Tuyen Tap Cac Bai Tap Hoa Hoc Hay Va Kho

gv: - 016 7576 8182

Cu 1. Hn hp X gm 1 anehit v 1 hirocacbon mch h (2 cht hn km nhau 1 nguyn

t- cacbon). t ch|y ho{n to{n 0,4 mol hn hp X thu c 1,3 mol CO2 v 0,4 mol H2O.

Nu cho 31,8 gam hn hp X t|c dng vi dung dchAgNO3/NH3 th khi lng kt ta thu

c ti a l{

A. 209,25 gam.

B. 136,80 gam.

C. 224,10 gam.

D. 216,45 gam.

B[I GII s nguyn t C trung bnh = 1,3/0,4 = 3,25 ==> C3 a mol v C4 b mol ,mol hh = a + b

= 0,4,mol CO2 = 3a + 4b = 1,3 ==> a = 0,3 v b = 0,1 .s nguyn t H trung bnh = 2.0,4/0,4 =

2 ==> mi cht u c 2 nguyn t H ===> C3H2On v C4H2Om

Nu n = 0 ==> hydrocarbon mch h CT C3H2 ==> loi. Vy hydrocarbon l{ C4H2 0,1 mol

c cu to CHC-CCH

v andehit C3H2O 0,3 mol CHC-CHO ==> m hh = 54a + 50b = 21,2

CHC-CCH ---> CAgC-CCAg

0,1------------------------0,1

CHC-CHO --> CAgC-COONH4 + 2 Ag

0,3-----------------------0,3---------------0,6 khi lng kt ta = 265.0,1 + 194.0,3 + 108.0,6 =

149,4 . Khi lng kt ta thu c khi cho 31,5 g X t|c dng AgNO3 l :===>

31,5.149,5/21,2 = 224,1 ==> cu C

Cu 2(Chuyn Bn Tre).Hn hp X gm Al, Fe2O3, Fe3O4, CuO trong oxi chim 25,39% khi lng hn hp. Cho m gam hn hp X tc dng vi 8,96 lt CO(iu kin tiu chun) sau 1 thi gian thu c cht rn Y v hn hp kh Z c t khi so vi hiro l{ 19. Cho cht rn Y tc dng vi dung dch HNO3 lo~ng d thu c dung dch T v{ 7,168 lt NO(iu kin tiu chun, sn phm kh duy nht). C cn dung dch T thu c 3,456m gam mui khan. Gi tr ca m l A. 41,13 B. 35,19 C. 38,43 D. 40,03 -X gm 0,2539m ca O v{ 0,7461m ca c|c kim loi. Dng pp ng cho v{ bo

ton C=0,4 => trong Z c CO=0,15 v CO2=0,25

=> Y cn 0,7461m Kim loi v{ O=0,2539m/160,25 (mol). Y cn lng HNO3= 4NO + 2O

=0,32.4+2(0,2539m/160,25)

=> dung dch T c NO3- = HNO3 - NO=0,32.3+2(0,2539m/160,25). Vy T

gm 0,7461m ca Kim Loi v{ 62.NO3- ca NO3-

=> 3,456m=0,7461m+62(0,32.3+2(0,2539m/160,25)). Dng my tnh => C

gv: - 016 7576 8182

CCH 2 hh Z (CO2 v{ CO d) c mol = 0.4 v{ MZ = 38 --> nCO2 = 0.25 = nO p. .

hhX {M=x molO=y mol hh Y{M=xOd=y0.25

bton e: nx = 2(y - 0.25) + 3.0,32 --> nx = 2y + 0,46 = mol NO3 trong mui

{Mx+62(2y+0.46)=3.456mmX=Mx+16y=m 108y+28.52=2.456m kt hp vi mO = 16y

= 0.2539m --> m = 38.43g

Cu 3(Chuyn Bn Tre).Hn hp X gm 1 mol amin no mch h A v{ 2 mol aminoaxit no

mch h B t|c dng va vi 4 mol HCl hay 4 mol NaOH. t a gam hn hp X cn 46,368

lt O2(ktc) thu c 8,064 lt kh N2 (ktc). Nu cho a gam hn hp trn t|c dng vi dung

dch HCl d thu c bao nhiu gam mui?

A. 75,52

B. 84,96

C. 89,68

D. 80,24

T c}u n{y: Hn hp X gm 1 mol amin no mch h A v{ 2 mol aminoaxit no mch h B t|c

dng va vi 4 mol HCl hay 4 mol NaOH.

=> A c dng CaH2a+4N2 v{ B c dng CbH2b-1O4N

}y mnh xin gii theo kiu CT tng qu|t nh sau:

X s c dng CnH2n+2+t-zOzNt .Do A:B=1:2 nn dng ng cho th c z=8/3 v t=4/3 => X=

CnH2n+2/3O8/3N4/3. Dng cng thc n{y c}n bng phn ng t ch|y, ta c ..... + 0,5(3n-

7/3)O2 -------------> 2/3 N2 +.......................2,07 -------------->

0,3=> 0,36.0,5(3n7/3)=2,07.2/3 => n=10/3

=> X= 0,36.3/2=0,54 mol v{ ph}n t khi X l{ 14.10/3+2/3+16.8/3+14.4/3=326/3

=> a=0,54.326/3=58,68

=> mui= 58,68+36,5.0,54.t=58,68+36,5.0,54.4/3 => B

Cu 4(Chuyn Bn Tre).Hn hp X gm Fe2O3 v{ Cu. Cho m gam hn hp X t|c dng vi dung dch HCl va thu c dung dch cha 122,76 gam cht tan.Cho m gam hn hp X t|c dng vi dung dch H2SO4 lo~ng va thu c dung dch Y cha 3 cht tan vi t l s mol 1:2:3. Dung dch Y l{m mt m{u ti a bao nhiu gam KMnO4 trong mi trng axit sunfuric? A. 6,004 B. 5,846 C. 5,688 D. 6,162

u tin i x|c nh t l Fe2O3: Cu trong X . Gi s X c Fe2O3=1 mol th Cu= a mol , X tan vo H2SO4 thu c Y c CuSO4=a, Fe2(SO4)3=1a v FeSO4=2a. D thy phi c a Fe2O3: Cu=4:1, Khi m gam X s c Fe2O3=4Cu=4x mol

gv: - 016 7576 8182

Ha tan X vo HCl th HCl= 3.2.Fe2O3=24x mol, => 122,76=56.2.4x+64.x+35,5.24x => x=0,09 Vy Y c FeSO4= 0,092=0,18 => 5.KMnO4= FeSO4 KMnO4=0,036 mol => C CCH 2 Gi a, b l{ s mol Fe2O3 v Cu, HCl ch phn ng vi a mol Fe2O3 ==> mol Cl- = 6a

Sau , Cu t|c dng ht vi mui Fe3+

khi lng cht tan = 56.2a + 64.b + 35,5.6a = 122,76 ===> 325a + 64b = 122,76 (1)

Tng t, khi t|c dng vi H2SO4 lo~ng, thu c 3 mui :

Fe2O3 ---> Fe2(SO4)3

a---------------------------a

Cu + Fe2(SO4)3 ---> 2 FeSO4 + CuSO4

b--------b-------------------2b---------b

CuSO4 b mol, FeSO4 2b mol v Fe2(SO4)3 (a-b) mol ==> a > b

==> t l mol: CuSO4 : FeSO4 : Fe2(SO4)3 = b : 2b : (a-b) = 1 : 2 : 3 ==> a = 4b ==> (1) ==> b =

0,09

==> mol FeSO4 = 2b = 0,18 mol

10 FeSO4 + 2 KMnO4 --> 5 Fe2(SO4)3 + ---

0,18 ------------0,036

Khi lng KMnO4 cn dng = 158&0,036 = 5,688 ==> c}u C

Cu 5: ( lng 1- NGh An) Khi nung butan vi xc t|c thch hp n khi phn ng xy ra ho{n to{n thu c hn hp T gm CH4,C3H6,C2H4,C4H8,H2,C4H6. t ch|y ho{n to{n hn hp T thu c 8,96 lt CO2 (ktc) v{ 9 gam H2O. Mt kh|c hn hp T l{m mt m{u va ht 19,2g Br2 trong dung dch nc Brm. % v s mol ca C4H6trong T l: A:16,67% B:22,22% C:9,091% D:8,333%

t ch|y hh T ging t ch|y C4H10 ban u : s mol CO2 = 0,4 v{ s mol H2O = 0,5 ==> mol C4H10 = 0,5 - 0,4 = 0,1 C4H10 CH4+C3H6 a---------------------a------------a C4H10 2H2+C4H6 b---------------------2b---------b C4H10 H2+2C2H4 c----------------------c--------2c s mol butan phn ng = a + b + c = 0,1. S mol Br2 = a + 2b + 2c = 0,12 ==> b + c = 0,02 ==> b < 0,02 ==> mol hh T = 2a + 3b + 3c= 0,22. ==> % v s mol ca C4H6=0,02.1000,22 cu D Cu 6: thi th trng THPT Lng 1- Ngh An. Dung dch X cha 14,6 gam HCl v 22,56gam Cu(NO3)2. Thm m gam bt st v{o dung dch

gv: - 016 7576 8182

X , sau phn ng xy ra ho{n to{n thu c hn hp kim loi c khi lng l{ 0,535m gam v{ ch to kh NO l{ spk duy nht. Gi| tr ca m l{: A: 9,28 B: 14,88 C:16 D:1,92

s mol H+ = mol Cl- = mol HCl = 0,4. S mol Cu(NO3)2=0,12 ==> mol Cu2+=0,12 v mol NO3=0,24, V cn d kim loi ==> c mui Fe2+ : Fe2e Fe2+ a------2a--------a NO3+3e+4H+ NO+2H2O 0,1--------0,3-------0,4 Cu2++2e Cu 0,12--------0,24-----0,12 .Bo to{n mol e : 2a = 0,3 + 0,24 = 0,54 ==> a = 0,27 khi lng rn = mCu+mFedu=640,12+m56a=0,535m===>m=16===> cu C Cu 7 ( Qunh Lu I ln II,2012-2013): X l{ hp cht hu c n chc C,H,O . Cho 1 lng cht X t|c dng ho{n to{n vi 500 ml dung dch KOH 2,4M ri c cn c 105(g) rn khan Y v{ m(g) ancol. Oxi ha m(g) ancol bng oxi c xc t|c c hn hp Z, chia Z th{nh 3 phn bng nhau : Phn 1: t|c dng vi AgNO3/NH3 d thu c 21,6(g) Ag Phn 2: t|c dng vi NaHCO3 d thu c 2,24(l) kh (ktc) Phn 3:t|c dng vi Na (va ), thu c 4,48(l) kh (ktc) & 25,8(g) rn khan. X|c nh CTPT ca X. Bit ancol un vi axit sunfuric c nng,170oC to olefin: A. C6H12O2 B. C3H6O2 C. C4H8O2 D. C5H10O2

KOH=0,4 v{ mY=35 gam tin l{m vic vi mi phn sau n{y ! . t ancol dng RCH2OH => Z gm H2O, RCOOH, RCHO, v RCH2OH (d) + phn 2: d c ngay RCOOH=CO2=0,1. + phn 3: H2=0,2 H2O, RCOOH, RCH2OH qua Na thu c NaOH, RCOONa, RCH2ONa . Bo to{n khi lng => mA = mH2O + mRCOOH + mRCH2OH =25,8-0,2*2*22=17. + phn 1: Ag=0,2 RCHO: Ag= 1:2 hoc 1:4 Xt RCHO: Ag=1:2 => RCHO=0,1 => H2O=RCOOH+RCHO=0,2 v RCH2OH=0,1. => R= (mA-mH2O-0,1*COOH-0,1*CH2OH)/0,2 =(17-0,2*18-0,1*45-0,1*31)/0,2=29 => R=C2H5 => C3H8OH l{ ancol thu c sau phn ng thy ph}n este X . Khi nX=0,3 mol => X=(35+0,3*60-0,4*56)/0,3=102 => C5H10O2 Xt RCHO: Ag=1:4=HCHO:Ag => HCHO=0,05 mol, H2O=0,15 mol v CH3OH=0,05 mol => mA= 8,9 < 17 => loi .Vy D

gv: - 016 7576 8182

Cu 8: X l{ hn hp FeBr3 v MBr2. Ly 0,1 mol X nng 25,84 gam t|c dng vi dung dch

AgNO3 d thu c 52,64 gam kt ta. X|c nh v khi lng FeBr3 trong X?

A. 91,64% B. 41,77% C. 51,63% D. 60,71%

Gi a, b l{ s mol FeBr3 v MBr2 ==> khi lng hh = 296a + b(M+160) = 25,84 (1). mol hh = a + b = 0,1 (2) v mol Br = 3a + 2b TH 1 : Nu kt ta l{ AgBr ==> mol Br- = mol kt ta = 3a + 2b = 0,28 (3). (2) v (3) ==> a = 0,08 v b = 0,02 . (1) ==> M < O ==> loi TH 2 : ==> Ngo{i AgBr (3a+2b) mol kt ta cn c Ag b mol kt ta ==> M l{ kim loi a ha tr. Khi lng kt ta = 188*(3a+2b) + 108b =52,64 (4). (2) v (4) ==> a = 0,053 v b = 0,047. (1) ==> M = 56 ==> khi lng FeBr3 = 296a = 15,688 ===> %m = 60,71 ==> cu D Cu 9: Hn hp X c 2 hirocacbon ng ng lin tip, c t khi hi so vi H2 bng 15,8. Ly 6,32 gam X li v{o 200 gam dung dch cha xc t|c thch hp th thu c dung dch Z v{ thy tho|t ra 2,688 lt kh kh Y iu kin tiu chun c t khi hi so vi H2 bng 16,5. Bit rng dung dch Z cha anehit vi nng C. (Bit phn ng ch xy ra theo hng to th{nh sn phm chnh). Gi| tr ca C l: A. 1,043% B. 1,305% C. 1,407% D. 1,208%

Ta c M=31,6. Suy ra 2 hidrocacbon t|c dng vi nc sinh ra andehit th phi l{ ankin, do ta c 2 hidrocacbon l{ C2H2 v C3H4 Khng kh thy rng nC2H2=0,12;nC3H4=0,08. Sau phn ng c hai hidrocacbon d: C2H2 v C3H4 d 0,06 mol. S mol C2H2 phn ng 0,06 mol, to ra 0,06 mol CH3CHO V C3H4 to ra axeton.. Khi lng dung dch sau phn ng bng: 6,32+2002,68822,4.16,5.2=202,36. Nng cht andehit l: C1,305

Bi 10.Dung dch X cha hn hp gm Na2CO3 1.5M v KHCO3 1M .Nh t t tng git cho n ht 200ml dung dch HCl1M v{o 100ml dung dch X, sinh ra V lt kh ktc. un nng c cn dung dch sau phn ng thu c m(g) mui khan . Gi| tr ca m l{ ?

s mol Na2CO3 = 0,15 ==> mol Na+ = 0,3 v CO32- = 0,15. mol KHCO3 = 0,1 ==> mol K+ = 0,1 v HCO3- = 0,1. mol HCl = mol H+ = mol Cl- = 0,2 H+ + CO32- ---> HCO3- 0,15--0,15--------0,15 H+ + HCO3- ---> CO2 + H2O 0,05--0,05--------0,05 dd sau phn ng c: Na+ = 0,3 mol, K+ = 0,1 mol , Cl- = 0,2 mol v HCO3- d 0,2 mol khi un nng : 2 HCO3- ---> CO2 + CO32- + H2O 0,2----------------------0,1 mui gm Na+ = 0,3 mol, K+ = 0,1 mol , Cl- = 0,2 mol v CO32- d 0,1 mol. ==> khi lng mui = 23*0,3 + 39*0,1 + 35,5*0,2 + 60*0,1 = 23,9

gv: - 016 7576 8182

Cu 11: Hn hp X gm 0,1 mol anehit acrylic v{ 0,3 mol kh hiro. Cho h n hp X qua

ng s nung nng c cha Ni l{m xc t|c, thu c hn hp Y gm propanal, propan-1-ol,

propenal v{ hiro. T khi hi ca hn hp Y so vi metan bng 1,55. S mol H2 trong hn

hp Y bng bao nhiu?

A. 0,05

B. 0,15

C. 0,20

D. 0,10 Ta c s mol trc l{ 0,4. Khi lng c|c cht trc phn ng bng:

0,1.56+0,3.2=6,2(g). Bo to{n khi lng nsau=6,21,55.160,25.

S mol gim chnh l{ s mol hidro tham gia phn ng bng 0,40,25=0,15. Nn s mol ca

hidro trong Y l 0,30,15=0,15. Chn B.

Cu 12/ Cho m gam hn hp X gm Andehit axetic , andehit acrylic v andehit oxalic tc dng vi dung dch AgNO3/NH3 d , thu c 69,12 gam Ag . Mc kh|c ch|y ho{n to{n m gam hn hp X , thu c 37,86 gam hn hpCO2 v{ hi nc c t khi so vi H2 l{ 16,46 . Ga tr ca m l{ A/17.22 B/14.02 C/19.42 D/13.62

Gi a, b, c l{ s mol C2H4O, C3H4O v C2H2O2 s mol Ag = 2a + 2b + 4c = 0,64 ===> a + b + 2c = 0,32 = s mol nguyn t O = z = 0,32. Gi x, y l{ s mol CO2 , H2O ==> x + y = 37,86/2*16,46 = 1,15. V{ 44x + 18y = 37,86 ==> x = 0,66 v y = 0,49 ==> m = 12x + 2y + 16z = 12*0,66 + 2*0,49 + 16*0,32 = 14,02 ==> cu B

Cu 13/Cho m gam Al v{o hn hp dung dch cha 0,15 mol FeCl3 v{ 0,12 mol CuCl2 . Sau

khi kt thc phn ng , thu c 2,25m gam cht rn gm hai kim loi. Gi| tr ca m l{

A/3.48 B/3.36 C/3.6 D/3.86

BL: Cht rn sau gm 2 KL Fe,Cu .==>DD gm Al3+=x;Fe2+=y mol \p dng nh lut bo

to{n in tch ta thu c HPT: 3x+2y=0,15.3+0,12.2 , 27x=m v

(0,15y).56+0,12.64=2,25m ==> m= 3,76258 .

Cu 14: Cho 18,3 gam hn hp X gm hai hp cht hu c n chc l{ dn xut ca benzen

c cng cng thc ph}n tC7H6O2t|c dng ht vi dung dch AgNO3 trong NH3 thu c

10,8 gam Ag. Vy khi cho 9,15 gam X ni trn t|c dng vi 300 ml dung dch NaOH 1M, ri

c cn dung dch sau phn ng s thu c bao nhiu gam cht rn khan ?

A. 16,4 gam. B. 20,8 gam. C. 19,8 gam. D. 8,0 gam s mol hh X = 18,3/122 = 0,15v{ s mol

Ag = 0,1

mol X : mol Ag = 0,15 : 0,1 < 1 : 2 ===> ch c 1 cht cho phn ng tr|ng gng l{ HO-C6H4-

gv: - 016 7576 8182

CHO hoc H-COO-C6H5 0,05 mol, cht cn li l{ C6H5-COOH 0,1 mol

s mol 9,15 gam X = 0,15*9,15/18,3 = 0,075

X + NaOH --> mui + H2O

0,075--------------------------0,075. Bo to{n khi lng : mran = mX + mNaOH - mH2O = 9,15 +

40*0,3 - 18*0,075 = 19,8 ==> cu C

Trch Trn Ph-ln I-2013

Cu 15.Cho m gam hn hp X (gm Mg, Al, Zn v{ Cu) t|c dng ht vi dung dch HNO3 thu

c dung dch Y (khng c mui amoni) v{ 11,2 lt (ktc) hn hp kh Z (gm N2, NO, N2O

v NO2, trong N2 v NO2 c phn trm th tch bng nhau) c t khi i vi heli bng 8,9.

S mol HNO3 phn ng l{

A. 3,4 mol.

B. 3,0 mol.

C. 2,8 mol.

D. 3,2 mol. Hn hp kh Z gm: N2, NO, N2O v NO2, trong mol N2 = mol NO2 => tng

ng hh Z c: a mol NO v{ b mol N2O

s mol Z = a + b = 0,5khi lng Z = 30a + 44b = 4*8,9*0,5 = 17,8 ==>a = 0,3 v{ b = 0,2

mol HNO3 phn ng = 10*mol N2O + 4*mol NO + = 10*0,2 + 4*0,3 = 3,2 ===> cu D

Cu 16: Cho a gam hn hp X gm mt axit no n chc v{ 14 este to ra bi axit no n chc B l{ ng ng k tip ca A v{ 1 ru no n chc. Cho a gam hn hp X t|c dng va NaHCO3, thu c 1,92 gam mui. Nu cho a gam hn hp X t|c dng vi mt lng va NaOH un nng thu c 4,38gam hn hp hai mui ca hai axit A v{ B v{ 0,03 mol ru, ru n{y c t khi so vi H2 l 23. t ch|y hai mui bng mt lng oxi thu c mui Na2CO3, hi nc v{ 2,128l CO2. Gi thit c|c phn ng xy ra ho{n to{n. Gi| tr ca a A 3,98g B 4,12g C 3,56g D 2,06g

S mol ru = mol este = 0,03, Ph}n t lng ru = 46 ==> ru etylic . X gm axit CnH2nO2 x mol v CmH2m+1 -COO-C2H5 0,03 mol Axit CnH2nO2 x mol + NaHCO3 cho 1,92 gam mui CnH2n-1O2Na x mol. HH X + NaOH cho hai mui: 1,92 gam CnH2n-1O2Na v 0,03 mol CmH2m+1 COONa . ==> khi lng mui CmH2m+1 -COONa = 4,38 - 1,92 = 2,46 Ph}n t lng mui CmH2m+1 -COONa = 14m + 68 = 2,46/0,03 = 82 ===> m = 1 ==> CH3 -COONa ==> axit B l CH3COOH ==> este CH3 -COO-C2H5 0,03. V axit A: CnH2nO2 ng ng lin tip ca B ==> n = 1 hoc n = 3 C2H3O2Na ----> 1,5 CO2 + 0,5 Na2CO3 0,03---------------0,045 CnH2n-1O2Na ---> (n-0,5) CO2 + 0,5 Na2CO3

gv: - 016 7576 8182

x------------------------x(n-0,5) mol CO2 = 0,045 + x(n-0,5) = 0,095 ===> x(n-0,5) = 0,05 Nu n = 1 ==> x = 0,1 ==> loi v khi lng mui HCOONa = 68*0,1 = 6,8 > 1,92 Nu n = 3 ==> x = 0,02 ==> khi lng mui C3H5O2Na =96*0,02 = 1,92 ==> nhn ===> a = 74x + 88*0,03 = 4,12 ==> cu B

Cu 17: Hn hp X gm etanial, axetilen, propanial v{ vinyl fomat (trong s mol ca

etanial v{ axetilen bng nhau). t ch|y ho{n to{n m gam hn hp X cn dng va V lt

O2 thu c 52,8 gam CO2. Mt kh|c t ch|y ho{n to{n 0,1V lt hn hp etan, propan

cn 0,455V lt O2 thu c a gam CO2. Tnh a. A. 14,344 B. 16,28 C. 14,256 D. 16,852

Mt b{i to|n hay...

*Hn hp ban u X : C2H2O2(xmol);C2H2(xmol);C3H4O2;C3H4O2

Ta quy hn hp trn tr th{nh Y: C2H2O(2xmol);C3H4O2(ymol)

Y= C2(H2O)(2xmol);C3(H2O)2(ymol)

Suy ra sau phn ng t ch|y nH2O=2x+2y=n (O trong Y). \p dng bo to{n O

==> nCO2=nO2=V=1,2 ( V cho tin tnh to|n)

* n phn ng t ch|y th 2: ===> Bo to{n O

: 2nCO2=2nO2H2O=2nO2(nankan+nCO2) ==> 3nCO2=2nO2nankan==>nCO2=0,324

==> a=14,256 (Chn C)

Cu 18 Cho hn hp kh CO,H2 qua m gam hn hp Fe3O4 v CuO (t l mol 1:2 )sau phn

ng thu c 10,4 gam hn hp rn A.Ha tan ho{n to{n A trong dung dch HNO3 d thu

c 0,05 mol kh NO l spk duy nht.

m c gi| tr A.11,638 B.11,368 C.14 ,2 D.12,4

Gi a, 2a l{ s mol Fe3O4 v{ CuO ==> trong rn A c 3a mol Fe, 2a mol Cu v{ b mol O

khi lng A = 56*3a + 64*2a + 16b = 10,4 ===> 296a + 16b = 10,4. Bo to{n mol e : 9a +

4a = 2b + 0,15 ===> 13a - 2b = 0,15

===> a = 029 v b = 0,1135. ==> m = 232*0,029 + 80*2*0,029 = 11,368 ==> cu B

Cu 19: Cho 33,6 g Fe vo dd H2SO4 c, nng sau khi phn ng xy ra ho{n to{n thu c 2,24 lt kh duy nht SO2 v{ 14,4 gam cht rn .S mol axit ~ tham gia phn ng l{ A.0,8 B.0.4 C.0,6 D.1,2 S mol Fe = 33,6/56 = 0,6 ==> c th cho Fe2+ hoc Fe3+ ==> mol e nhn t 2*0,6 = 1,2 n 3*0,6 = 1,8 S mol SO2 = 0,1 ===> mol e nhng = 2*0,1 = 0,2 TH 1: Nu 14,4 gam rn ch l{ S ==> mol S = 0,35 ==> mol e nhng = 6*0,35 = 2,1 > 1,8 ==> loi TH 2 : Nu 14,4 gam rn ch l{ Fe d ==> mol Fe d = 0,257 ==> mol Fe phn ng = 0,342 ==> mol e nhn = 0,684 > 0,2 ===> loi TH 3 : Trong 14,4 gam rn c : Fe d (0,6-x) mol v{ a mol S, vi x l{ s mol Fe phn ng

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khi lng rn = 33,6 - 56x + 32a = 14,4 ====> 56x - 32a = 19,2 (1) Fe - 2e ---> Fe2+ x-------2x--------x S6+ + 2e + ----> S4+ ------0,2--------0,1 S6+ + 6e + ----> S--------6a---------a Bo to{n mol e : 2x = 6a + 0,2 ===> x - 3a = 0,1 (2). (1) v (2) ==> x = 0,4 v a = 0,1 Bo to{n s mol S ==> mol H2SO4 = mol SO4/mui + mol SO2 + mol S = 0,4 + 0,1 + 0,1 = 0,6 ==> cu C

Chuyn LQ-QUNG TR Cu 20. ha tan ht hn hp X gm Cr2O3, CuO ; Fe3O4 cn va 550 ml HCl 2M, sau phn ng thu c dung dch Y.Mt na dung dch Y ha tan ht ti a 2,9 gam Ni. C cn na dung dch Y cn li thu c bao nhiu gam mui khan ? A. 30,8 gam B. 30,525 gam C. 61,6 gam D. 61,05 gam P/s : c th gii ngn hn nu : MCr2O3 + MCuO = MFe3O4 = 232

Cu 21: Ly 21,33 gam cao su isopren ~ c lu ha em t chy hon ton bng oxi va , sau phn ng ngng t ht hi nc th cn li 34,272 lt kh (ktc). Trung bnh c bao nhiu mt xch isopren th c 1 cu ni isufua (S-S)?

Cng thc ca cao su isopren ~ c lu ha l{:C5nH8n2S2.Khi t ch|y ta c phng trnh: C5nH8n2S2+O2 CO2+H2O+2SO2. S mol ca C5nH8n2S2 =21,33/68n+62. S mol kh CO2 v SO2 l:34,272/22,4=1,53 mol \p dng nh lut bo to{n nguyn t ,ta c phng trnh: 5n.21,3368n+62+2.21,3368n+62=1,53 n=20

Cu 22: Hn hp X gm andehit fomic, andehit oxalic, axit axetic, etilenglycol, glyxerol. Ly 4,52 gam X t chy hon ton ri cho sn phm i qua bnh 1 ng H2SO4 (c, d), bnh 2 ng 600 ml dung dch Ba(OH)2 0,2M thy bnh 1 tng 2,88 gam, bnh 2 xut hin m gam kt ta. X|c nh m? A. 23,64 gam B. 17,73 gam C. 15,76 gam D. 19,70 gam

Khi vit cng thc ca c|c hp cht n{y nhn thy s C bng s O cng thc chung ca X l:CxHyOx \p dng nh lut bo to{n nguyn t H ta c phng trnh: 4,52.y/28x+y=2.0,16 x/y=1532. Chn x=15, y=32 Bo to{n nguyn t C:nCO2=4,52.x/28x+y=0,15. nBa(OH)2=0,12. Suy ra to hai mui nCO32=0.09 mBaCO3=17,73 Chn B

C}u 23: Mt hn hp gm 2 ancol X v{ Y no, n chc hoc c mt lin kt i. Bit 16,2g

hn hp l{m mt m{u ho{n to{n 500 g dung dch brom 5,76%. Khi cho 16,2g hn hp trn

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t|c dng vi Na d th th tch H2 ti a l{:

A. 2,016 lt B.. 4,032 lt C. 8,064 lt D. 6,048 lt.

lng kh ln nht===> s mol ancol ln nht===> M ca mi ancol phi nh nht c th. * Bit 16,2g hn hp l{m mt m{u ho{n to{n 500 g dung dch brom 5,76%==> c 1 ancol c ni i M nh nht ==> C3H6O =0,18 mol ===> Kl ancol cn li =5,76 ==> M nh nht CH3OH =0,18 mol ==> V H2 ti a =4,032 l Chn B Cu 23: Hn hp X gm ancol propylic, ancol metylic, etylen glicol v sobitol. Khi cho m gam X tc dng vi Na d th thu c 5,6 lit H2(ktc). Nu t chy m gam X cn va 25,76 lt kh O2( ktc) , sau phn ng thu c 21,6 gam H2O. % Khi lng ca ancol propylic c trong hn hp X l: A.70% B. 45% C.67,5% D.30%

Chia lm 2 nhm

(CO)nH2n+2>OH=x(mol). C3H7OH=ymol ==> x+y=0,5 (1) T }y bin i v

nCO2 = x+3y

\p dng bo to{n nguyn t O ==> 2(x+3y)+1,2=1,15.2+x+y ==> x+5y=1,1 (2)

T (1) v{ (2) ==> x=0,35,y=0,15 ==> m=20g ==> %C3H7OH =45% Chn B

Cu 24: Hn hp X gm axit fomic, axit propionic, axit oxalic v axit axetic. Cho m gam X

phn ng va vi dung dch NaHCO3 thu c 1,68 lt CO2(ktc) v{ 5,63 gam hn hp

mui. t chy hon ton m gam X , ri cho ton b sn phm chy hp th ht vo dung

dch Ba(OH)2 d th thy khi lng phn dung dch gim bt 15,795 gam. % Khi lng

ca axit oxalic c trong X l:

A. 22,61%. B. 33,92%. C. 45,23%. D. 39,575%.

RCOOH+NaHCO3==>RCOONa+CO2+H2O . nCO2=nH2O=NaHCO3=0.075mol,

BTKL==>maxit=3.98g, nCO2=n[OO]trongaxit=0.075mol

nC=x mol, nH=y mol. ==>mC+mH=1.58==>12x+y=1.58(1)

MDDgim=mkt-mH2O-mCO2==>mH2O+mCO2=mkt-mddgiam44x+9y=197x-15.795(2) .

gpt(1),(2)==> x=0.115, y=0.2==>nH2O=0.1mol

noxalic=nH2O-nCO2=0.115-0.1=0.015 mol ==>33.91% Cu 25: t a mol X l{ trieste ca glixerol v{ axit n chc, mch h thu c b mol CO2 v c mol H2O, (bit b-c=4a). Hidro ha m g X cn 6,72 lt kh H2( ktc) thu c 39 g X. Nu un m g X vi dung dch cha 0,7 mol NaOH n phn ng sau c cn dung dch th thu c s g cht rn l{? A 57,2 B 53,2 C 61,48 D 52,6 Vi hp cht cha C,H,O ta c : nX=(nCO2-nH2O)/(k-1) vi k l{ s lin kt pi. \p dng vi b{i ny ta c a=4a/(k-1)==>k=5 ==>trong X c 5 lin kt pi m{ gc -COO- c 3 gc ~ cha 3 lin kt pi==>gc H-C c 2 lin

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kt pi X+0.3mol H2 ==>nX=0.3/2=0.15 mol. m=39-0.3*2=38.4g, X+NaOH==>rn +glixerol ta c nglyxerol=nX=0.15 mol. BTKL==>mran=38.4+0.7*40-0.15*92=52.6

Cu 26: in ph}n vi in cc tr dd cha 0,2 mol AgNO3 vi cng dng in 2,68 A,

trong thi gian t gi thu c dd X (Hiu sut qu| trnh in ph}n l{ 100% ) Cho 16,8 g bt

Fe v{o X thy tho|t ra kh NO (spk duy nht) v{ sau c|c phn ng xy ra ho{n to{n thu

c 22,7g cht rn . Gi| tr ca t l{ A.0,25 B.2 C.1 D.0,5

Khi lng Crn sau phn ng > m Ag. Cht rn sau phn ng c Ag v{ Fe d

nHNO3=xnAgNO3(pu)=x . nAgNO3(du)=0,2x

2AgNO3+H2O2HNO3+12O2+2Ag . Dung dch sau cng l{ Fe(NO3)2 Ta

c 2.nFepu=nAg++nHNO3/4.3 nFe(pu)=((0,2x)+3.x/4)/2=0,10,125x .

mFedu+mAg=22,716,8.56.(0,10,125x)+108(0,2x)=22,7x=0,1t=96500.0,1/2,68=3

600s=1h.

|p |n C

Cu 27: Ha tan ht 10,24 gam Cu bng 200 ml dd HNO3 3M thu c dd A. Thm 400 ml dd NaOH 1M v{o dd A .Lc kt ta , c cn dd ri nung cht rn n khi lng khng i thu c 26,44 gam cht rn.S mol HNO3 ~ phn ng vi Cu l{: A. 0,568 mol B.0,48 mol C.0,4 mol D.0,56 mol

s mol Cu = 0,16 v{ mol HNO3 dng = 0,6 ==> Dd A c: mol Cu(NO3)2 0,16 mol v{ HNO3 d x mol (nu c) TH 1 : nu khng c HNO3 ht : Cu(NO3)2 + 2 NaOH --> Cu(OH)2 + 2 NaNO3 nung ----> NaNO2 0,16------------0,32------------------------0,32--------------------0,32 rn thu c sau khi c cn v{ nung l{ 0,32 mol NaNO2 v{ 0,08 mol NaOH d ==> khi lng rn = 69*0,32 + 40*0,08 = 25,28 ==> loi TH 2 : nu c HNO3 d : HNO3 + NaOH ----> H2O + NaNO3 ---> NaNO2 x------------x-----------------------x-------------x Cu(NO3)2 + 2 NaOH --> Cu(OH)2 + 2 NaNO3 nung ----> NaNO2 0,16------------0,32------------------------0,32--------------------0,32 rn thu c sau khi c cn v{ nung l{ 0,32+x mol NaNO2 v{ (0,08 - x ) mol NaOH d ==> khi lng rn = 69*(0,32+x) + 40*(0,08-x) = 26,44 ==> x = 0,04 ==> mol HNO3 phn ng = 0,6 - 0,04 = 0,56 ==> cu D Cu 28.Ha tan ht 2,72g hh X gm FeS2, FeS, Fe, CuS, Cu trong 500ml dd HNO31M, sau phn ng thu c dd Y v 0,07 mol 1 kh thot ra. Cho Y tc dng vi lng d dung dch BaCl2 thu c 4,66g kt ta. Mt khc, Y c th ha tan ti a m g Cu. Bit trong cc qa trnh trn , sn phm kh duy nht ca N5+ l NO. Gi tr ca m l: A 9,76 B. 5,92 C. 9,12 D, 4,96 nS=nBaSO4=0,02 . ta c 64.nCu+56.nFe=2,720,02.32 v 2.nCu+3nFe=0,07.30,02.6nCu=0,015;nFe=0,02.

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Dung dch sau phn ng ho tan ti a Cu. Nn dung dch sau phn ng gm Fe2+:0,02;Cu2+:x;NO3:y;SO24:0,02 . nHNO3=0,5;nNO=z .(NO: cho ton qu trnh). Ta c 2x=y, 0,02.2+2x+0,02.6=3zv y+z=0,5 . Vy x = 0,1675 ,mCu=(0,16750,015).64=9,76(g) Cu 29. Hn hp X gm 1 anehit v 1 hirocacbon mch h (2 cht hn km nhau 1 nguyn t- cacbon). t ch|y ho{n to{n 0,4 mol hn hp X thu c 1,3 mol CO2 v 0,4 mol H2O. Nu cho 31,8 gam hn hp X t|c dng vi dung dchAgNO3/NH3 th khi lng kt ta thu c ti a l{ A. 209,25 gam. B. 136,80 gam. C. 224,10 gam. D. 216,45 gam.

s nguyn t C trung bnh = 1,3/0,4 = 3,25 ==> C3 a mol v C4 b mol mol hh = a + b = 0,4 mol CO2 = 3a + 4b = 1,3 ==> a = 0,3 v b = 0,1 s nguyn t H trung bnh = 2*0,4/0,4 = 2 ==> mi cht u c 2 nguyn t H ===> C3H2On v C4H2Om Nu n = 0 ==> hydrocarbon mch h CT C3H2 ==> loi Vy hydrocarbon l{ C4H2 0,1 mol c cu to CHC-CCH v andehit C3H2O 0,3 mol CHC-CHO ==> m hh = 54a + 50b = 21,2 CHC-CCH ---> CAgC-CCAg 0,1------------------------0,1 CHC-CHO --> CAgC-COONH4 + 2 Ag 0,3-----------------------0,3---------------0,6 khi lng kt ta = 265*0,1 + 194*0,3 + 108*0,6 = 149,4 Khi lng kt ta thu c khi cho 31,5 gam X t|c dng AgNO3 l :===> 31,5*149,5/21,2 = 224,1 ==> cu C

Cu 30: Hn hp X gm 1 s aminoaxit no mch h. 0,43 mol hn hp X t|c dng va vi 0,55 mol NaOH hay 0,51 mol HCl. t m gam hn hp X cn 98,672 lt O2 (ktc) thu c 80,192 lt CO2 (ktc). Gi| tr ca m l{ A. 99,39 B. 100,14 C. 101,17 D. 99,71

T gi thit ==> CTPT trung bnh (NH2)51/43R(COOH)55/43 thc hin phn ng t ch|y. (NH2)51/43R(COOH)55/43+O2===>CO2+H2O+N2 --------------------------------------- x (mol)---------------------------------- 4,405 mol --- 3,58 mol Bo to{n O ==> nH2O=220x/43+3,3 mol ==> s nguyn t H trong aa = 220/43+3,3/x Ta cng c s nguyn t C= 3,58/x. Mt kh|c aa no ==> lin kt pi l{ s nhm COOH=55/43 \p dng cng thc tnh lin kt k= 2C+NH2=55/43 ==> x=0,86 ==> m= 100,14

MINH SU TM THI NH CHC C\C BN HC TT.TH]N MN!

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