x 3xR R n+ + ney ny

N+5 + e N+4 (NO2)0,063 0,063 0,063

S+6 + 2e S+4 (SO2)0,042 0,021 0,021

Ta c : 3x + ny = 0,105 (3)

R R

R

2x +ny = 0,09 x 0,01556x +M y +0,09.35,5 = 4,575 ny 0,06 M 27 RlAl

M 9n3x +ny = 0,105

=

= = =

Bi 6.Cho 23,6g hn hp Cu, Ag tc dng ht vi V lt dung dch HNO3 1M thuc hn hp mui v 0,5 mol kh NO2 bay ra. Thnh phn % khi lngmi kim loi trong hn hp lA. 52,25% ; 47,75% B. 54,23% ; 45,77%C. 54,00% ; 46,00% C. 52,34% ; 47,66% Hng dn

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3 2

3 2 2

3

Cu Cu(NO )HNO NO H O

Ag AgNO+ + +

Cht cho e Cu : x ; Ag : y(mol) Cht nhn e : HNO32 5 4

2

3 2 2

Cu Cu 2e N e N NOx 2x 0,5 0,5(mol)Ag Ag e (hoc NO e 2H NO H O)y y

+ + +

+ +

+ +

+ + + +

e(cho) e(nhn)n 2x y 0,5 n= + = = Ta c h Cu

Ag

%m 54,23%64x 108y 23,6 x 0,2%m 45,77%2x y 0,5 y 0,1

=+ = = =+ = =

Bi 7. 10,08 g bt st trong khng kh sau mt thi gian thu c hn hpc khi lng m g gm Fe, FeO, Fe3O4, Fe2O3. Cho A tc dng hon tonvi dung dch HNO3 thy gii phng ra 2,24 lt kh NO duy nht ( iu

kin tiu chun). Khi lng m ca hn hp A lA. 11 g B. 12gC. 13g D. 14g Hng dnS phn ng

2 2 3 3 3 3 2

3 4

Fe, FeO

Fe + O Fe O HNO Fe(NO ) NO H OFeO + + +

- Cht cho e : Fe 0,18 mol

Fe Fe3+ + 3e

0,18 0,54 e(cho)n 0,54(mol) =

Cht nhn e : O2 :m 10,08

(mol)32

; HNO3

57


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O + 2e O2-

m 0,18 m 0,1816 8

N+5 + 3e N+2 NO0,3 0,1 0,1

e(nhn)m 10,08n 0,3

8

= +p dng LBT e :m 10,08 0,3 0,54 m 12(gam)8

+ = =

4. Phng php ng cho trong bi ton trn ln hai dung dch hoc hhp hai khC s

p dng nh lut bo ton khi lng trong qu trnh trn ln cc dundch ca cng mt cht tan, ta lun c : Khi lng dung dch thu c bng tng khi lng ca cc dung dthnh phn. Khi lng cht tan thu c cng bng tng khi lng cht tan c trotng dung dch thnh phn .

58


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Phm vi p dng Pha long hay c cn dung dch Pha trn cc dung dch ca cng mt cht, cng loi nng

Pha trn cc khKhi trn ln 2 dung dch c nng khc nhau hay cho thm cht tannguyn cht vo dung dch cha cht tan , hoc qu trnh c cn dundch. tnh c nng dung dch trng thi cui ta c th gii b phng php bo ton khi lng, tuy nhin ta nn dng phng phng cho th gii bi ton s nhanh hn.Sau y gii thiu mt s s hay c s dng :

Nu trn dung dch 1 c khi lng l m1(g) v nng C1% vi dung dch2 c khi lng m2(g) v nng C2% (gi s C1 < C2) thu c dungdch mi c nng C% (vi C1 < C < C2) ta s dng s :

Ch :

Ta coi H2O c C% = 0.Ta coi cht tan nguyn cht c C = 100%.

Nu trn dung dch 1 c th tch V1 (lt) v nng CM(1) vi dung dch 2

c th tch V2 (lt) v nng CM(2) (gi s CM(1)< CM(2)) ta thu c dungdch mi c nng CM (vi CM(1)< C < CM(2)) ta s dng s sau :

59

=

m (g)......... C C Cm C CCm C Cm (g).........C C C

1 1 2

1 2

2 12 2 1

=

M ( ) M ( ) M

M( ) MM

M M ( )

M( ) M M ( )

V (lit)...... C C CC CVC

V C CV (lit)...... C C C

1 1 2

21

2 1

2 2 1


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Nu trn mt th tch V1 (lt) kh A c phn t khi MA vi mt th tch khB c phn t khi MB (gi s MA < MB) ta thu c hn hp kh c phn tkhi trung bnh lM (vi MA < M< MB) ta s dng s sau :

Bi ton minh ho

Bi 1. Cn trn V1 ml dung dch HCl 2M vi V2 ml dung dch HCl 0,5M thuc 300 ml dung dch HCl 1M. Gi tr V1, V2 ln lt l

A. V1 = V2 = 150 B. V1 = 100, V2 = 200

C. V1 = 200, V2 = 100 D. V1 = 50, V2 = 250

Li gii

Ta c1

2 1 2

1 2

V 1V 2 V 100ml ; V 200mlV V 300

= = =

+ =

Bi 2. Cn cho s g H2O vo 100 g dung dch H2SO4 90% c dung dchH2SO4 50% lA. 90 g B. 80 gC. 60 g D. 70 gLi gii

60

m 0

40

50 100 90

50

A B

B

A

B A

V (lit)...... M M Mn V M MMn V M M

V (lit)...... M M M

= =

1

1 1

2 2

2

V1 20,5

1 V2 0,5

1


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= =m 40 m 80(gam)100 50

Bi 3 . Lm bay hi 500 ml dung dch cht A 20% (D = 1,2 g/ml) ch cn 300 gdung dch. Nng % ca dung dch ny lA. 30% B. 40%C. 50% D. 60%Li gii

mdd = 500.1,2 = 600 (g)

y l bi ton c cn nn s :

600 x x 40%300 x 20 = =

Bi 4. Trn V1 ml dung dch NaOH (d = 1,26 g/ml) vi V2 ml dung dch NaOH(d = 1,06 g/ml) thu c 1lt dung dch NaOH (d = 1,16 g/ml). Gi tr V1, V2ln lt l

A. V1 = V2 = 500 B. V1 = 400, V2 = 600

C. V1 = 600, V2 = 400 D. V1 = 700, V2 = 300

Li gii

= = =1 1 22

V 0,1 V V 500mlV 0,1

61

dung d ch A : 60020 x

x

H2O: 300x 20

V1 1,260,1

1,16

V2 1,06

0,1


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Bi 5. T 200g dung dch KOH 30% c dung dch 50% cn thm vo s gKOH nguyn cht lA. 70 g B. 80 g

C. 60 g D. 90 gLi gii

= =m 20 m 80g200 50

Bi 6. Mt dung dch HNO3 nng 60% v mt dung dch HNO3 khc c nng 20%. c 200g dung dch mi c nng 45% th cn phi pha ch khi lng gia 2 dung dch HNO3 60%, 20% ln lt lA. 75g ; 125g. B. 125g ; 75g.C. 80g ; 120g. D. 100g ; 100g.Li gii

= =+ =

1

2

1 2

m 15 3m 25 5m m 200

= =1

2m 75gm 125g

Bi 7 . Mt hn hp 104 lt (ktc) gm H2 v CO c t khi hi i vi metan bng1,5 th 2HV v VCO trong hn hp l

A. 16 lt v 88 lt. B. 88 lt v 16 lt.

C. 14 lt v 90 lt. D. 10 lt v 94 lt.

62

m 10020

50

200 3050

m1 2015

45

m2 6025


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Li gii

12

V 2V 11

= ==

1

2

V 16ltV 88lt

Bi 8. Cho 6,12g Mg tc dng vi dung dch HNO3 thu c dung dch X ch cmt mui v hn hp kh Y gm NO v N2O c t khi hi i vi hiro bng 16,75. Th tch NO v N2O ( ktc) thu c ln lt lA. 2,24 lt v 6,72 lt. B. 2,016 lt v 0,672 lt.C. 0,672 lt v 2,016 lt. D. 1,972 lt v 0,448 lt.Li gii

Qu trnh cho electron : Mg Mg2+ + 2e

0,225 0,51

Qu trnh nhn electron : N+5 + 3e N+2 (NO)3x x

N+5 + 4e N+ (N2O)

8y 2y y

= =2N ONO

V 1 xV 3 y

3x 8y 0,51 x 0,093x y 0 y 0,03

+ = = = =

63

V 1 H2 2

4

24

V2 CO 2822

V1

NO 3010,5

33,5

V2

N2O 44 3,5


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Bi tp vn dngBi 1. Trn hai th tch metan vi mt th tch hirocacbon X thu c hn h

kh (ktc) c t khi so vi H2 bng 15. Cng thc phn t ca X l

A. C2H6 B. C3H8

C. C4H10 D. C5H12

Bi 2. Cho hn hp X gm 2 este c CTPT l C4H8O2 v C3H6O2 tc dng vi NaOH d thu c 6,14g hn hp 2 mui v 3,68g ancol B duy nht c khi so vi oxi l 1,4375. S g ca C4H8O2 v C3H6O2 trong A ln lt lA. 3,6g v 2,74g. B. 3,74g v 2,6g.C. 6,24g v 3,7g. D. 4,4g v 2,22g. Hng dn

MB = 1,4375.32 = 46ancol B l C2H5OH.

nB = nmui=3,6846 = 0,08 (mol)

= =mui 6,14M 76,75(g/mol)0,08p dng quy tc ng cho:

=== = =4 8 2

3 6 2

C H O

C H O

m 4,4gx 0,05x 5y 3 y 0,03 m 2,22g

Bi 3. T 1 tn qung hematit (A) iu ch c 400kg st. T 1 tn qumanhetit (B) iu ch c 500kg st. c 1 tn qung hn hp m

64

y HCOONa 685,25

76,75

x CH3COONa 82

8,75


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1 tn qung hn hp ny iu ch c 460kg st th phi trn 2 qung Avi t l v khi lng lA. 2 : 3 B. 3 : 5

C. 3 : 4 D. 1 : 3 Hng dn

= =AB

m 40 2m 60 3

Bi 4.Mt hn hp kh X gm SO2 v O2 c t khi so vi metan bng 3. Thm Vlt O2 vo 20 lt hn hp X thu c hn hp Y c t khi so vi metan bn2,5. Gi tr ca V lA. 20 B. 30C. 5 D. 10 Hng dnTrong 20 lt X ban u th th tch mi kh :

= = =1 1 22

V 16 V V 10(lit)V 16

Khi thm V lt O2 vo 20 lt hn hp X thu c hn hp Y :

= =+10 1 V 20(lit)

10 V 3

65

m A 40040

460

mB 500 60

V1 SO2 6416

48

V2 O2 3216

10 SO 2 648

40

10 + V O 2 3224


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Bi 5. S ml H2O cn thm vo 1 lt dung dch HCl 2M thu c dung dchmi c nng 0,8M lA. 1,5 lt B. 2 lt

C. 2,5 lt D. 3 ltBi 6.Trn 1 lt dung dch KCl C1 M (dung dch A) vi 2 lt dung dch KCl C2 M

(dung dch B) c 3 lt dung dch KCl (dung dch C). Cho dung dch C tcdng va vi dung dch AgNO3 thu c 86,1 g kt ta. Nu C1 = 4C2th C1 c gi tr lA. 1 M B. 1,2 M

C. 1,4 M D.1,5 MHng dn

Ag Cl AgCl0,6 0,6(mol)

+ +

1 2

2 1

1lt C 0,6 C0,6

2lt C C 0,6

Z]

Z ]

1 2

21 2 2 2 1

1

C 4C0,6 C1 C 0,6 1,2 2C 6C 1,8 C 0,3M ; C 1,2M

2 C 0,6

== = = = =

Bi 7. T khi hi ca hn hp kh C3H8 v C4H10 i vi hiro l 25,5. Thnh phn % th tch ca hn hp lA. 50% ; 50%. B. 25% ; 75%.C. 45% ; 55%. D. 20% ; 80%.

5. Phng php dng phng trnh ion rt gnC s Bn cht ca cc phn ng trao i xy ra trong dung dch cht in li l phn ng ca cc ion vi nhau to ra cht kt ta, bay hi hay cht ili yu,...

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Cch p dngKhi cho dung dch hn hp (X) phn ng vi dung dch hn hp (Y) thay vvic vit nhiu phng trnh phn ng gia cc phn t ta vit cc ph

trnh dng ion rt gn. Sau y l mt s s minh ha :

Cho s : 2 43

HClH SOHNO

+2

NaOHBa(OH) hn hp 6 mui + H2O

Bn cht l : H+(axit)+ OH (baz) H2O v 2 24 4Ba SO BaSO+ +

Khi mi trng trung tnh th :H (axit)

n + =OH ( )

n baz

Cho s 2 3

4 2 3

2 3

Na CO(NH ) COK CO

+ 22

BaClCaCl

Bn cht l :+

+ + +

2 23 3

2 23 3

Ca CO CaCO

Ba CO BaCO

Cho s :FeMgZn

+2 4

HClH SO (l) Hn hp mui + H2

Bn cht lFeMg

Al

+ H+ Hn hp mui + H2

Bi tp minh ha

Bi 1. Cho dung dch X cha ng thi 2 axit H2SO4 1M v HCl 2M vo 200mldung dch Y cha NaOH 1,5 M v KOH 1M. Khi mi trng dung dchtrung tnh th th tch dung dch X cn lA. 120 ml B. 125 ml

C. 200 ml D. 150 ml

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Li gii

Bn cht cc phn ng trn l 2H OH H O+ +

+

= + == + =

H

OH

n V.(2.1 2) 4V (mol)n 0,2.(1,5 1) 0,5(mol)

Khi mi trng trung tnh : 4V = 0,5V= 125 ml

Bi 2. Cho 100ml dung dch A cha ng thi 2 axit HCl 1M v HNO3 2M vo200ml dung dch B cha NaOH 0,8 M v KOH x M thu c dung dch CBit rng trung ho 100ml dung dch C cn 60ml dung dch HCl 1M. c gi tr lA. 1,2M B. 2,2MC. 3,3M D. 2,5MLi giiC 3 axit phn ng vi 2 baz. Bn cht cc phn ng l

2H OH H O+ +

+

= + + =

= +

H

OH

60.500n 0,1(1 2) .1 0,6(mol)100.1000

n 0,2(0,8 x)(mol)

Mi trng trung tnh: 0,6 = 0,2(0,8+x)x = 2,2M.

Bi 3. Trn 100ml dung dch X gm KHCO3 1M v K 2CO3 1M vi 100ml dungdch Y gm NaHCO3 1M v Na2CO3 1M thu c dung dch Z. Nh t t100ml dung dch T gm H2SO4 2M v HCl 1M vo dung dch Z thu c Vlt CO2 (ktc). Gi tr ca V l

A. 2,24 lt B. 8,96 ltC. 6,72 lt D. 4,48 ltLi giiBn cht ca cc phn ng gia cc cht trong T v Z l :

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23 3

3 2 2

CO 2H HCO

HCO H H O CO

+

++ + +

Khi cho dung dch X vo Y thu c dung dch Z c3HCO 0,2 (mol) v 23CO 0,2 (mol).

+ = + =2 4HCl H SOH (ddT)n n 2n 0,5(mol)

Nh t t dung dch T vo dung dch Z, phn ng xy ra theo th t : + + 23 3CO H HCO

0,2 0,2 0,2Tng s mol = + = 3HCOn 0,2 0,2 0,4(mol)

+ = = < =3H (cnl i) HCO

n 0,5 0,2 0,3(mol) n 0,4(mol)

3H (d- ) HCOn 0,5 0,2 0,3 n 0,2 0,2 0,4(mol)+ = = < = + =

3 2 2HCO H CO H O0,3 0,3 0,3

+

+ +

Bi 4. Tnh th tch dung dch cha HCl 0,5M v H2SO4 0,75M cn thit hotan hon ton 23,2g Fe3O4.A. 200 ml B. 300 mlC. 350 ml D. 400 ml

Li giiBn cht phn ng gia hai axit v Fe3O4 l :

Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O.

0,1 0,8Gi th tch dung dch l V : 0,5V + 2V.0,75 = 0,8 V = 400 ml

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Bi 5. Mt dung dch A cha HCl v H2SO4 theo t l mol l 3 : 1. Cho 100 mldung dch A trung ho va vi 500ml dung dch NaOH c cha 20g NaOH/lt. Tng khi lng mui thu c sau phn ng l

A. 31,175g B. 15,875gC. 42,113g D. 23,175gLi giiGi s mol ca H2SO4 l x (mol) th s mol HCl l 3x mol

Trong 100ml dung dch A c Hn 5x(mol)+ =

Trong 500 ml NaOH20 500

n . 0,25(mol) 5x 0,25 x 0,05(mol)40 1000= = = =

24

mui cation anion

Na Cl SO

m m mm m m 0,25.23 3.0,05.35,5 0,05.96 15,875g+

= += + + = + + =

Bi 6. Cho 100ml dung dch A cha NaCl 1,5M v HCl 3M vo 100ml dung dchB cha AgNO3 1M v Pb(NO3)2 1M thu c dung dch C v kt ta D.Khi lng kt ta D lA. 56,72 g B. 49,13 gC. 34,48 g D. 50,10 gLi giiBn cht cc phn ng xy ra gia A v B l

Ag+ + Cl AgCl

Pb2+ + 2Cl PbCl2

+ +

+

= + = == = = + =2

( )Cl

( )Ag Pb

n 0,1.(1,5 3) 0,45(mol) nn 0,1mol ; n 0,1mol n 0,1 0,1.2 0,3(mol)

t+(A) t (B)n n> ion Cl d :

mmui= 108.0,1 + 0,1.207 + 0,3.62 = 50,10 (g).

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Bi 7. Dung dch A cha axit HCl a M v HNO3 b M. trung ho 100 ml dungdch A cn dng 200 ml dung dch hn hp B cha NaOH 0,05M vBa(OH)2 0,15M. Mt khc kt ta hon ton ion Cl c trong 50ml dung

dch A cn 100ml dung dch AgNO3 0,1M. Cc gi tr a, b ln lt lA. 0,2M; 0,1M. B. 0,2M; 0,2M.C. 0,2M; 0,3M. D. 0,1M; 0,2M.Li gii

Bn cht cc phn ng xy ra gia A v B l : 2H OH H O+ +

+

= += + HOH

n 0,1.(a b) mol

n 0,2.(0,05 0,15)mol a b 0,04 + =

Bn cht phn ng xy ra gia A v AgCl l

Ag Cl AgCl+ + a 0,2M0,05a 0,01b 0,1M

== =Bi 8. Cho 2 kim loi Fe, Mg tc dng vi 200ml dung dch A gm HCl 0,1M,

H2SO4 0,2M thu c dung dch B v kh C. Cho t t dung dch D gm NaOH 0,3M, KOH 0,1M vo B tc dng va vi cc cht trong B thth tch dung dch D cn dng lA. 0,15 lt. B. 0,25 lt.C. 0,35 lt. D. 0,45 lt.

Li gii+

+ +

++ + +

22

2 22

2

Fe Fe Fe(OH)

H H H OH H OMg Mg Mg(OH)

nh lut bo ton in tch :

+ ( )n trong B = + Hn trong A = OHn trong D.

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Dung dch trung tnh khi :

+ = + = + = H OHn n 0,2(0,1 0,2.2) V(0,3 0,1) V 0,25 (lt)

Bi tp vn dngBi 1. tc dng va vi 0,96g hiroxit ca 2 kim loi kim hai chu k li

tip trong bng tun hon, phi dng 20ml dung dch HCl 0,4M v H2SO40,3M. Cc kim loi kim lA. Na, K B. Li, NaC. K, Rb D. Na, Rb Hng dn

Gi cng thc chung ca hai hiroxitROHBn cht cc phn ng :

2OH H H O++

ROHH OHn 0,02(0,4 0,3.2) 0,02(mol) n n+ = + = = =

ROH R R0,96M 48 M 17 48 M 310,02= = + = =

Hai km loi kim l Na, K.Bi 2. Ha tan hn hp hai kim loi Ba v Na (dng ht rt nh) vo nc thc dung dch A v 672 ml kh (ktc). Nh t t dung dch FeCl3 vo dung dchA cho n d, lc kt ta, ra sch, sy kh v nung n khi lng khng

c m g cht rn. Gi tr m lA. 3,2 g B. 6,4 gC. 1,6 g D. 4,8 g

Bi 3. Cho 4,64g hn hp A gm FeO, Fe2O3, Fe3O4 ( 2 3FeO Fe On :n 1:1= ) hotan hon ton trong V lt dung dch H2SO4 0,2M v HCl 0,6M. V c gi trl

A. 1,60 lt B. 1,22 lt

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C. 1,90 lt D. 1,56 lt Hng dn

Do FeO.Fe2O3 = Fe3O4, vy A xem hn hp ch l Fe3O4

= = =3 4A Fe O4,64n n 0,02232 (mol)

Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O0,02 0,16

0,16 = 0,1V V =0,16 1,60,1 = (lt)

Bi 4.Hn hp cht rn X gm Fe, Fe2O3, Fe3O4 v FeO (c s mol bng nhau l0,1 mol). Ha tan ht X vo dung dch Y gm HCl v H2SO4 long (d), thuc dung dch Z v 1,12 lt kh H2 (ktc). Nh t t dung dch Cu(NO3)20,5M vo dung dch Z cho ti khi ngng kh NO thot ra th dng li. Thtch dung dch Cu(NO3)2 dng lA. 158,3 ml. B. 140,0 ml.

C. 100,0 ml. D. 160,5 ml. Hng dn

Ta c: FeO + Fe2O3 Fe3O40,1 0,1 0,1

Hn hp X coi nh gm: 0,2 mol Fe3O4; 0,1 mol Fe + dung dch Y:

Fe3O4 + 8H+

Fe2+

+ 2Fe3+

+ 4H2O (1)0,2 0,2 0,4

Fe + 2H+ Fe2+ + H2 (2)0,05 0,05 0,05

Dung dch Z cha Fe2+ (0,35 mol), Fe3+ (0,35 mol), H+ d, Cl , SO2

4 .

Nh t t dung dch Cu(NO3)2 vo dung dch Z :

3Fe2+ + NO3 + 4H+ 3Fe3+ + NO + 2H2O (3)

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0,35 0,353

3 23 3

3 2

Cu(NO ) NO NO

Cu(NO )

0,35 1n (mol) n n 0,05 mol3 20,05V 0,1(lit)

0,5

= = =

= =

Bi 5. Cho 12,15 g bt Al vo 100 ml dung dch hn hp NaNO3 1,5M v NaOH3M, khuy u cho n khi ngng kh thot ra th dng li. Th tch kthot ra ktc lA. 5,04 lt B. 7,56 ltC. 6,72 lt D. 4,48 lt Hng dn

nAl = 0,45 (mol) ; 33 NaNO NOn n 0,15 mol = = ;

OHn = n NaOH= 0,3 mol

3 2 4 38Al 3NO 5OH 18H O 8[Al(OH) ] 3NH (1): 0,45 0,15 0,3: 0,4 0,15 0,25 0,4 0,15

: 0,05 0 0,05

Ban uPhn ngD-

+ + + +

2 4 22Al 2OH 6H O 2 Al(OH) 3H (2): 0,05 0,05

Ph : 0,05 0,05 0,075D : 0 0

[ ]Ban u

n ng-

+ + +

(1) v (2) nKh = 0,15 + 0,075 = 0,225 (mol)Vkh = 0,225.22,4 = 5,04 (lt)

Bi 6. Cho 6,4 g Cu tc dng vi 60 ml dung dch hn hp gm HNO3 2M vH2SO4 1M, thu c V lt kh NO duy nht (ktc), phn ng xy ra hon

ton. Gi tr ca V lA. 0,672 B. 0,896

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C. 1,344 D. 2,24 Hng dn

nCu = 0,1 mol; 3HNOn 0,12 mol= ; 2 4H SO2n 0,06.1 0,06 (mol)= =

Hn 0,12 2.0,06 0,24 (mol)+ = + =3Cu + 2NO

3 + 8H+ 3Cu2+ + 2NO + 4H2O

Ban u : 0,1 0,12 0,24Phn ng : 0,09 0,06 0,24 0,06D : 0,01 0,06 0 V NO= 0,06.22,4 = 1,344 (lt)

Bi 7. Dung dch A th tch 200ml cha ng thi hai mui MgCl2 0,4M vCu(NO3)2 0,2M. Dung dch B cha ng thi KOH 0,16M v Ba(OH)20,02M. Th tch dung dch B cn lm kt ta ht hai ion Mg2+, Cu2+ lA. 1 lt B. 1,2 ltC. 1,5 lt D. 1,7 lt Hng dnS phn ng

2 2

3 2 2 2 2

MgCl KOH Mg(OH) KCl(I)

Cu(NO ) Ba(OH) Cu(OH) BaCl+ +

Phng trnh ion thu gn22

22

Mg Mg(OH)OH (II)

Cu Cu(OH)

+

+

+

t(+)

OH

t(+) t( )

n 2.0,08 2.0,04 0,24(mol)n V.0,16 V.0,04 0,2V (mol)

n n 0,24 0,2V V 1,2(lt)

= + == + =

= = =

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77/119

2HH

2

H (d-)

OH

3,92n 0,5.1 0,5.0,5 0,75(mol) ; n 0,175(mol)22,4

2H 2e H

0,35 0,175n 0,75 0,35 0,4(mol)n 0,2V 2.0,1V 0,4V (mol)

+

+

+

= + = = =

+

= == + =

22

33

2

Mg 2OH Mg(OH)Al 3OH Al(OH)H OH H O

+

+

+

+ + +

Trong B : 2 3dt( ) Mg Al Hn 2n 3n n H 0,75(mol)+ + ++

+ = + + = =Trong C : dt( ) OHn n 0,4V = =

Mi trng trung tnh khi 0,4V = 0,75 V = 1,875(lt).Bi 10.Ho tan hon ton mg hn hp X gm 2 kim loi kim A, B v kim lo

kim th M vo nc thu c dung dch C v 5,6 lt kh H2 (ktc). Dungdch D gm H2SO4 0,1M v HCl 0,3M. trung ho dung dch C cn s ltdung dch D lA. 0,5 lt B. 1,5ltC. 1,5 lt D. 2 lt Hng dn

2H5,6

n 0,25 (mol)2,4= =

Kim loi kim kh nc gii phng H2, na phn ng nc b oxi ho sau

2 22H O 2e H 2OH0,25 0,5

+ +

Khi mi trng trung tnh thH OHn n 0,2V 0,3V 0,5 V 1lit+ = = + = =

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Bi 11. Ha tan hon ton 9,65 g hn hp 2 kim loi Fe, Al trong dung dch hnhp HCl v H2SO4 long, kt thc phn ng thu c 7,28 lt H2 (ktc).Phn trm khi lng ca st trong hn hp ban u l

A. 50,32%. B. 35,53%.C. 51,81 %. D. 56,48%. Hng dn

2Hn = 0,325 mol ; t x, y ln lt l s mol ca Al, Fe cha trong hn hTa c: 27x + 56y = 9,65 (I)Phng trnh ion rt gn ca cc phn ng

2Al + 6H+ 2Al3+ + 3H2 (1)x 1,5x

Fe + 2H+ Fe2+ + H2 (2)y y

T (1) v (2) 2HV = 1,5 x + y = 0,325 y = 0,325 1,5x (II)

T (I) v (II) ta tnh c : x = 0,15 mol ; y = 0,1 mol

%mAl =0,15.27.100% 41,19%

9,65= %Fe = 100 41,19 =58,81%

Bi 12. Cho m g hn hp Mg, Al vo 250 ml dung dch X cha hn hp axitHCl 1 M v axit H2SO4 0,5 M, thu c 5,32 lt H2 ( ktc) v dung dch Y(coi th tch dung dch khng i). Dung dch Y c pH l

A. 7 B. 1C. 2 D. 6

Bi 13.Dung dch A cha NaOH 1 M v Ca(OH)2 0,01 M. Sc 2,24 lt kh CO2 vo 400 ml dung dch A ta thu c mt kt ta c khi lng lA. 1,5g B. 10gC. 4g D. 0,4g

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6. Phng php xc nh cng thc cht ho hcC s Dng 1.Tm Z (in tch ht nhn) ca nguyn t . Dng 2.Tm c M (nguyn t khi) ca nguyn t . Bit ho tr ca nguyn t cn tm ta ch cn xc nh nguyn t khi cnguyn t . Khng bit ho tr ca nguyn t cn tm, ta phi lp c biu thc lh gia M v ho tr n ca nguyn t :

M = k.n

xc nh cng thc ca mt hp cht da vo nh lut thnh phkhng i. Xc nh cng thc ho hc ca mt hp cht.

+ Nu bit cc nguyn t thnh phn cu to nn cht AxByCx, ta ch vicxc nh x, y, z hoc x : y : z suy ra cng thc phn t (v i vi hcht v c th cng thc thc nghim thng trng vi cng thc phn t)

+ Nu cha bit mt nguyn t thnh phn no th ta phi xc nnguyn t khi ca nguyn t v ho tr tng ng ca n.+ Nu bi ton yu cu xc nh cng thc ca oxit kim loi m kim loi cho tr khng i, ta t CT ca oxit kim loi l M2On. Trng hp khc ta phi t cng thc ca oxit dng MxOy. T tm cng thc thc nghimv xc nh cng thc phn t.Bi tp minh ho

Bi 1. Tng s ht proton, ntron v electron trong nguyn t ca nguyn t Xl 22. Nguyn t X lA. N B. OC. F D. ClLi gii.

a)Theo gi thit : 2Z + N = 22N = 22 - 2Z

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Mt khc ng v bn c : N1 1,5Z

22 2Z 1Z

22 2Z 1,5

Z

6,2 Z

7,3V Z phi l nguyn dng nn chn gi tr Z = 7. Vy X l Nit

Bi 2. Tng s cc ht proton, ntron v electron trong nguyn t ca nguyn tX thuc nhm VA ca bng tun hon l 47. Nguyn t X lA. Si B. PC. Cl D. SLi giiTheo gi thuyt ta c : 2Z + N = 47 N = 47 2Z

Ta c : 47 2Z1 1,5Z 13,4 Z 15,6

Z phi nguyn dng nn Z c th l : Z = 14 hoc Z = 15

Nu Z = 14 c cu hnh electron nguyn t : 1s22s22p63s23p2

X thuc nhm IVA (Loi, v tri vi u bi)Nu Z = 15 c cu hnh electron nguyn t : 1s22s22p63s23p3.

X thuc nhm VA vy X l photpho (P)Bi 3. Ha tan hon ton mt kim loi (ha tr 3) trong 100 ml dung dch H2SO4

1M. trung ha lng axit d phi dng ht 20 ml dung dch NaOH 1M

Dung dch thu c cho tc dng vi dung dch NH3 d, lc kt ta, emnung n khi lng khng i thu c 3,06 g cht rn. Kim loi lA. B B. AlC. Cr D. GaLi giiGi kim loi cn tm l A.

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A. PCl3 B. FeCl3

C. CrCl3 D. AuCl3Li gii

3+ 3+

X X

Al + X Al + XM 27 M =M 27

gim khi lng thc ca3,87 0,14(mol)27 = Al l

m = 0,14(MX 27) = 4,06 MX = 56 ; X l Fe.

Bi 6.Ho tan 3,12 g hn hp A gm FeSO4 v XSO4 (c s mol bng nhau) vonc thu c 100 ml dung dch A. Cho lng A trn tc dng vi dundch BaCl2 d thu c 4,66 g kt ta v dung dch B. X l nguyn t nosau y ?A. Zn B. CuC. Ca D. BaLi gii

4 2

2 4

4 2

FeSO FeCl

BaCl BaSOXSO XCl

+ +

chnh lch khi lng ca 1 mol A BaSO4:M= 2.137 56 MX = 218 MX

Vi s mol FeSO4 = XSO4 = 1 4,66. 0,01(mol)2 233 = th chnh lch khi

lngm = 0,01(218 MX ) = 4,66 3,12 =1,54 MX = 64 : Cu

Bi 7. Ho tan 9,6 g mt kim loi A trong dung dch HNO3 d, thu c dungdch cha mt mui B v 2,24 lt kh NO (ktc). Kim loi lA. Cu B. Pb

C. Zn D. Fe

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Li giit s mol v ho tr ca A ln lt l x v n

3A + 4nHNO3 3A(NO3)n + nNO + 2nH2O

NOA

3.n 3.0,1 9,6nx ; M 32nn n 0,3= = = =

gi tr ph hp n = 2 v MA = 64 : Cu

Bi 8. Ho tan hon ton 5,3g mui cacbonat ca mt kim loi kim R 2CO3 trong500ml dung dch HCl 0,1M, H2SO4 0,2M. Sau phn ng thu c dung dchX v th tch kh thot ra l V = 1,12 lt (ktc). R l nguyn t no sau y ?A. Na B. K C. Li D. CsLi giiBn cht cc phn ng l

23CO + 2H

+ 2CO + H2O

0,05 0,1 0,05

R R5,32M 60 106 M 230,05+ = = = ; M l Na.

Bi 9. Ho tan va hn hp gm kim loi M v oxit MO (M c ho tr khni v MO khng phi l oxit lng tnh) trong dung dch HNO3 d thuc dung dch A v kh NO. Cho A tc dng va vi 240ml dung dc NaOH 0,5M thu c kt ta. Lc thu kt ta ri nung ti khi lng khi c 2,4g cht rn. M l nguyn t no sau y ?A. Ca B. CuC. Ba D. MgLi gii

NaOHn 0,12mol= . S phn ng :

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2+3 2

MHNO ... M + 2OH M(OH) MO

MO 0,12 0,06

+ +

Theo nh lut bo ton in tch :2M OH1n n 0,06(mol)2+ = =

MM + 16 = M2,4 M 240,06 = ; M l Mg.

Bi 10.Ho tan 32g kim loi M trong dung dch HNO3 d thu 8,96 lt hn hp haikh (ktc) NO v N2O4 c t khi hi so vi khng kh l 1,34 v dung dch

X. Kim loi M lA. Zn B. MgC. Cu D. AlLi gii

3 3 2 22 4

NOM HNO M(NO ) H O

N O+ + +

2 4

NO

N O

a b 0,4 a n 0,3mol30a 92b b n 0,05mol34

a b

+ = = = + = ==+

Cc bn phn ng :n +5 +2

5 42 4

M M ne N + 3e N (NO)1

1 0,9 0,3 0,03nN e N (N O )

0,1 0,1 0,05

+

+ +

+

+

32nM 32n1= = . Gi tr ph hp n = 2 th M

M 64(Cu)= .

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Bi tp vn dng

Bi 1.Ho tan hon ton 2,16g mt oxit kim loi A trong dung dch HNO3 long,thu c dung dch B (khng cha mui amoni) v 0,224 lt kh NO (ktc).

Cng thc ca oxit kim loi lA. CuO B. FeO

C. Fe3O4 D. CrO

Hng dn

t cng thc ca oxit MxOy ta c s phn ng

x y 3 3 a 2M O HNO M(NO ) NO H O+ + +

x y x y

2ya 5 2x

M O M O

Ch t cho e ch t nhn e2yM M (a ) N 3e Nx

0,03 0,03 0,010,03 0,03n m (x.M 16y) 2,16ax 2y ax 2y

2,16ax 4,8y 2yM 72a 160y (a )0,03x x

+ + + + + +

= = + =

= = >

Gi tr ph hp a=3 ; y = 1; M = 56 ; x = 1 vy cng thc ca oxit : FeOBi 2 . Cho mt lung kh CO i qua 16g mt oxit st nguyn cht nung nn

trong ng s. Sau khi phn ng kh ht oxit ta thy khi lng cht rgim 4,8g. Cng thc ca oxit st l

A. FeO B. Fe2O3C. Fe3O4 D. 2FeO. Fe2O3 Hng dn

FexOy + CO Fe + CO2

Fe16 4,8n 0,2

56= = mol ; O

4,8n 0,316= = mol

x : y = nFe : nO = 0,2 : 0,3 = 2 : 3 Cng thc ca oxit l 2 3Fe O

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Bi 3. Hp cht A c cng thc MxSy (M l kim loi). t chy hon ton A thuc oxit MnOm v kh B. Dn B vo dung dch cha Br 2 v Ba(NO3)2 dthy to thnh 69,9 g kt ta. Mt khc, kh hon ton MnOm bng CO d

thu c 8,4 g kim loi. Ho tan hon ton lng kim loi ni trn bndung dch HNO3 long thu c mui M3+ v 3,36 lt (ktc) kh NO. Cngthc phn t ca A lA. FeS B. CuS.

C. FeS2 D. Cu2S Hng dn

22 2 2O Br H O Ba2x y 2 4 4M S ... SO SO BaSO++ +

Theo s bo ton nguyn t S :

x y 4S(M S ) S(BaSO )69,9n n 0,3(mol)233= = =

3 3 3 2

M Fe

Fe S 2

M 4HNO M(NO ) NO H O

0,15 0,158,4M 56 (Fe) n 0,15mol0,15

x : y n : n 0,15: 0,3 1:2 A :FeS

+ + +

= = =

= = = Bi 4. A l hn hp dng bt gm Fe v kim loi M (c ho tr khng i). Ch

8,64g hn hp A tc dng va vi 80ml dung dch CuSO4 1,5M. Mtkhc ly mt lng A ng nh trn ho tan ht trong dung dch HNO3 thuc 3,136 lt kh NO duy nht ktc (khng c NH4 NO3). M l nguyn tno sau y ?A. Zn B. AlC. Cu D. Ag Hng dn

nFe = x (mol); nM = y (mol) Nu M y c Cu ra khi dung dch :

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2

2

n

Fe FeCu Cu

M M

+

+

++ +

p dng s bo ton electron, ta c :2x + ny = 2.0,12 = 0,24 (mol) (1)

Trong phn ng vi HNO3 :

3 3

3 2

3 n

Fe Fe(NO )

HNO NO H O

M M(NO )

+ + +

3x + ny = 3.3,136 0,4222,4 = (mol) (2)

Gii h (1) v (2). H ny v nghim. Vy M khng y c Cu.

Fe + Cu2+ Fe2+ + Cux = 0,12 ny = 0,06

mM = 1,92 MM =1,92.n 32n M :Cu0,06 =

Bi 5. Ho tan 3,06g ho hc Na2CO3 v MCO3 (t l 2 3 3Na CO MCOn :n 1:2= )trong H2SO4 long d, kh thu c hp th hon ton trong 100ml dung dchCa(OH)2 0,2M thu c 1,97g kt ta. Kim loi M l

A. Ba B. CaC. Zn D. Cu Hng dn

2 3 2 4

2 4 2 2

3 4

Na CO Na SOx H SO x CO H O

MCO MSOy y

+ + +

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cng iu kin. CTPT A l:

A: C2H6O2 B: C3H6O3

C: C3H4O3 D: C2H4O2 Hng dnVi mi m ta u xc nh c CTPT ca A, chn m = 15 g.

2

2

C(A) CO

H(A) H O O

C H O 2 n

3 6 3

n n 0,5 (mol)9 15 0,5.12 1n 2n 2. 1(mol);n 0,5(mol)18 16

3,6n : n :n 1:2:1 (CH O) 90 n 30,04CTPT :C H O

= =

+= = = = =

= = = =

7. Phng php gii cc bi ton cc i - cc tiu Cch tm khong gii hn ca mui : Hn hp kim loi (A, B) tc dng vi h

hp axit (HNO3 v H2SO4) to ra hn hp mui sunfat v mui nitrat.

+) Do 1 mol 24SO (nng 96 gam) tng ng vi 2 mol3NO (nng 124

gam). Vi cng mt hn hp kim loi nu to mui nitrat th khi lngnng hn mui sunfat. Khi lng mui cc i khi hn hp ch to ra mnitrat v cc tiu khi hn hp ch to mui sunfat. Vy khi lng thc t

2- -4 3thc tmuiSO mui NO

m m m

Nhung Bt Co Nhieu Pp Giai

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