Giai Chi Tiet a Va B 3 Nam 08 09 Va 10

8/3/2019 Giai Chi Tiet a Va B 3 Nam 08 09 Va 10

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GII THI TUYN SINH I HC, CAO NG NM 2009Mn thi : HO, khi B - M : 637

Cu 1 : Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO3)2 0,2M v H2SO4 0,25M. Sau khi ccphn ng xy ra hon ton, thu c 0,6m gam hn hp bt kim loi v V lt kh NO (sn phm kh duy nht, ktc). Gi tr ca m v V ln lt l

A. 17,8 v 4,48. B. 17,8 v 2,24. C. 10,8 v 4,48. D. 10,8 v 2,24.Gii chi tit

nCu(NO3)2 = 0,16 mol, nH2SO4 = 0,2 molSau phn ng thu c hn hp kim loi l Cu v Fe d.Phn ng theo th t u tin : cht kh mnh nht tc dng vi cht oxi ha mnh nht trc ri n cc phn

ng khc.Fe + 3NO

+ 4H+ Fe3+ + NO + 2H2O0,1 0,1 0,4 0,1 0,1 molFe + Cu2+ Fe2+ + Cu0,16 0,16 0,16 0,16 molFe + 2Fe3+ 3Fe2+

0,05 0,1 molGi s mol Fe d l x.

(0,1 + 0,16 + 0,05 + x)56 = m56x + 0,16.64 = 0,6mGii h 2 phng trnh trn ta c m = 17,8Theo phn ng (1) VNO = 0,1.22,4 = 2,24 lt

Cu 4: t chy hon ton mt hp cht hu c X, thu c 0,351 gam H2O v 0,4368 lt kh CO2 ( ktc). BitX c phn ng vi Cu(OH)2 trong mi trng kim khi un nng. Cht X l

A. CH3COCH3. B. O=CH-CH=O. C. CH2=CH-CH2-OH. D. C2H5CHO.Gii chi tit

nH2O = nCO2 = 0,0195 mol X c mt lin kt i

X tc dng c vi Cu(OH)2/OH-, to=> X l anhit n chc.

Cu 7: Cho 61,2 gam hn hp X gm Cu v Fe3O4 tc dng vi dung dch HNO3 long, un nng v khuy u.Sau khi cc phn ng xy ra hon ton, thu c 3,36 lt kh NO (sn phm kh duy nht, ktc), dung dch Yv cn li 2,4 gam kim loi. C cn dung dch Y, thu c m gam mui khan. Gi tr ca m l

A. 151,5. B. 97,5. C. 137,1. D. 108,9.Gii chi titKim loi cn d l Cu v mui st thu c s l mui st (II) (Cu kh Fe3+)3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + 14H2O

x 3x 3

x

mol

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

y y2

3

ymol

Cu + 2Fe(NO3)3 Cu(NO3)2 + 2Fe(NO3)23

2

x3x

3

2

x3x mol

Khi X phn ng l: 61,2 2,4 = 58,8g

232x + 64(y +3

2

x) = 58,8 (1)

nNO =3,36

22,4= 0,15 mol =>

3

x+

2

3

y= 0,15 (2)


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T (1) v (2) => x = 0,15, y = 0,15Mui thu c gm Cu(NO3)2 v Fe(NO3)2 c khi lng l:

(0,15 +3.0,15

2)188 + 3.0,15.180 = 151,5g

Cu 9: Hn hp X gm axit Y n chc v axit Z hai chc (Y, Z c cng s nguyn t cacbon). Chia X thnhhai phn bng nhau. Cho phn mt tc dng ht vi Na, sinh ra 4,48 lt kh H2 ( ktc). t chy hon ton phnhai, sinh ra 26,4 gam CO2. Cng thc cu to thu gn v phn trm v khi lng ca Z trong hn hp X ln ltl

A. HOOC-CH2-COOH v 70,87%. B. HOOC-COOH v 60,00%.C. HOOC-CH2-COOH v 54,88%. D. HOOC-COOH v 42,86%.Gii chi tit

X tc dng vi Na:

RCOOH Na1

2H2

x1

2x mol

R(COOH)2 Na H2y y mol

nH2 =4,48

22,4 = 0,2 mol =>

1

2x + y = 0,2 (1)

t chy X:CnH2nO2 2O nCO2x nx molCnH2n-2O4 2O nCO2y ny mol

nCO2 =26,4

44

= 0,6 mol =>

nx + ny = 0,6 (2)

Bin i (1) v (2) ta c: y =0,4 0,6n

n

Mt khc: 0 < y < 0,2 => 0 y = 0,1 x = 0,2Z l HOOC-COOH

%Z =0,1.90

.1000,1.90 0,2.60+

= 42,86%

Cu 11: Cho dung dch cha 6,03 gam hn hp gm hai mui NaX v NaY (X, Y l hai nguyn t c trong tnhin, hai chu k lin tip thuc nhm VIIA, s hiu nguyn t ZX < ZY) vo dung dch AgNO3 (d), thu c8,61 gam kt ta. Phn trm khi lng ca NaX trong hn hp ban u l

A. 58,2%. B. 52,8%. C. 41,8%. D. 47,2%.Gii chi tit

Gi X l nguyn t trung bnh ca X v YS dng phng php tng gim khi lng.

Na X Ag X

nNa X =8, 61 6, 03

108 23

= 0,03 mol

M(Na X ) =6,03

0,03= 201

23 + X = 201 => X = 178 khng hp l v khng c halogen no c nguyn t khi ln hn 178 (trnguyn t phng x At).


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Chng t c mt halogenua khng to kt ta vi ion bc. Vy X l F cn Y l Cl

nNaCl = nAgCl =8,61

143,5= 0,06 mol

mNaCl = 0,06.58,5 = 3,51 gmNaF = 6,03 3,51 = 2,52g%NaF = 41,8%

Cu 12: Ha tan hon ton 20,88 gam mt oxit st bng dung dch H 2SO4 c, nng thu c dung dch X v

3,248 lt kh SO2 (sn phm kh duy nht, ktc). C cn dung dch X, thu c m gam mui sunfat khan. Gitr ca m lA. 52,2. B. 48,4. C. 54,0. D. 58,0.Gii chi tit

2FexOy + (6x 2y) H2SO4 xFe2(SO4)3 + (3x 2y)SO2 + (3x y)H2O

nSO2 =3,248

22,4= 0,145 mol

nFexOy =2.0,145

3x-2y

(56x + 16y)

2.0,145

3x-2y = 20,88Gii phng trnh trn ta c x = y

nFe2(SO4)3 =.0,145

3x-2y

x

mFe2(SO4)3 =.0,145

3x-2y

x400 = 58g

Cu 15: Cho 0,02 mol amino axit X tc dng va vi 200 ml dung dch HCl 0,1M thu c 3,67 gam muikhan. Mt khc 0,02 mol X tc dng va vi 40 gam dung dch NaOH 4%. Cng thc ca X l

A. (H2N)2C3H5COOH. B. H2NC2C2H3(COOH)2.C. H2NC3H6COOH. D. H2NC3H5(COOH)2.

Gii chi titnX = nHCl X c mt nhm aminnNaOH = 0,04, gp hai ln X X c 2 nhm COOHt cng thc ca X l: RNH2(COOH)2

MX =3,67 0,02.36,5

0,02

= 147

R + 16 + 90 = 147R = 41

=> R l C3H5.Vy X l C3H5NH2(COOH)2

Cu 16: Cho 2,24 gam bt st vo 200 ml dung dch cha hn hp gm AgNO3 0,1M v Cu(NO3)2 0,5M. Saukhi cc phn ng xy ra hon ton, thu c dung dch X v m gam cht rn Y. Gi tr ca m l

A. 2,80. B. 4,08. C. 2,16. D. 0,64.Gii chi tit

nFe = 0,04 mol, nAg+ = 0,02 mol, nCu2+ = 0,1 molCht kh mnh nht s tc dng vi cht oxi ha mnh nht trc:

Fe + 2Ag+ Fe2+ + 2Ag0,01 0,02 0,02molFe + Cu2+ Fe2+ + Cu0,03 0,03 0,03 molm = 0,02.108 + 0,03.64 = 4,08g


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Cu 17: Cho X l hp cht thm; a mol X phn ng va ht vi a lt dung dch NaOH 1M. Mt khc nu cho amol X phn ng vi Na (d) th sau phn ng thu c 22,4a lt kh H 2 ( ktc). Cng thc cu to thu gn caX l

A. CH3-C6H3(OH)2. B. HO-C6H4-COOCH3.C. HO-CH2-C6H4-OH. D. HO-C6H4-COOH.Gii chi titnX = nNaOH X c 1 nhm COOH hoc 1 nhm COO, hoc 1 nhm OH (phenol).X tc dng vi Na cho ra s mol H 2 bng s mol X. Suy ra X c 2 nhm chc tc dng c vi Na trong c mt nhm khng tc dng vi NaOH. Vy nhm ny l OH ca ancol, nhm cn li c th l chc axithoc phenol. X phi l HO-CH2-C6H4-OH

Cu 18: Hn hp kh X gm H2 v mt anken c kh nng cng HBr cho sn phm hu c duy nht. T khi caX so vi H2 bng 9,1. un nng X c xc tc Ni, sau khi phn ng xy ra hon ton, thu c hn hp kh Ykhng lm mt mu nc brom; t khi ca Y so vi H2bng 13. Cng thc cu to ca anken l

A. CH3-CH=CH-CH3. B. CH2=CH-CH2-CH3. C. CH2=C(CH3)2. D. CH2=CH2.Gii chi tit

t cng thc ca anken l CnH2n v s mol ca n l x ; s mol ca H2 l y.MX l: 14nx + 2y = 18,2(x+y) (1)

MY l: (14n + 2)x + 2(y x) = 26(x + y x ) (2)(1) 14nx = 18,2x + 16,2y (3)(2) 14nx = 24y (4)(4) (3) 7,8y = 18,2x

y =7

3x

Thay y vo (4) c n = 4Anken cng HBr ch cho 1 sn phm duy nht nn anken phi l CH3-CH=CH-CH3

Cu 25: Nung nng m gam hn hp gm Al v Fe3O4 trong iu kin khng c khng kh. Sau khi phn ng xyra hon ton, thu c hn hp rn X. Cho X tc dng vi dung dch NaOH (d) thu c dung dch Y, cht rnZ v 3,36 lt kh H2 ( ktc). Sc kh CO2 (d) vo dung dch Y, thu c 39 gam kt ta. Gi tr ca m l

A. 48,3 B. 57,0 C. 45,6 D. 36,7Gii chi titX tc dng vi NaOH to H2 chng t Al d, Fe3O4 ht.nH2 = 0,15 molnAld = 0,1 molKt ta thu c l Al(OH)3, nA(OH)3 = 0,5 molnAl phn ng nhit nhm: 0,5 0,1 = 0,4 molnFe3O4 = 0,15 molm = 0,5.27 + 0,15.232 = 48,3g

Cu 26: Cho hn hp X gm CH4, C2H4 v C2H2. Ly 8,6 gam X tc dng ht vi dung dch brom (d) th khilng brom phn ng l 48 gam. Mt khc, nu cho 13,44 lt ( ktc) hn hp kh X tc dng vi lng d dungdch AgNO3 trong NH3 , thu c 36 gam kt ta. Phn trm th tch ca CH4 c trong X l

A. 40% B. 20% C. 25% D. 50%Gii chi tit

Kh ca bi ton ny l cc d kin khng ng nht.Gi s mol ca cc cht trong X khi tc dng vi AgNO3/NH3 ln lt l x, y, zx + y + z = 0,6 (1)nC2H2 = nC2Ag2 = 0,15 mol z = 0,15

x + y = 0,45 (2)Gi k l h s t l ca X trong trng hp tc dng vi brom so vi X tc dng vi AgNO3/NH3(16x + 28y + 26z)k = 8,6 (3)nBr2 = 0,3(y + 2z)k = 0,3 (4)(3) (4).13 (16x + 15y)k 4,7 (5)


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Kt hp (5) v (2) bin i ta c :4,7

7,2yk

= (6)

Mt khc : thay z vo (4) v bin i c :0,3

0,3yk

= (7)

T (6) v (7) tm c k =5

7,5

Thay k v y vo (2) c x = 0,3

%CH4 =0,3

1000,6

= 50%

Cu 27: Cho cht xc tc MnO2 vo 100 ml dung dch H2O2, sau 60 giy thu c 3,36 ml kh O 2 ( ktc) . Tc trung bnh ca phn ng (tnh theo H2O2) trong 60 giy trn l

A. 2,5.10-4 mol/(l.s) B. 5,0.10-4 mol/(l.s) C. 1,0.10-3 mol/(l.s) D. 5,0.10-5 mol/(l.s)Gii chi tit

1 2 1 2

.

C C n nv

t V t

= =

nO2 = 1,5.10-3 nH2O2 = 3.10-3

33.10

0,1.60v

= = 5.10-4 mol/(l.s)

Cu 28: Trn 100 ml dung dch hn hp gm H2SO4 0,05M v HCl 0,1M vi 100 ml dung dch hn hp gmNaOH 0,2M v Ba(OH)2 0,1M thu c dung dch X. Dung dch X c pH l

A. 1,2 B. 1,0 C. 12,8 D. 13,0Gii chi titnH+ = 0,1(2.0,05 + 0,1) = 0,02 mol

nOH-

= 0,1(0,2 + 2.0,1) = 0,04 mol nOH d = 0,02 mol [OH-] = 0,1 = 10-1

[H+] = 10-13 pH = 13

Cu 29: in phn c mng ngn 500 ml dung dch cha hn hp gm CuCl2 0,1M v NaCl 0,5M (in cc tr,hiu sut in phn 100%) vi cng dng in 5A trong 3860 giy. Dung dch thu c sau in phn ckh nng ho tan m gam Al. Gi tr ln nht ca m l

A. 4,05 B. 2,70 C. 1,35 D. 5,40Gii chi tit

Th t u tin in phn: in phn CuCl2 ht ri mi in NaClnCuCl2 = 0,05 mol

Cng thc tnh khi lng cht in phn:96500.

AItm

n=

S mol cht in phn:96500.

It

n

t =.96500.mol n

I

Thi gian in phn CuCl2: t =0,05.96500.2

5= 1930 giy

Thi gian in phn NaCl: 3860 1930 = 1930 giynNaOH = nH =

5.1930

96500= 0,1 mol

nAl = 0,1 mol mAl = 2,7g


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Cu 30: Cho hn hp X gm hai hp cht hu c no, n chc tc dng va vi 100 ml dung dch KOH0,4M, thu c mt mui v 336 ml hi mt ancol ( ktc). Nu t chy hon ton lng hn hp X trn, sau hp th ht sn phm chy vo bnh ng dung dch Ca(OH)2 (d) th khi lng bnh tng 6,82 gam. Cngthc ca hai hp cht hu c trong X l

A. HCOOH v HCOOC2H5 B. CH3COOH v CH3COOC2H5C. C2H5COOH v C2H5COOCH3 D. HCOOH v HCOOC3H7Gii chi tit

nKOH = 0,04 molnROH = 0,015 mol 0,04

Chng t X gm 1 axit no n chc v mt este no n chc c gc axit ca axit t do:CnH2n +1COOH v C(n + m + 1) H2(n + m+ 1)O2n.este = nROH = 0,015 moln.ax = 0,04 0,015 = 0,025 molt chy X ta thu c nc v CO2 theo phng trnh:18[(n + m +1)0,015 + (n + 1)0,025] + 44[(n + m +1)0,015 + (n + 1)0,025] = 6,82 (18 + 44)[ (n + m +1)0,015 + (n + 1)0,025] = 6,82 8n + 3m = 14 n v m ch c th l n = 1, m = 2 CH3COOH v CH3COOC2H5

Cu 32: in phn nng chy Al2O3 vi anot than ch (hiu sut in phn 100%) thu c m kg Al catot v67,2 m3 ( ktc) hn hp kh X c t khi so vi hiro bng 16. Ly 2,24 lt ( ktc) hn hp kh X sc vo dungdch nc vi trong (d) thu c 2 gam kt ta. Gi tr ca m l

A. 54,0 B. 75,6 C. 67,5 D. 108,0Gii chi titin phn Al2O3 c oxi. Oxi t in cc than ch thu c cc kh CO 2, CO v O2 d. Gi s mol ca cc

kh ny trong 2,24 lt X ln lt l x, y, z.nX = 0,01nCO2 = nCaCO3 = 0,02 mol x = 0,02

y + z = 0,08 (1)Da vo t khi kh X ta c : 44.0,02 + 28y + 32z = 16.2.0,1 (2)T (1) v (2) gii ra c : y = 0,06, z = 0,02Ta c s mol O2 trong Al2O3 bng s mol O2 trong thnh phn cc cht trong XnO2 trong 2,24 lt X bng 0,02 + 0,02 + 0,03 = 0,07 molnO2 trong 67,2 m3 X bng 0,07.30.103

=> nAl =0,28

3.30.103.27 = 75,6.103g = 75,6 kg

Cu 34: Hn hp X gm hai este no, n chc, mch h. t chy hon ton mt lng X cn dng va 3,976 lt kh O2 ( ktc), thu c 6,38 gam CO2. Mt khc, X tc dng vi dung dch NaOH, thu c mt mui

v hai ancol l ng ng k tip. Cng thc phn t ca hai este trong X lA. C2H4O2 v C3H6O2 B. C3H4O2 v C4H6O2C. C3H6O2 v C4H8O2 D. C2H4O2 v C5H10O2Gii chi tit

t cng thc trung bnh ca 2 este l C n H2 n O2

C n H2 n O2 +3 2

2

n O2 n CO2 + n H2O

0,1775 0,145 mol

3 2

2

n 0,145 = 0,1775 n

Gii ra c n = 3,6Hai este l C3H6O2 v C4H8O2


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Cu 35: Hai hp cht hu c X v Y l ng ng k tip, u tc dng vi Na v c phn ng trng bc. Bitphn trm khi lng oxi trong X, Y ln lt l 53,33% v 43,24%. Cng thc cu to ca X v Y tng ng l

A. HO-CH2-CH2-CHO v HO-CH2-CH2-CH2-CHOB. HO-CH(CH3)-CHO v HOOC-CH2-CHOC. HO-CH2-CHO v HO-CH2-CH2-CHOD. HCOOCH3 v HCOOCH2-CH3Gii chi tit

t cng thc ca X v Y l: ROx v RCH2Ox. (R l thnh phn cn li ca phn t)

Theo bi ta c:16x

16xR + 100 = 53,33 (1)16x

16x+14R +100 = 43,24 (2)

Gii h phng trnh (1) v (2) ta c x = 2, R = 28 (C2H4)Vy X, Y phi l : HO-CH2-CHO v HO-CH2-CH2-CHO

Cu 36: Hp cht hu c X tc dng c vi dung dch NaOH un nng v vi dung dch AgNO 3 trong NH3.Th tch ca 3,7 gam hi cht X bng th tch ca 1,6 gam kh O2 (cng iu kin v nhit v p sut). Khi tchy hon ton 1 gam X th th tch kh CO2 thu c vt qu 0,7 lt ( ktc). Cng thc cu to ca X l

A. CH3COOCH3 B. O=CH-CH2-CH2OH C. HOOC-CHO D. HCOOC2H5

Gii chi titX tc dng vi NaOH un nng vy X c chc este tc l trong phn t X phi c t nht 2 nguyn t oxi.

t cng thc ca X l CnRO2 (R l thnh phn cn li)nX ca 3,7g = nO2 = 0,05 molMX = 74CnRO2 nCO2

1

74

1

74n

1

74n >

0,7

22,4

n > 2,3 n 3

=> 3 n n = 3R = 74 32 36 = 6=> X l C3H6O2X tc dng c vi AgNO3/NH3 suy ra X l este ca axit HCOOH.X l HCOOC2H5

Cu 38: Ho tan hon ton 2,9 gam hn hp gm kim loi M v oxit ca n vo nc, thu c 500 ml dungdch cha mt cht tan c nng 0,04M v 0,224 lt kh H2 ( ktc). Kim loi M lA. Ca B. Ba C. K D. NaGii chi tit

M v oxit ca n M2On u tc dng vi nc to dd cha mt cht tan. Chng t M v oxit ca n u tcdng vi H2O.

nM(OH)n = 0,02

nM =2

nH2 =

0,02

n

M + nH2O M(OH)n +2

nH2

0,02n

0,02

nmol

M2On + nH2O 2M(OH)n

(0,01 -0,01

n) (0,02 -

0,02

n) mol


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0,02

nM + (0,01 -

0,01

n)(2M + 16n) = 2,9

0,16n + 0,02M = 3,06 8n + M = 153 n = 2, M = 137 M l Ba

Cu 42:Nhng mt thanh st nng 100 gram vo 100 ml dung dch hn hp gm Cu(NO 3)2 0,2M v AgNO3

0,2M. Sau mt thi gian ly thanh kim loi ra, ra sch lm kh cn c 101,72 gam (gi thit cc kim loi tothnh u bm ht vo thanh st). Khi lng st phn ng lA. 2,16 gam B. 0,84 gam C. 1,72 gam D. 1,40 gamGii chi tit

nFe = 0,5 mol, nCu(NO3)2 = 0,02 mol, nAgNO3 = 0,02 molNh vy Fe dNu Ag+, Cu2+ phn ng ht th thanh st s c khi lng l:100 0,03.56 + 0,02.108 + 0,02.64 = 101,76101,76 > 101,72

Nu Ag+ phn ng ht v Cu2+ cha phn ng th thanh st s c khi lng l:100 0,01.56 + 108.0,02 = 101,6

101,6 < 101,72Chng t Ag+ phn ng ht v Cu2+ phn ng mt phnGi s mol Cu2+ phn ng l x.100 (0,01 + x)56 + 0,02.108 + 64x = 101,72=> x = 0,015Khi lng Fe phn ng : (0,01 + 0,015)56 = 1,4g

Cu 43: Hir ho hon ton m gam hn hp X gm hai anehit no, n chc, mch h, k tip nhau trong dyng ng thu c (m + 1) gam hn hp hai ancol. Mt khc, khi t chy hon ton cng m gam X th cn va 17,92 lt kh O2 ( ktc). Gi tr ca m l

A. 10,5 B. 17,8 C. 8,8 D. 24,8Gii chi titt cng thc trung bnh ca 2 anhit l : C n H2 n O

C n H2 n O + H2 C n H n +2O2m (m + 1) g

nX = nH2 =1

2= 0,5 mol

nO2 = 0,8

C n H2 n O +3 1

2

n O2 n CO2 + n H2O

0,5 0,8

3 1

2

n 0,5 = 0,8

n = 1,4

m = (14 n + 16)0,5 = (14.1,4 + 16)0,5 = 17,8g


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Cu 46: Cho hirocacbon X phn ng vi brom (trong dung dch) theo t l mol 1 : 1, thu c cht hu c Y(cha 74,08% Br v khi lng). Khi X phn ng vi HBr th thu c hai sn phm hu c khc nhau. Tn gica X l

A. but-1-en B. but-2-en C. propilen D. XiclopropanGii chi tit

Nu X tc dng vi brom l phn ng th (cc hp cht phenol phn ng th vi dd brom) :RH + Br2 RBr + HBr

8010080R + = 74,08

R = 28 R l C2H4iu ny khng hp lVy tc dng vi dd brom l phn ng cng :R + Br2 RBr2

160100

160R += 74,08

R = 56R l CxHy12x + y = 56 x = 4, y = 8 X l C4H8X cng HBr cho 2 sn phm khc nhau. Vy X khng phi but-2-en

Cu 48: Este X (c khi lng phn t bng 103 vC) c iu ch t mt ancol n chc (c t khi hi sovi oxi ln hn 1) v mt amino axit. Cho 25,75 gam X phn ng ht vi 300 ml dung dch NaOH 1M, thu cdung dch Y. C cn Y thu c m gam cht rn. Gi tr m l

A. 29,75 B. 27,75 C. 26,25 D. 24,25Gii chi tit

nX = 0,25 mol, NaOH = 0,3 mol X phn ng ht, X l este n chc, NaOH d 0,05 molt cng thc ca X l: (H2N)xRCOOR16x + 44 + R + R = 10316x + R + R = 59R > 15 16x + R < 43 x = 1 R + R = 43 R > 15 suy ra R = 29 R = 14R l CH2, R l C2H5Y gm 0,25 mol H2NCH2COONa v 0,05 mol NaOH dm = 0,25.97 + 0,05.40 = 26,25g

Cu 49: Ho tan hon ton 24,4 gam hn hp gm FeCl2 v NaCl (c t l s mol tng ng l 1 : 2) vo mtlng nc (d), thu c dung dch X. Cho dung dch AgNO3 (d) vo dung dch X, sau khi phn ng xy rahon ton sinh ra m gam cht rn. Gi tr ca m l

A. 68,2 B. 28,7 C. 10,8 D. 57,4Gii chi tit

im mu cht ca bi ton ny l y khng ch c phn ng trao i m cn c phn ng Ag + oxi ha

Fe2+

. Vy cht rn thu c gm AgCl v AgGi s mol ca FeCl2 l x th s mol ca NaCl l 2x.127x + 58,5.2x = 24,4 x = 0,1nAgCl = nNaCl + 2nFeCl2 = 0,2 + 0,2 = 0,4 molnAg = nFeCl2 = 0,1 mol


10/33

m = 0,4.143,5 + 0,1.108 = 68,2g

Cu 50: t chy hon ton 1 mol hp cht hu c X, thu c 4 mol CO 2. Cht X tc dng c vi Na, thamgia phn ng trng bc v phn ng cng Br2 theo t l mol 1 : 1. Cng thc cu to ca X l

A. HOOC-CH=CH-COOH B. HO-CH2-CH2-CH=CH-CHOC. HO-CH2-CH2-CH2-CHO D. HO-CH2-CH=CH-CHOGii chi tit

t chy 1 mol X cho 4 mol CO2 => X c 4 nguyn t C trong phn t.X tham gia phn ng trng bc v cng Br2 theo t l 1 : 1 => X c 1 lin kt i, c nhm CHOX tc dng vi Na => X c OH hoc COOH=> X l HO-CH2-CH=CH-CHO

Cu 52: Cho cc th in cc chun :0 0 0

3 2 2Al / Al Zn / Zn Pb / Pb

E 1,66V;E 0,76V;E 0,13V+ + +

= = = ;0

2Cu / Cu

E 0,34V+

= + . Trong cc pin sau y, pin no c sut in ng chun ln nht?

A. Pin Zn - Pb B. Pin Pb - Cu C. Pin Al - Zn D. Pin Zn CuGii chi tit

Sut in ng chun ca pin: Eopin = Eo(+) Eo(-)Trong cc pin cho th pin Zn Cu c sut in ng chun ln nht.

Cu 54: Cho dung dch X cha hn hp gm CH3COOH 0,1M v CH3COONa 0,1M. Bit 250C Ka caCH3COOH l 1,75.10-5 v b qua s phn li ca nc. Gi tr pH ca dung dch X 25o l

A. 1,00 B. 4,24 C. 2,88 D. 4,76Gii chi tit

CH3COONa CH3COO- + Na+

0,1 0,1CH3COOH CH3COO

- + H+

x x xCng c 2 qu trnh trn ta c nng ca CH3COO- l 0,1 + x

Nng ca CH3COOH cn li l 0,1 - x(0,1 )0,1

x x

x

+

= 1,75.10-5 (*)

V ka nh nn x rt nh so vi 0,1

(*) c vit li l0,1

0,1

x= 1,75.10-5

x = 1,75.10-5

pH = 5 log1,75 = 4,76Cu 55: Khi ho tan hon ton 0,02 mol Au bng nc cng toan th s mol HCl phn ng v s mol NO (sn

phm kh duy nht) to thnh ln lt l

A. 0,03 v 0,01 B. 0,06 v 0,02 C. 0,03 v 0,02 D. 0,06 v 0,01Gii chi titAu + 3HCl + HNO3 AuCl3 + NO + 2H2O

0,02 0,06 0,02

Cu 56: Cho 0,04 mol mt hn hp X gm CH2=CH-COOH, CH3COOH v CH2=CH-CHO phn ng va vi dung dch cha 6,4 gam brom. Mt khc, trung ho 0,04 mol X cn dng va 40 ml dung dch NaOH0,75 M . Khi lng ca CH2=CH-COOH trong X l

A. 1,44 gam B. 2,88 gam C. 0,72 gam D. 0,56 gamGii chi tit

nX = 0,4 nBr 2 = 0,04

Ngoi phn ng cng brom ca axit khng no v anhit khng no cn phn ng oxi ha ca anhit.Gi s mol ca cc cht trong X ln lt l x, y, z.x + y + z = 0,4x + 2z = 0,4

nNaOH = 0,03 => S mol ca 2 axit l 0,03 => S mol ca anhit l 0,04 0,03 = 0,01 mol z = 0,01 =>x = 0,02


11/33

mCH2=CH-COOH = 0,02.72 = 1,44g

Cu 57: Ngi ta iu ch anilin bng s sau+ +

HNO ac Fe HCl3

0H SO ac2 4 tBenzen Nitrobenzen Anilin

Bit hiu sut giai on to thnh nitrobenzen t 60% v hiu sut giai on to thnh anilin t 50%. Khilng anilin thu c khi iu ch t 156 gam benzen l

A. 186,0 gam B. 111,6 gam C. 55,8 gam D. 93,0 gamGii chi tit

nC6H6 = 2 mol

nC6H5NH2 =60 50

. 2 0,6100 100

= mol

mC6H5NH2 = 0,6.93 = 55,8g

Cu 59: Hn hp X gm hai ancol no, n chc, mch h, k tip nhau trong dy ng ng. Oxi ho hon ton0,2 mol hn hp X c khi lng m gam bng CuO nhit thch hp, thu c hn hp sn phm hu c Y.Cho Y tc dng vi mt lng d dung dch AgNO3 trong NH3, thu c 54 gam Ag. Gi tr ca m l

A. 15,3 B. 8,5 C. 8,1 D. 13,5Gii chi tit

nX = 0,2 => nY = 0,2nAg = 0,5 mol Trong Y c mt anhit l HCHO Anhit cn li l CH3CHO 2 ancol l CH3OH v CH3CH2OHGi s mol ca 2 ancol ln lt l x v y.x + y = 0,24x + 2y = 0,5x = 0,05, y = 0,15 => m = 0,05.32 + 0,15.46 = 8,5g

Cu 60: Ho tan hon ton 1,23 gam hn hp X gm Cu v Al vo dung dch HNO3 c, nng thu c 1,344 lt

kh NO2 (sn phm kh duy nht, ktc) v dung dch Y. Sc t t kh NH 3 (d) vo dung dch Y, sau khi phnng xy ra hon ton thu c m gam kt ta. Phn trm v khi lng ca Cu trong hn hp X v gi tr ca mln lt l

A. 21,95% v 0,78 B. 78,05% v 0,78 C. 78,05% v 2,25 D. 21,95% v 2,25Gii chi titnNO2 = 0,06 molS mol e nhn = 0,06`Gi s mol ca Cu v Al l x v yS mol e m kim loi nhng bng s mol e nhn nn ta c :

2x + 3y = 0,06Mt khc : 64x + 27y = 1,23x = 0,015, y = 0,01=> %Cu = 78,05%Kt ta ch gm Al(OH)3 : m = 0,01.78 = 0,78%

Gii chi tit ho khi B nm 2009M : 475Cu Ni dung p n1 t a, b, c ln lt l s mol CH4, C2H4, C2H2 16a+28b+26c=8,6(1)

C2H2+2Br2C2H2Br4 C2H4+Br2C2H4Br2c 2c b b b+2c=0,3(2)

C2H2+Ag2OC2Ag2ck ck=0,15(3) v k(a+b+c)=0,6(4)Kt hp 1,2,3,4 ta c a=0,2; b=0,1; c=0,1 % th tch =% s mol =50%CH4

B

2 S i peptit l: ala-ala; gly-gly, ala-gly; gly-ala D3 Ch cho NH3 vo dung dch AlCl3 mi to kt ta sau phn ng

3NH3+3H2O+AlCl3 Al(OH)3+3NH4Cl


12/33

4 Cc phn ng trong HCl th hin tnh kha. 4HCl-1+PbO2PbCl2+Cl20+2H2Oc.2HCl-1+2HNO32NO2+Cl20+H2O Clo tng s oxi ho t -1 ln 0

C

5 X: RCOOH amol Y: R(COOH)2 b mol a/2+b=0,2(1)Gi n l s cc bon ta c n(a+b)=s mol CO2=0,6 Ta nhn thy a+b>a/2+b=0,2 nnn


13/33

X tc dng vi Na s mol H2=s mol X X c thm 1 nhm OH ancolX l HO-CH2-C6H4-OH

20 Trt t nhit si tng dn CH3CHO, C2H5OH, HCOOH, CH3COOH D21 KMnO4 v NaNO3 ln lt l X v Y B22 B B23 1, 3, 4, 6 B24 AlAl3+, FeFe3+ v dung dch gm Al3+, Fe3+, SO42- tc dng vi Ba(OH)2 d

Al3++4OH-[Al(OH)4]- Fe3++3OH-Fe(OH)3 v Ba2++SO42-BaSO4 khi nhit phnkt ta cht rn thu c l BaSO4 v Fe2O3

B

25 CnH2nO2+(3n-2)/2O2nCO2+nH2O0,1775 0,145 suy ra n=3,625 nn ta chn A

A

26 X+NaOH nX/nNaOH=0,02/0,04 nn X c 2 nhm COOHX +HCl nX=nHCl X c 1 nhm NH2H2HR(COOH)2+HClClH3NR(COOH)20,02 0,02 suy ra R=41(C3H5)

A

27 Khi in phn ht Cu2+ ng vi thi gian = (0,05.2.96500)/5=1930(s) thi gian cnli 1930s in phn NaClnH2 thot ra = 0,05(mol)2NaCl+2H2O2NaOH+H2+Cl2

0,25 0,1 0,052Al+2NaOH+2H2O2NaAlO2+3H20,1 0,1 khi lng Al=27.0,1=2,7(g)

C

28 s mol H+=0,4 s mol NO3-=0,32 s mol Cu2+=0,16Fe+4H++NO3-Fe3++NO+2H2O0,1 0,4 0,32 0,1 0,1 VNO=2,24(lt)Fe+2Fe3+=3Fe2+

0,05 0,1 0,15Fe+Cu2+=Fe2++Cu0,16 0,16 0,16m-(0,1+0,05+0,16).56+64.0,16=0,6m suy ra m=17,8(g)

D

29 nAl d=2/3nH2=0,1(mol)AlAl(OH)30,5 39/78 suy ra s mol Alp=0,48Al+3Fe3O44Al2O3+9Fe0,4 0,15m=0,5.27+0,15.232=48,3(g)

A

30 Xt trong 1 mol X (H2 v CnH2n) mX=9,1.2=18,2(g)=mYnY=9,1:14=0,7(mol) suy ra s mol H2 p=s mol anken=0,3(mol) s mol H2 banu=0,3+0,4=0,7(mol)Ta c 0,7.2+0,3.14n=18,2 n=4 (X +HBr cho 1 sn phm duy nht X l But-2-en

A

31 B(1,1,2,2-tetra floeten, propilen, stiren, vinyl clorua) B32 NaXAgX

x x x=(8,61-6,03)/(108-23)=0,03(mol)(Phng php tng gim khilng) X=178(khng tho mn) X, Y ln lt l F v Cl v kt ta l ca AgClnAgCl=8,61:143,5=0,06(mol) mNaCl=58,5.0,06=3,51(g) %m=58,2% suy ra%NaF=41,8%

D

33 V=(33,6.10-3):(60:0,1.22,4)=2,5.10-4 C34 KClO3 C35 Pht biu C ng C36 MX=3,7: (1,6:32)=74 s mol X=1:7a

CxHyOzxCO2 theo bi ra x>2 v trng c gng, tc dng c vi NaOH X leste ca axit fomic

A

37 X chc chn c CO2CO2+CaCO3CaCO3+H2O0,02 0,02 67,2m3=30000 .2,24s mol CO2 trong X=600(mol) Mx=32 nn X c O2 a mol, CO b mol, CO2 600mol

C


14/33

(32.a+28b+44.600)/(a+b+600)=32(1)a+b+600=3000(2) gii 1 v 2 ta c a=600; b=18002C+O22CO C+O2CO2 Tng s mol O2=600+900+600=2100(mol)

900 1800 600 6002Al2O34Al+3O2

2800 2100 mAl=2,8.27=75,6(kg)38 Theo bi ra X gm mt axit no n chc v mt este no n chc cng c to ra

t axit trnRCOOH (A)v RCOOR(B)nB=nancol=0,015(mol) suy ra nA=0,014-0,015=0,025(mol)CnH2n+1COOH(n+1)CO2+(n+1)H2O0,015 0,015(n+1) 0,015(n+1)CnH2n+1COOCmH2m+1(n+m+1)CO2+(n+m+1)H2O0,025 0,025(n+m+1) 0,025(n+m+1)Khi lng bnh tng =khi lng nc v CO244[0,015(n+1)+0,025(n+m+1)]+18[0,015(n+1)+0,025(n+m+1)]=6,82suy ra n=1, m=2

A

39 A.03 B

40 A.1,2,3,6 A41 B. iu ch oxi B42 a[127+58,5.2]=24,4 a=0,1 Tng s mol Cl-=0,4(mol)

Cl-AgCl Fe2++Ag+Fe3++Ag0,4 0,4 0,1 0,1m=143,5.0,4+108.0,1=68,2(g)

C

43 D.Amilopectin c cu trc mch nhnh D44 s mol KOH=0,15(mol)

s mol H3PO4=0,1(mol)t s nKOH/naxit=1,5 nn xy ra 2 phn ngKOH+H3PO4KH2PO4+H2O

2KOH+H3PO4K2HPO4+2H2O

B

45 S mol H2=s mol anehit=1:2=0,5(mol)CnH2nO+(3n-1)/2O2nCO2+nH2O0,5 0,8 suy ra n=1,4 m=(14.1,4+16).0,5=17,8(g)

D

46 RCOORm RH>32 nn cng thc ca este l H2NCH2COOC2H5 duy nhtH2NCH2COOC2H5+NaOHH2NCH2COONa+C2H5OH0,25 0,3 0,25 0,25mmui=25,75+0,3.40-46.0,25=26,25(g)

A

47 Cr(OH)3+KOHKCrO2(X)+H2O2KCrO2+3Cl2+8KOH2K2CrO4+6KCl+4H2OK2CrO4+2H+Cr2O72-+2OH-

..

D

48 Hirocacbon tho mn l but-1-en D49 t chy 1 mol X thu c 4 mol CO2 chng t X c 4 cacbon loi A v D

tc dng c vi Na c OH hoc COOHCng Br2 theo t l 1:1 c 1 ni pi nhnhtrng bc c CHO X l HO-CH2-CH=CH-CHO

C

50 Fe+2AgNO3Fe(NO3)2 +2Agx 2x 2xFe+Cu(NO3)2Fe(NO3)2+Cu

y y y2x=0,02 suy ra x=0,01 v 108.2x+64y-56(x+y)=1,72 suy ra y=0,015Khi lng Fe phn ng =0,025.56=1,4(g)

51 CH3C OOHCH3COO-+H+

0,1 0,1 0x x x

D


15/33

0,1-x x+0,1 xx(x+0,1)/(0,1-x)=1,75.10-5 x nh hn nhiu 0,1 ta cx2+0,1x-1,75.10-6=0 x=1,75.10-5 pH=4,76=-log [x ]

52 Benzennitrobenzen78 123156 x do hi u su t l 60%x=(156.123.60): (78.100)=147,6(g)

N itro benzenAnilin123 93246 y do hi u su t 50%

y=(147,6.93.0,5): (123.100)=55,8(g)

B

53 64a+27b=1,23(1)CuCu2++2e AlAl3++3e N+5+1eNO2a a 2a b b 3b 0,06 0,06 2a+3b=0,06(2)T 1 v 2 suy ra a=0,015; b=0,01Kt ta l Al(OH)3=0,01 mol v Cu2+ to phcM=78.0,01=0,78 gam %Cu=(64.0,015.100): (1,23)=78,05%

C

54 Au+3HCl+HNO3AuCl3+NO+2H2O0,02 0,06 0,02

B

55 Phn bn lm tng chua l NH4NO3NH4H++NH3H+ lm tng chua ca t

C

56 B.CH3-CH(MgBr)-CH2-CH3 B57 Khi glucoz dng mch vng tt c cc nhm OH u phn ng vi CH3OH l sai C58 Hai ancol no n chc s to ra anehit n chc

nAg/nancol>2 chng t 2 ancol l CH3OH v C2H5OHCH3OHHCHO4Aga 4aC2H5OHCH3CHO2Ag

b 2b

a+b=0,2 v 4a+2b=0,5 a=0,05; b=0,15.m=32.0,05+46.0,15=8,5(g)

C

59 CH2=CH-COOH amol, CH3COOH b mol, CH2=CH-CHO c mola+b+c=0,04 (1)CH2=CH-COOH+Br2CH2Br-CHBr-COOHa aCH2=CH-CHO+2Br2+H2OCH2Br-CHBr-COOH+2HBrc 2ca+2c=0,04(2)CH2=CH-COOH+NaOHa aCH3COOH+NaOH

b b suy ra a+b=0,03(3) kt hp 1,2,3 ta ca=0,02; b=0,01; c=0,01Khi lng ca CH2=CH-COOH=0,01.144=1,44(g)

60 A. pin Zn-Cu c sut in ng chun ln nht A

Gii chi tit thi khi A mn Ho nm hc 2007-2008

( M 794)

Cu 1: Cho cc phn ng:

(1) Cu(NO3)2 (2) NH4NO2

(3) NH3 + O2 (4) NH3 + Cl2

t0 t0

T8500C,Pt t0

t0 t0


16/33

(5) NH4Cl (6) NH3 + CuO

Cc phn ng u to kh N2 l:

A. (2), (4), (6). B. (1), (2), (5). C. (1), (3), (4). D. (3), (5),

(6).Gii : Cc phn ng:

(1) 2Cu(NO3)2ot 2CuO + 4NO2 + O2 (2) NH4NO2

ot N2 + 2H2O

(3) 4NH3 + 5O2oPt , 850 C 4NO + 6H2O (4) 2NH3 + 3Cl2

ot N2 + 6HCl

(5) NH4Clot NH3 + HCl (6) 2NH3 + 3CuO

ot 3Cu + N2 + 3H2O

Vy p n A l p ng.

Cu 2: Cho cc cht : Al, Al2O3, Al2(SO4)3, Zn(OH)2, NaHS, , (NH4)2CO3 . S

cht u phn ng c vi dung dch HCl, dung dch NaOH l:

A. 7. B. 6. C. 4. D. 5.Gii: Cc cht ln lt l: Al, Al2O3, Zn(OH)2, NaHS, (NH4)2CO3.

Vy p n D l p ng.

Cu 3: Hp th hon ton 4,48 lt kh CO2 ( ktc) vo 500 ml dung

dch hn hp gm NaOH 0,1 M v Ba(OH)2 0,2 M , sinh ra m gam kt ta.

Gi tr ca M l:

A. 9,85. B. 11,82. C.17,73. D.

19,70

Gii: nCO2 = 0,2 ; nOH- = 0,25 => ta c :2nCO

nOH > 1 => P to ra 2 mui

CO2 + 2OH- CO2

3

+ H2O (1)

B: 0,2 0,25

P: 0,125 0,25 0,125

sau P: 0,075 0 0,125

CO2 + CO2

3

+ H2O 2HCO 3 (2)

B : 0,075 0,125

P: 0,075 0,075 0,15

Sau p : 0 0,05 0,15

CO32- + Ba2+ BaCO3

0,05 0,05 0,05 => m = 0,05 .197 = 9,85 g.

Vy p n A l p ng.

Cu 4: ho tan hon ton 2,32 gam hn hp gm FeO, Fe 2O3 v Fe3O4

( trong s mol FeO bng s mol Fe2O3), cn dng va V lt dung

dch HCl 1M. Gi tr ca V l:

A. 0,16. B.0,18. C.0,08.


17/33

D.0,23.Gii: Fe3O4 l hn hp ca FeO v Fe2O3

V nFeO = nFe2O3 => cho nn ta coi hn hp chc Fe3O4.

Vy n3 4Fe O = 2,32 : 232 = 0,01 mol.

Phn ng : Fe3O4 + 8HCl FeCl2 + 2FeCl3 + 4H2O

0,01 mol 0,08 mol

Vy p n C l p ng.

Cu 5: Cho Cu vo dung dch H2SO4 long tc dng vi cht X ( mt loi

phn bn ho hc), thy kh thot ra khng mu ho nu trong khng

kh. Mt khc, khi X tc dng vi dung dch NaOH th kh c mi khai

thot ra. Cht X l:

A. Ure. B. amoni nitrat. C. amophot.

D. Natri nitrat

Gii: X + Cu + H2SO4 NO => X c gc nitrat.

Mt khc X + NaOH NH3 => X c gc amoni.

Vy p n B l p ng.

Cu 6: Pht biu khng ng l:

A. Trong dung dch, H2N-CH2-COOH cn tn ti dng ion lng cc H3N+-CH2-

COO-

B. Aminoaxit l nhng cht rn, kt tinh, tan tt trong nc va c v ngt.

C. Aminoaxit l hp cht hu c tp chc, phn t cha ng thi nhm

amino v nhm cacboxyl.

D. Hp cht H2N-CH2-COOH3N-CH3 l este ca glyxin (glixin).Gii: H2N-CH2-COOH3N-CH3 l sn phm ca phn ng gia glyxin v metyl amin.

H2N CH2 COOH + CH3-NH2 H2N-CH2-COOH3N-CH3

Vy p n D l p ng.

Cu 7: Trong phng th nghim, ngi ta iu ch oxi bng cch :

A. Nhit phn KClO3 c xc tc MnO2 B. Nhit phn

Cu(NO3)2

C. in phn nc. D. Ch ng ct phn

on khng kh lngGii : Vy p n A l p ng.

Cu 8: :Cho m gam hn hp X gm hai ru (ancol) no, n chc, k tip nhau trong dy

ng ng tc dng vi CuO (d) nung nng, thu c mt hn hp rn Z v mt hn hphi Y (c tkhi hi so vi H2 l 13,75). Cho ton b Y phn ng vi mt lng d Ag2O

(hoc AgNO3) trong dung dch NH3un nng, sinh ra 64,8 gam Ag. Gi tr ca m l

A. 7,8. B. 7,4 C. 9,2 D. 8,8


18/33

Gii : Ru n chc tc dng CuO c tl mol 1:1

Cng thc tnh M 2 cht1 1 2 2 1 2

1 2

1 22

n M n M M M M khi n n M

n n

+ += = =

+

th

Hay M l trung bnh cng.V ngc li khi M l trung bnh cng th s mol 2

cht bng nhau.

Phn ng trng gng HCHO 4Ag

Gi cng thc phn t ca ru l CnH2n + 1OH x mol

Phn ng: CnH2n+1OH + CuO CnH2nO + Cu + H2O

x x x x x

Vy hn hp Z gm CnH2nO (x mol) v H2O x (mol). S mol bng nhau

Z

18 +(14n +16)M =

2= 13,75.2 = 27,5.(s mol bng nhau th M l trung bnh cng). => n =

1,5. Vy 2 ru l CH3OH v C

2H

5OH v n = 1,5 l trung bnh cng ca 1 v 2 nn s mol

ca 2 ru phi bng nhau = x/2 mol. v anehit l HCHO v CH3CHO

- Phn ng vi Ag2O. HCHO 4 Ag ; v CH3CHO 2 Ag

x/2 2x x/2 x (mol)

Vy nAg = 2x + x = 3x = 0,6 => x = 0,2 (mol).=> m = 0,2.(14n + 18) = 0,2.(14.1,5+18)

= 7,8g.

Vy p n A l p ng.

Cu 9: S ng phn hirocacbon thm ng vi cng thc phn t

C8H10 l:

A. 5. B. 3. C. 2.

D. 4.Gii:

Cc ng phn ln lt l

CH2 CH3 CH3

CH3

CH3

CH3

CH3

CH3

Vy p n D l p n ng

Cu 10: Trn ln V ml dung dch NaOH 0,01M vi V ml dung dch HCl 0,03 M c 2V ml

dung dch Y. Dung dch Y c pH l

A.4. B.2 C. 3 D.1.

Gii: S mol OH = 0,01V mol

S mol H+ = 0,03V mol

H+ + OH- = H2O


19/33

0,03V 0,01V H+ d 0,02 mol => [H+] =V

V

2

02,0

20,02[ ] 0,01 102

H M+

= = = pH = 2

Vy p n B l p n ng.

Cu 11:un nng V lt hi anehit X vi 3V lt kh H2 (xc tc Ni) n khi phn ng xy ra

hon ton chthu c mt hn hp kh Y c th tch 2V lt (cc th tch kho cng iukin nhit , p sut). Ngng t Y thu c cht Z; cho Z tc dng vi Na sinh ra H2 c s

mol bng s mol Z

phn ng. Cht X l anehit

A. khng no (cha mt ni i C=C), hai chc.

B. khng no (cha mt ni i C=C), n chc

C. no, hai chc.

D.no, n chc.

Gii:

Hn hp ban u c th tch l 4V, sau phn ng hn hp Y c th tch 2V gim 2V

chnh l th tch H2 phn ng.

Th tch andehit l 1V , th tch th tch H2 phn ng l 2V andehit c 2 lin kt .

Ru Z + Na H2 c s mol H2 = s mol Z Z c 2 nhm OH. andehit c 2 nhm

CHOVy p n C l p n ng.

Cu 12: Cho cn bng h hc : 2SO2 (k) + O2 (k) 2SO3 (k); phn

ng thun l phn ng to nhit. Pht biu ng l:

A. Cn bng chuyn dch theo chiu thun khi tng nhit .

B. Cn bng chuyn dch theo chiu thun khi gim p sut h phn

ng.

C. Cn bng chuyn dch theo chiu nghch khi gim nng O2D. Cn bng chuyn dch theo chiu nghch khi gim nng SO3.Gii:

Theo nguyn l L-sa-t-lie khi gim nng mt cht cn bng dch chuyn theo chiu

lm tng nng cht .

Vy p n C l p n ng.

Cu 13: Cho hn hp bt gm 2,7 gam Al v 5,6 gam Fe vo 550 ml dung dch AgNO3 1M.

Sau khi cc phn ng xy ra hon ton, thu c m gam cht rn. Gi tr ca m l (bit th

t trong dy thin ha : Fe3+/Fe2+ng trc Ag+/Ag)

A. 64,8 B. 54,0 C. 59,4 D. 32,4


20/33

Gii: sau phn ng vi Al, Ag+ cn 0,55 0,3 = 0,25 mol dng phn ng vi Fe

Fe + 2 Ag+ Fe2+ + 2Ag

0,1 0,2 0,1 0,2

D Ag+ = 0,25 0,2 = 0,05 mol

Fe2+ + Ag+ Fe3+ + Ag

0,05 0,05

Sau cc phn ng cht rn l Ag c s mol 0,3 + 0,2 + 0,05 = 0,55 mol.

Khi lng Ag = 0,55. 108 = 59,4 gam

Vy p n C l p n ng.

Cu 14: Este c c im sau:

- t chy hon ton X to CO2 v H2O c s mol bng nhau;

-Thu phn X trong mi trng axit c cht Y ( tham gia phn ng trng g-

ng) v cht Z (c s nguyn t cac bon bng mt na s nguyn t

cacbon trong X )

Pht biu khng ng l :

A. Cht X thuc loi este no, n chc.

B. Cht Y tan v hn trong nc.

C. un Z vi dung dch H2SO4 c 1700C thu c anken.

D. t chy hon ton 1 mol X sinh ra s phm gm 2 mol CO2 v 2

mol H2O.Gii: Da vo cc d kin ca u bi

- t chy hon ton X to thnh CO2 v H2O c s mol bng nhau => X l este no

n chc

- Thy phn X thu c Y phn ng trng gng => Y phi l axit fomic. => E l este

ca axit fomic. Z c s C bng mt na ca X vy s C ca Z phi bng ca axit

fomic => Z l CH3OH. Tch nc t CH3OH khng thu c anken.

Vy p n C l p n ng.

Cu 15: Khi in phn NaCl nng chy ( in cc tr) ti catt xy ra :

A. S kh ion Na+ B. S kh ion Cl- C. S oxi ha ion Cl- D.

S oxi ho ion Na+

Gii : p n A l p n ng.

Cu 16: Trung ho 5,48 gam hn hp gm axit axetic, phenol v axit benzoic, cn dng600 ml dung dch NaOH 0,1M. C cn dung dch sau phn ng, thu c hn hp cht rn

khan c khi lng l

A. 6,84 gam. B. 4,9 gam. C. 6,8 gam. D. 8,64 gam.


21/33

Gii: V phn ng chxy ra nhm -OH nn c th thay hn hp trn bng ROH.

Ta c. ROH + NaOH RONa + H2O

Phn ng : H+ + OH- = H2O

=>S mol NaOH = s mol OH- = 0,06 mol = s mol H2O

TheoLBTKL:

Axt + NaOH = Mui + H2O

5,48 + 0,06.40 = Mui + 0,06.18

Mui = 6,8 gam .

Vy p n A l p n ng.

Cu 17: Cho 3,6 gam anehit n chc X phn ng hon ton vi mt lng d Ag2O

(hoc AgNO3) trong dung dch NH3 un nng, thu c m gam Ag. Ho tan hon ton m

gam Ag bng dung dch HNO3 c, sinh ra 2,24 lt NO2 (sn phm kh duy nht, ktc).

Cng thc ca X l

A. C3H7CHO. B. HCHO. C. C2H5CHO. D. C4H9CHO.

Gii:

Dng phng php bo ton electron cho phn ng oxi ha kh:

C+1 2e C+3

v N+5 + 1e N+4

Bo ton e : nanehit = 0,05 mol =>3,6

720,05RCHO

M = = => R = 72 29 = 43.

12x + y = 43

x 1 2 3y 31 19 7

loi loiVy anehit l C3H7CHO => p n A l p n ng.

Cu 18: S ng phn este ng vi cng thc phn t C4H8O2 l:

A. 6. B. 4. C. 5. D.

2.Gii: C4H8O2 l este no n chc. Cc ng phn l

HCOOCH2CH2CH3 ; HCOOCH(CH3)2 ;

CH3COOCH2CH3 ; CH3CH2COOCH3

Vy p n B l p n ng.

Cu 19: Dy gm cc cht c xp theo chiu nhit si tng dn t

tri sang phi l:

A. CH3CHO, C2H5OH, C2H6, CH3COOH. B. C2H6, C2H5OH, CH3CHO,CH3COOH.

C. C2H6, CH3CHO, C2H5OH, CH3COOH D. CH3COOH, C2H6,

CH3CHO, C2H5OH.


22/33

p n C l p n ng.

Cu 20: Cho 11,36 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung dch

HNO3 long (d), thu c 1,344 lt kh NO (sn phm kh duy nht, ktc) v dung dch

X. C cn dung dch X thu c m gam mui khan. Gi tr ca m l

A. 49,09. B. 34,36. C. 35,50 D. 38,72

Gii:

DngLBTKL cho cht v cho nguyn t.

Bit : S mol NO = 0,06 mol

M (Fe(NO3)3) = 242

t x = s mol mui khi lng mui = 242x

S mol N (trong HNO3) = 3x + 0,06 = s mol HNO3 khi lng HNO3 = (3x + 0,06)63

S mol H2O =3 0,06

2

x + khi lng H2O =

3 0,0618

2

x +

TheoLBTKL:

11,36 + (3x + 0,06)63 = 242x +3 0,06

182

x +

x = 0,16 khi lng mui = 242. 0,16 = 38,72 gam

Vy p n D l p n ng.

Cu 21: Bn knh nguyn t ca cc nguyn t: Li3 , O8 , F9 , Na11 c

xp theo th t tng dn t tri sang phi l:

A. Li, Na, O, F. B. F, O, Li, Na. C. F, Li, O, Na. D. F,

Na, O, Li.Gii : Ta c cu hnh e ca : 3Li : 1s22s1 ;

8O: 1s22s22p4 ;

9F: 1s22s22p5 ;

11Na : 1s22s22p63s1

T cu hnh e nhn thy : Na chu k 3 nn bn knh ln nht. Li, O, F cng chu k 2 nnkhi in tch ht nhn tng th bn knh nguyn t gim. => th t: F, O, Li, Na

Vy p n B l p n ng.

Cu 22: Cho hn hp gm Na v Al c tl s mol tng ng l 1 : 2 vo nc (d). Sau

FeO

2O

3

+ HNO

Fe


23/33

khi cc phn ng xy

ra hon ton, thu c 8,96 lt kh H2 (ktc) v m gam cht rn khng tan. Gi tr ca m

l

A. 43,2 B. 5,4. C. 7,8. D. 10,08.

Gii:

Gi s mol cua Na v Al ln lt l x v 2x. Phn ng

Na + H2O NaOH + 1/2H2 (1)

x x 0,5x (mol)

Al + NaOH + H2O NaAlO2 + 3/2H2 (2)

x x 1,5x (mol)

Sau phn ng cn cht rn chng t sau phn ng (2) Al vn cn d.

=> nH2= 0,5x + 1,5x = 2x =8,9622,4

= 0,4 (mol) => x = 0,2 mol. => nNa = 0,2 mol v nAl

= 0,4 mol

Theo (2) s mol Al phn ng l x =0,2 mol => s mol Al d l 0,4 0,2 = 0,2 mol

Khi lng Al (cht rn sau phn ng) = 0,2.27 = 5,4 gam

Vy p n B l p n ng.

Cu 23: Khi lng ca mt on mch t nilon-6,6 l 27346 vC v ca mt on mch

t capron l 17176 vC. S lng mt xch trong on mch nilon-6,6 v capron nu trn

ln lt l

A. 113 v 152. B. 113 v 114. C. 121 v 152. D. 113 v

114.

Gii:

T nilon-6,6: [- HN (CH2)6 NH CO (CH2)4 CO-]n

c M = 226n = 27346 => n =27346226

= 121

T capron : [ - HN-(CH2

)5-CO - ]

mc M = 113m = 17176 => m = 152

Vy p n C l p n ng.

Cu 24: T 2 mui X v Y thc hin cc phn ng:

X ot X1 + CO2 X1 + H2O X2

X2 + Y X+ Y1 + H2O X2 + 2Y X+ Y2 + 2H2O

Hai mui tng ng X v Y l:

A. CaCO3, NaHCO3. B. MgCO3, NaHCO3. C. CaCO3, NaHSO4 D.

BaCO3, Na2CO3

Gii : CaCO3ot CaO + CO2

(X) (X1)


24/33

CaO + H2O Ca(OH)2

(X1) (X2)

Ca(OH)2 + NaHCO3 CaCO3 + NaOH + H2O

(X2) (Y) (X) (Y1)

Ca(OH)2 + 2NaHCO3 CaCO3 + Na2CO3 + 2H2O

(X2) (Y) (X) (Y2)Vy p n A l p n ng

Cu 25: C cc dung dch ring bit sau:

C6H5-NH3Cl( phenylamoni clorua), H2N-CH2-CH2-CH(NH2)-COOH, ClH3N-CH2-

COOH, HOOC-CH2-CH(NH2)-COOH, H2N-CH2-COONa.

S lng cc dung dch c pH < 7 l:

A. 2. B. 3. C. 5.

D. 4.Gii : Cc cht l: C6H5-NH3Cl, ClH3N-CH2-COOH, HOOC-CH2-CH2-CH(NH2)-COOH.

Vy p n B l p n ng.

Cu 26: Cho V lt dung dch NaOH 2M vo dung dch cha 0,1 mol Al2(SO4)3 v 0,1 mol

H2SO4 n khi phn ng hon ton, thu c 7,8 gam kt ta. Gi tr ln nht ca V thu

c lng kt ta trn l

A. 0,35. B. 0,25. C. 0,45. D. 0,05.

Gii:V max khi Al(OH)3 to thnh ti a ri b ha tan mt phn.

Theo bi ra ta c:2 4

2 4 3

H SO

Al (SO )

n =0,1 mol

n =0,1 mol

=>

+

3+

2-4

H

Al

SO

n =0,2 mol

n =0,2 mol

n =0,4 mol

v n3Al(OH)

=0,1 mol

Khi cho NaOH vo phn ng xy ra theo th t:

H+ + OH- H2O

0,2 mol 0,2 mol

v Al3+ + 3OH- Al(OH)3 ; Al3+ + 4OH- AlO-

2+ H2O

0,1 mol 0,3 mol 0,1 mol 0,1 mol 0,4 mol

nNaOH = -OHn = 0,2 + 0,3 + 0,4 = 0,9 mol => Vmax = 0,9 : 2 = 0,45 lt

Cu 27: Hn hp X c tkhi so vi H2 l 21,2 gm propan, propen v propin. Khi t chy

hon ton 0,1 mol

X, tng khi lng ca CO2 v H2O thu c l

A. 18,60 gam. B. 18,96 gam. C. 20,40 gam. D. 16,80 gam.

Gii:


25/33

Bin i Phn ng: C3Hy 3CO2 + y/2 H2O

0,1 0,3 0,1.0,5y

XM = 21,2. 2 = 42,4 = 12.3 + y => y = 6,4

=> mCO2 + mH2O = 0,3.44 + 0,1.0,5.6,4.18 =18,96g

Vy p n B l p n ng.Cu 28: Ho tan hon ton 0,3 mol hn hp gm Al v Al4C3 vo dung dch KOH (d), thu

c a mol hn hp kh v dung dch X. Sc kh CO2 (d) vo dung dch X, lng kt ta

thu c l 46,8 gam. Gi tr ca a l

A. 0,60. B. 0,55. C. 0,45. D. 0,40.

Gii:

Gii theo bo ton khi lng nguyn t.

: Cc phn ng : Al + KOH + H2O KAlO2 + 3/2 H2

Al4C3 + HOH 3CH4 + Al(OH)3

Al(OH)3 + KOH KAlO2 + 2H2O

Ton b lng Al ban u u nm trong KAlO2

Sc kh CO2 vo CO2 + KAlO2 + 2H2O Al(OH)3 + KHCO3.

Ton b lng Al ban u u nm trong kt ta

S mol kt ta78

8,46 = 0,6 mol

t x = s mol Al ; y = s mol Al4C3

Ta c: S mol hn hp: x+ y = 0,3

S mol Al : x+ 4y = 0,6

S mol H2 = 0,3 mol , s mol CH4 = 0,3 mol S mol 2 kh 0,6 mol

Vy p n A l p n ng.

Cu 29: Khi tch nc t ru (ancol) 3-metylbutanol-2( hay 3-metylbutan-2-ol), sn phm chnh thu c l:

A. 2-metylbuten-3( hay 2-metylbut-3-en) B. 3-metylbuten-2

(hay 3-metylbut-2-en)

C. 3-metylbuten-1( hay 3-metylbut-1-en) D. 2-metylbuten-2

(hay 2-metylbut-2-en)Gii: Tch nc theo quy tc Zai xep

CH3 CH CH CH3 CH3 C CH CH3 + H2O

CH3 OH CH3

1234 1 2 3 4

3-metylbutanol-2 2-metylbuten-2

=> x = 0,2 mol ; y = 0,1mol


26/33

Vy p n D l p n ng.

Cu 30: Hp cht c lin kt ion l:

A. HCl. B. NH3. C. H2O.

D. NH4Cl

Gii: lin kt gia ion Cl- v NH 4+

Vy p n D l p n ng.

Cu 31: Cho V lt hn hp kh (ktc) gm CO v H2 phn ng vi mt lng d hn hp

rn gm CuO v Fe3O4 nung nng. Sau khi cc phn ng xy ra hon ton, khi lng hn

hp rn gim 0,32 gam. Gi tr ca V l

A. 0,112. B. 0,560 C. 0,448. D. 0,224

Gii:

Theo bi: Khi lng rn gim 0,32 gam chnh l khi lng O trong oxit

S mol O trong oxt = 0,32 / 16 = 0,02 mol = s mol hn hp (CO + H2)

Vy th tch hn hp (CO + H2) = 0,02 . 22,4 = 0,448 lt

Vy p n C l p n ng.

Cu 32: Cho cc phn ng:

4HCl + MnO2 MnCl2 + Cl2 + 2H2O .

2HCl + Fe FeCl2 + H2.

14HCl + K2Cr2O7 2KCl + 2CrCl3 +3Cl2 + 7H2O.

6HCl + 2KMnO4 2kCl +2MnCl2 + 5Cl2 + 8 H2O.

S phn ng th hin tnh xi ho l:

A. 2. B.1. C. 4. D. 3.Gii : 2HCl + Fe FeCl2 + H2 v 6HCl + 2Al 2AlCl3 + 3H2

Vy p n A l p n ng

Cu 33: Tinh bt, xenluloz, saccaroz, mantoz u c kh nng tham

gia phn ng:

A. Ho tan Cu(OH)2. B. thu phn. C. trng ngng.

D. trng gngp n B l p n ng.

Cu 34: Pht biu ng l:

A. tnh axit ca phenol yu hn tnh axit ca ru (ancol).

B. Cao su thin nhin l sn phm trng hp ca isopren.

C. tnh baz ca anilin mnh hn tnh baz ca amoniac.

D. Cc cht eilen, toluen v stiren u tham gia phn ng trng hp.

p n B l p n ng.Cu 35: Cho glixerin trioleat ( hay triolein) ln ,lt vo mi ng nghim

cha ring bit : Na, Cu(OH)2, CH3OH, dung dch Br2, dung dch NaOH.

Trong iu kin thch hp, s phn ng xy ra l:


27/33

A. 5. B. 4. C. 3.

D. 2.Gii: glixerin trioleat l este ca glixerin v axit oleic ( axit bo khng no c mt lin kt

i). c cu to:

CH2 - COO - (CH2)7 - CH=CH -(CH2)7-CH3

CH - COO - (CH2)7 - CH=CH -(CH2)7-CH3

CH2 - COO - (CH2)7 - CH=CH -(CH2)7-CH3

. => c phn ng vi Br2 v NaOH.

Vy p n D l p n ng.

Cu 36: Cho 3,2 gam bt Cu tc dng vi 100 ml dung dch hn hp gm HNO3 0,8M v

H2SO4 0,2M. Sau khi cc phn ng xy ra hon ton, sinh ra V lt kh NO (sn phm kh

duy nht, ktc). Gi tr ca V l

A. 0,746 B. 0,448 C. 0,672 D. 1,792

Gii:

S mol Cu = 0,05 mol (d)

S mol H+ = 0,08 + 0,04 = 0,12 mol (ht)

S mol NO3 = 0,08 mol (d)

Phn ng : 3Cu + 8H+ + 2NO3 3Cu 2+ + 2 NO + 4 H2O

Ban u : 0,05 0,12 0,08 0 0 0

P : 0,015 0,12 0,03 0,03

=> VNO = 0,03 . 22,4 = 0,672 lt

Vy p n C l p n ng.

Cu 37: Cho s chuyn ha CH4 C2H2 C2H3Cl PVC. tng hp 250 kg

PVC theo s trn th cn V m3 kh thin nhin (ktc). Gi tr ca V l (bit CH4 chim

80% th tch kh thin nhin v hiu sut ca c qu trnh l 50%)

A. 224,0 B. 448,0 C. 286,7 D. 358,4

Gii:

Tnh theo phng trnh phn ng c cng hiu sut qua nhiu phn ng ch quan

tm n cht u v cht cui..

Phn ng 2n CH4 CH2 - CH2

H= 50%n


28/33

2n mol 62,5 n kg

X mol 250 kg

S mol CH4 = molx

x16100

505,62

2502=

Th tch CH4 = 16.. 22,4 = = 358,4

Th tch kh thin nhin = 44810080

4,358 =

Vy p n B l p n ng.

Cu 38: Cho 2,13 gam hn hp X gm ba kim loi Mg, Cu v Al dng bt tc dng hon

ton vi oxi thu c

hn hp Y gm cc oxit c khi lng 3,33 gam. Th tch dung dch HCl 2M va

phn ng ht vi Y l

A. 90 ml. B. 57ml. C. 75 ml. D. 50 ml.

Gii:

3 kim loi + O 3 oxit

2,13 gam 3,33 gam

Khi lng sau P lch 3,33 2,13 = 1,2 gam = gam O (O trong oxit)

1, 2( ) 0,075

16On trong oxit mol = =

Theo phn ng : 2H+ + O2- = H2O s mol H+ = 0,075.2 = 0,15 mol

V HCl = = 0,075 lt = 75 ml

Vy p n C l p n ng.

Cu 39: Gluxit ( cacbohirat) ch cha hai gc glucoz trong phn t l:

A. xenluloz. B. tinh bt. C. saccaroz.

D. Mantoz.p n ng l :D

Cu 40:un nng hn hp kh gm 0,06 mol C2H2 v 0,04 mol H2 vi xc tc Ni, sau

mt thi gian thu c hn hp kh Y. Dn ton b hn hp Y li t t qua bnh ng dung

dch brom (d) th cn li 0,448 lt hn hp kh Z (ktc) c tkhi so vi O2 l 0,5. Khi

lng bnh dung dch brom tng l

A. 1,20 gam. B. 1,04 gam. C. 1,64 gam. D. 1,32

gam.

Gii:

Khi lng bnh brom tng l khi lng hirocacbon khng no b hp th.

Theo nh lut bo ton khi lng ta c:

0,152


29/33

mhn hp u = mY = mhirocacbon khng no + mZ

=> mhirocacbon khng no = mhn hp u mZ

=> mhirocacbon khng no = 0,06.26 + 0,04.2 0,5.32.0,44822,4

= 1,32g.

Vy khi lng bnh brom tng 1,32g

p n ng l :D

Cu 41: X l kim loi phn ng c vi dung dh H2SO4 long, Y l kim loi

tc dng c vi dung dch Fe(NO3)3. Hai kim loi X, Y ln lt l ( bit th t

trong dy in ho Fe3+/Fe2+ ng trc Ag+/Ag).

A. Mg, Ag. B. Fe, Cu. C. Cu, Fe.

D. Ag, Mg.Gii:

Y phn ng c vi Fe(NO3)3 Y khng th l Ag => loi A v D.

X phn ng vi H2SO4 long => X ch c th l Fe

p n ng l :B

Cu 42: Pht biu ng l:

A. Phn ng thu phn este trong mi trng axit l phn ng thun

nghch.

B. khi thu phn cht bo lun thu c C2H4(OH)2.

C. phn ng gia axit v ru khi c H2SO4 c l phn ng mt chiu.

D. Tt c cc este phn ng vi dung dch kim lun thu c sn phm

cui cng l mui v ru ( anol).p n ng l :A.

Cu 43: Khi phn tch thnh phn mt ru (ancol) n chc X th thu c kt qu: tng

khi lng ca cacbon v hiro gp 3,625 ln khi lng oxi. Sng phn ru (ancol)

ng vi cng thc phn t ca X l

A. 2. B. 4. C. 1. D. 3.

Gii:

t cng thc ru n chc CxHyO

Theo bi 12x + y = 3,625.16 = 58 CxHy = 58

x 1 2 3 4y 46 34 22 10

loi loi loi CxHy l C4H10 ru l C4H10 O => Ru C4H10O c 4 ng phn

p n ng l :B

Cu 44: Cho s chuyn ho sau:


30/33

C3H4O2 + NaOH X + Y

X + H2SO4long Z + T

Bit Y v Z u c phn ng trng gng. Hai cht Y, Z tng ng l:

A. HCOONa, C3CHO. B. HCHO, CH3CHO.

C. HCHO. HCOOH. D. CH3CHO, HCOOHGii :

X + H2SO4 long Z + T => X phi l mui natri ca axit v Z c phn ng trng

gng nn X l HCOONa. => Cng thc ca este l HCOOCH=CH2. v Y l CH3CHO

p n ng l :D

Cu 45: Nung nng m gam hn hp Al v Fe2O3 (trong mi trng khng c khng kh)

n khi phn ng xy ra hon ton , thu c hn hp rn Y. Chia Y ta thnh hai phn

bng nhau:

Phn 1 tc dng vi dung dch H2SO4 long ( d) sinh ra 3,08 lt kh H2 ( ktc).

Phn 2 tc dng vi dung dch NaOH (d) , sinh ra 0,84 lt kh H2 ( ktc).

Gi tr ca m l

A. 22,75 B. 21,40 C. 29,40 D. 29,43

Gii:

Fe p1 0,1375 mol (H2) (1)

Al Al2O3

Fe2O3 Al d

Phn ng : 2Al + Fe2O3 2Fe + Al2O3

Sau phn ng c Al cn d v Al d phn ng (1) v (2) u to cng s mol H2 .

Theo (2) Al 2

3H2 s mol Al d =

0,03752 0,025

3mol=

Theo (1) s mol H2 do Fe sinh ra = 0,1375 0,0375 = 0,1 mol = s mol Fe sau phn ng

Vy sau phn ng nhit nhm c 0,1.2 = 0,2 mol Fe = s mol Al phn ng

C 0,025.2 = 0,05 mol Al d

Vy hn hp ban u c s mol Al = 0,2 + 0,05 = 0,25 mol Al hay 0,25.27 = 6,75 gam

C Fe2O3 0,1 mol hay 0,1.160 = 16 gam

Khi lng hn hp ban u = 16 + 6,75 = 22,75 gam

H2SO

4long d-

P2 NaOH d- 0,0375 mol H2 (2)


31/33

p n ng l :A

Cu 46: Bit rng ion Pb2+ trong dung dch oxi ho c Sn. Khi nhng hai

thanh kim loi Pb v Sn c ni vi nhau bng dy dn in vo mt

dung dch cht in li th:

A. Ch c Sn b n mn in ho. B. c Pb v Sn u khng

b n mn in ho.

C. c Pb v Sn u b n mn in ho. D. ch c Pb b n mn

in ho.Gii:

-Da vo iu kin c s n mn in ho

-Sn l kim loi ng trc Pb trong dy in ho => Sn b n mn theo kiu

in ho.

p n ng l :A

Cu 47: Khi crackinh ho ton mt th tch ankan X thu c 3 th tchhn hp Y ( cc th tch o cng iu kin nhit v p sut); t

khi ca Y so vi H2 bng 12. Cng thc phn t ca X l:

A. C6H14. B. C3H8 C. C4H10.

D. C5H12.Gii : p dng LBTKL

V cng iu kin nn tl th tch coi l tl s mol

Crackinh 1 mol A c 3 mol hn hp kh Y.

YM = 12.2 = 24. => mY = 24.3 = 72g. Theo nh lut bo ton khi lng : mX = mY =

72 g

=> MX = 72:1 = 72 . X l CnH2n + 2 Vy : 14n + 2= 72 => n = 5 => X l C5H12

p n ng l :D

Cu 48: Cho cc cht sau: CH2=CH-CH2-CH2- CH=CH2, CH2=CH-CH=CH-

CH2-CH3, CH3-CH3-C(CH3)=CH-CH3, CH2=CH-CH2-CH=CH2, s cht c ngphn hnh hc l:

A. 4. B.1. C. 2. D.

3.Gii : iu kin c ng phn hnh hc l:

=>

V vy chc cht CH2=CH-CH=CH-CH2-CH3 c ng phn hnh hc

C=C

a

b

c

d

a bc d


32/33

CH2 CH CH2CH3

H

C C

H

CH2 CH

CH2CH3H

C C

H

cis- pentaien-1,3 trans- pentaien-1,3

p n ng l :A

Cu 49: Trong cc loi qung st , qung c hm lng st cao nht l:

A. hematit . B. xierit. C. hematit nu.

D. manhetit.p n ng l :D

Cu 50: iso-pentan tc dng vi Cl2 theo t l s mol 1:1, s sn phm

monoclo ti a thu c l:

A. 3. B. 5. C. 4.

D. 2.p n ng l :C

Cu 51: Lng glucoz cn to ra 1,82 g sobitol vi hiu sut 80% l:

A. 2,25gam. B. 1,82 gam. C. 1,44 gam.

D. 1,8 gam.Gii:

HOCH2-[CH(OH)]5-CHO + H2oNi, t HOCH2-[CH(OH)]5-CH2OH

C th tnh nhanh:

C6H12O6 C6H14O6

180 182

m 1,82

Khi lng glucoz thc t cn dng l: 100.80.182

.180.82,1= 2,25g

p n ng l :A

Cu 52: oxi ho hon ton 0,01 mol CrCl3 thnh K2CrO4 bng Cl2 khi c

mt KOH, lng ti thiu Cl2 v KOH tng ng l:

A. 0,015 mol v 0,04 mol. B. 0,03mol v 0,08 mol.

C. 0,03 mol v 0,04 mol. D. 0,015 mol. V 0,08

mol.Gii: Phng trnh phn ng:

2CrCl3 + 3Cl2 + 16KOH 2K2CrO4+ 12KCl + 8H2O0,01 0,015 0,08

p n ng l :D

Cu 53: Mt pin in ho c0s in cc Zn nhng trong dung dch ZnSO4


33/33

v in cc Cu nhng trong dung dch CuSO4. Sau mt thi gian pin

phng in th khi lng :

A. in cc Zn gim cn in cc Cu tng .

B. c hai in cc Zn v Cu u gim.

C. c hai in cc Zn v Cu u tng

D. in cc Zn tng cn in cc Cu gim.

p n ng l :ACu 54: tc nhn ch yu gy ma axit l:

A. CO v CO2. B. SO2 v NO2 C. CH4 v NH3 .

D. CO v CH4p n ng l: B

Cu 55: Cho s chuyn ho qung ng thnh ng:

CuFeS2 X Y

Cu

Hai cht ln lt l:

A. Cu2S, Cu2O. B. Cu2O, CuO. C. CuS, CuO.

D. Cu2S, CuOp n ng l: A

Cu 56: S ng phn xeton ng vi cng thc phn t C5H10O l:

A. 5. B. 4. C. 3. D. 6.

Gii:CH3 C CH2 CH2 CH3 CH3 C CH CH3 CH3 CH2 C CH2 CH3

CH3O O O

p n ng l: C

-------------------------Ht ---------------------

+O2, t0 + O2,t0 + X, t0

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