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gii chi tit thi th i hc - m 292Cu 1. p n A

Gi a l s mol MCln dng, ta c:

MCln dpdd M + 2n

Cl2

Mol: a aan

2

Theo trn v ta c:aM = 16

an 5,6=

2 22,4

T h trn cho ta: M = 32nLp bng:

n 1 2 3M 32 64 96

Ch c gi tr n = 2; M = 64 l hp l. Vy M l CuCu 2. p n A

Gi ru th nht l CxHyO, ancol k tip l Cx+1Hy+2OPhn ng t chy ru:

CxHyO + O2ot xCO2 +

y

2H2O

Cx+1Hy+2O + O2ot (x + 1)CO2 +

y + 2

2H2O

Theo trn v ta c:x + 1 x

> y > 2xy + 2 y

2 2

Ta bit, ancol c s H > 2C l ancol noCu 3. p n D

t cng thc amin l CnH2n+1NH2Phn ng ca amin vi dung dch FeCl3:

3CnH2n+1NH2 + 3H2O + FeCl3 Fe(OH)3 + 3CnH2n+1NH3+Cl-

Mol: 0,3 0,1

T trn xc nh: Mamin =17,7

= 590,3

T cng thc tng qut ca amin cho ta: 14n + 17 = 59 n = 3Suy ra cng thc ca amin l C3H7NH2

Cu 4. p n C

t cng thc ca 2 axit l R(COOH)

2 axit 16M = = 91,420,175

Hay: R + 45m = 91,42 (1)Phn ng ca axit vi Na2CO3:

R(COOH)m 2 3Na CO R(COONa) m

Mol: 0,175 0,175Theo trn v ta c: 0,175( R + 67 m ) = 22,6Hay: R + 67 m = 129,1 (2)T (1) v (2) ta c m = 1,71

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V s nhm COOH 2 (theo ) nn: 1 < m = 1,71 < 2Do hn hp A c: mt axit no n chc v mt axit no hai chcThay m = 1,71 vo (1) hoc (2) c R = 14,47 14 < R = 14,47 < 15Vy 2 axit trong hn hp l CH3COOH

COOHv CH2

COOH

Cu 5. p n Dt x, y, z l s mol ca Fe, Fe2O3, Fe3O4, theo c:

56x + 160y + 232z = 5,60 (1)Phn ng ca hn hp vi CO v CuSO4

Fe2O3 + 3CO 2Fe + 3CO2 Mol y 2y

Fe3O4 + 4CO 3Fe + 4CO2 Mol z 3zTheo trn v ta c: 56 (x + 2y + 3z) = 4,48Hay x + 2y + 3z = 0,08

Fe + CuSO4 Cu + FeSO4Mol x x

Cng theo trn v c:64x + 160y + 232z = 5,84

56x + 160y + 232z = 5,60

x = 0,03

T (1) v (2) c:160y + 232z = 3,92

2y + 3z = 0,05

z = 0,01; y = 0,01

Vy trong hn hp c: mFe = 56.0,03 = 1,68 (gam)

2 3 3 4Fe O Fe Om = 160.0,01 = 1,60 (gam) ; m = 232.0,01 = 2,32 (gam)

Cu 6. p n Di vi phn ng c cn bng, ta nhit, khi nhit tng ln s lm chuyn dch cn bng

theo chiu nghch (Nguyn l L satli); xc tc khng lm chuyn dch cn bng v xc tc lmtng tc phn ng c hai chiu nh nhauCu 7. p n B

Cu 8. p n D

in tch ht nhn ca A = -18-19

3,2.10= 20 (Ca)

1,6.10

Suy ra A1 l CaO Theo , B chu k 2, nhm 4 (B l cacbon) nn B1 l CO2Vy M l CaCO3

Cu 9. p n A

Ha tan hn hp trong axit:- SiO2 khng tan trong axit, lc tch c SiO2- Dung dch cn li c ion Al3+ v Fe3+;

Al2O3 + 6H+ 2Al3+ + 3H2O

Fe2O3 + 6H+ 2Fe3+ + 3H2O

- Cho dung dch kim vo dung dch trn:Fe3+ + 3OH- Fe(OH)3 Al3+ + 4OH- AlO2

- + 2H2O Lc tch Fe(OH)3 ri nung to cao c Fe2O3

2Fe(OH)3ot Fe2O3 + 3H2O

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Cui cng axit ha nc lc trn c cha AlO2-AlO2

- + H+ + H2O Al(OH)3 Nung kt ta c Al2O3:

2Al(OH)3ot Al2O3 + 3H2O

Cu 10. p n CCc cht cho trong c tnh baz do nit cn d electron t do nn c kh nng nhn protonAmin c gc y electron (R-) s lm tng tnh baz. Ngc li gc ht electron (Ar-) s lm

gim tnh bazCu 11. p n D

Cng thc tng qut ca ancol: CnH2n+2-2aOmPhn ng t chy: CnH2n+2-2aOm 2o

O

t nCO2 + (n + 1 a)H2O

Theo trn v :n + 1 - a

= 2n

hay : n + a = 1

Ch c gi tr hp l l a = 0 n = 1V cng do m ch c th l 1Vy ancol l CH4O hay CH3OH

Cu 12. p n C

Cu 13. p n CCu 14. p n D

2 4H SO NaOHn = 0,005 (mol); n = 0,005.4 = 0,02 (mol)

+2 4H SOH

n = 2n = 2.0,005 = 0,01 (mol)

- NaOHOHn = n = 0,02 mol

Phn ng trung ha:H+ + OH- H2O

Mol: Ban u 0,01 0,02Phn ng: 0,01 0,01Sau phn ng: 0 0,01

nOH d = 0,01 [OH-] = 0,01 = 0,0024 + 1

pOH = -lg[OH-] = -lg0,002 = -lg(2.10-3) = (3 0,3) = 2,7pH = 14 2,7 = 11,3

Cu 15. p n C

Theo :2CO NaOH

11,2n = = 0,5 (mol); n = 0,8.1 = 0,8 (mol)

22,4

T s gia 2 s mol2

NaOH

CO

n 0,8= < 2

n 0,5

V th dung dch tn ti hai loi mui theo cc phng trnh phn ng:

CO2 + NaOH

NaHCO3Mol x x xCO2 + 2NaOH Na2CO3 + H2O

Mol y 2y yGi x, y l s mol ca NaHCO3 v Na2CO3, theo trn ta c:

2CO NaOHn = x + y = 0,5; n = x + 2y = 0,8

V th,3M(NaHCO )

0,2C = .1000 = 0,25 (M)

800

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2 3M(Na CO )

0,3C = .1000 = 0,375 (M)

800

Cu 16. p n Din phn dung dch KNO3 thc cht l in phn nc

H2Odp H2 + 2

1O

2

2On anot =

280

= 0,0125 (mol)22400 Khi lng sn phm catot l khi lng hiro

2 2H On = 2n v = 0,0125.2 = 0,025 (mol)

Vy2H

m = 2.0,025 = 0,05 (gam)

Cu 17. p n BCc phn ng:

C4H8O2 + 5O2 4CO2 + 4H2O (1)

CO2 + Ca(OH)2 CaCO3 + H2O (2)

CO2 + H2O + CaCO3 Ca(HCO3)2 (3)

T (1) 2COn = 4n cht hu c = 4. 1,32 = 0,06 (mol)88

T (2) 2CO

n to KT =2 3Ca(OH) CaCO KT

3,7n = n = = 0,05 (mol)

74

T (3) 2CO

n ha tan KT =3CaCO

n tan = 0,06 0,05 = 0,01 (mol)

Do :3CaCO

n cn = 0,05 0,01 = 0,04 (mol)

mKT =3CaCO

m cn = 0,04.100 = 4 (gam)

Cu 18. p n Dt cng thc ca X l CxHyClzTheo : MX = 56,5.2 = 113 (gam)

Cl 35,5z%m = .100% = 62,83%113 z = 2

Suy ra: 12x + y + 71 = 113Hay: 12x + y = 42Ch c gi tr hp l l x = 3; y = 6Vy cng thc phn t ca X l C3H6Cl2

Cu 19. p n DPhn ng t chy:

CxHyO2 + 24x + y - 4

O4

ot xCO2 + 2

yH O

2

Mol: 0,1 0,1x

Theo trn v : 0,1x < 35,2 = 0,844

x < 8

CxHyO2 + Na H2 + Mol: 1 1

Vy trong A c 2 nguyn t hiro linh ngCxHyO2 + NaOH

Mol: 0,2 0,2Vy trong A c mt nguyn t hiro c tnh axit

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V phn t A c 2 nguyn t oxi, nn trong A phi c 2 nhm OH (mt nhm c tnh axit v 1 nhmthuc ancol)Vy A c cu to:

OH OH OHCH2OH ; ;

CH2OH CH2OH

Cu 20. p n Ct cng thc ca oxit kim loi l AxOy c a mol; khi lng mol ca A l M v ha

tr ca A l n:

AxOy + yCOot xA + yCO2 (1)

Mol: a ya xa yaCO2 + Ca(OH)2 CaCO3 + H2O (2)

Mol: ya 7

= 0,07100

p dng nh lut bo ton khi lng, t (1):4,06 + 28.0,07 = mA + 44.0,07 mA = 2,94 = M. xa

Phn ng ca A vi dung dch HCl:2A + 2nHCl 2ACln + nH2:Mol: xa

nxa

a

Theo :n

xaa

=1,176

= 0,052522,4

xa =0,105

n

Do M = 28nNghim thch hp khi n = 2 M = 56 A l Fe

Khi n = 2 ta c: xa = 0,105 = 0,05252

Suy ra:xa 0,0525 x 3

= =ya 0,07 y 4 Cu 21. p n C

Cu 22. p n DTa bit: nguyn t trung ha lun c p = e = z v s khi A = p + n

Theo , ta c:p + e + n = 2p + n = 52

p + n = 35

Gii h trn c p = 17Cu 23. p n CCu 24. p n D

Cho H2SO4 long tc dng vi c 5 kim loi:

Kim loi no khng tan l Ag (v Ag khng phn ng vi H2SO4 long); 4 kim loikia c phn ng:

2Al + 3H2SO4 Al2(SO4)3 + 3H2

Fe + 3H2SO4 FeSO4 + H2

Mg + H2SO4 MgSO4 + H2

Ba + 3H2SO4 BaSO4 + H2

Trng hp no c kt ta l Ba. Lc tch kt ta. Dung dch nc lc c chaBa(OH)2 (sau khi Ba phn ng vi H2SO4, cho thm Ba vo, Ba s phn ng vi H2O):

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Ba + 2H2O Ba(OH)2 + H2

Cho dung dch Ba(OH)2 vo 3 dung dch cn li: Trng hp no c kt ta trng, khng tan trong Ba(OH)2 d th kim loi ban u

l Mg:MgSO4 + Ba(OH)2 BaSO4 + Mg(OH)2

Trng hp no c kt ta mu trng xanh, trong khng kh dn dn ha nu thkim loi ban u l Fe:FeSO4 + Ba(OH)2 BaSO4 + Fe(OH)2

4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3 nu

Trng hp no c kt ta tan dn trong Ba(OH)2 d n khi hon ton khng tan,kim loi ban u l Al:

Al2(SO4)3 + 3Ba(OH)2 3BaSO4 + 2Al(OH)3

2Al(OH)3 + Ba(OH)2 Ba(AlO2)2 + 4H2OCu 25. p n D

Cu 26. p n D

Cu 27. p n B

Cc amin R-NH2 (amoniac chnh l amin khi R l H) c nhm amino nh vi gc y electron(R-) lm tng tnh baz, cng nhiu gc y, amin c tnh baz cng mnh. Ngc li, cc amin cnhm amino nh vi gc thm ht electron (Ar-) lm cho amin c tnh baz gim. Amin c nhmth y electron gc thm c tnh baz mnh hn amin c nhm th ht electron (NO2

-) cng gc thmCu 28. p n D

- T CH4 iu ch ru metylic:CH4 + Cl2 askt CH3Cl + HCl

CH3Cl + NaOHot CH3-OH + NaCl

- Ru etylic iu ch t CH4:2CH4

o1500 C

2C2H2 + 3H2C2H2 + HOH

o

+

80 C

H CH3-CHO

CH3-CHO + H2 Ni CH3-CH2OH- Etilenglicol t CH4:

C2H5OH 2 4oH SO d

170 C C2H4 + H2O

C2H2 + Br2 CH2 - CH2Br Br

CH2-CH2 + 2NaOH CH2 - CH2 + 2NaBrBr Br OH OH

- Cng t CH4 iu ch ru isopropylic:2C2H5OH

oxt, t C4H6 + 2H2O + H2

C4H6 + 2H2 Ni C4H10

C4H10cracking CH2=CH-CH3 + CH4

CH2=CH-CH3 + HOH+H CH3-CH-CH3

OHCu 29. p n C

Theo : %A =A.100 200

=A + 60 7

A = 24 A chnh l Mg

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Cng theo : %B =B.100

= 40B + 60

B = 40 B chnh l Ca

Vy hai mui l MgCO3 v CaCO3Cu 30. p n D

Phn ng trung ha:NaOH + HCl NaCl + H2O

Mol: 0,003 0,03.0,10

nNaOH ban u = 0,09.0,04 = 0,0036 (mol)Theo phng trnh phn ng trn, th nNaOH phn ng l 0,003 mol nNaOH d = 0,0036 0,0030 = 0,0006 (mol)

Nh vy dung dch mi d NaOH v c th tch l 130 ml

Do nng OH- = -30,0006

= 4,6.100,13

pOH = -lg[OH-] - lg 4,6.10-3 = 2,34Vy pH = 14 pOH = 14 2,34 = 11,66Cu 31. p n D

S gam KCl c trong 450 gam dung dch 8% l 450.8

= 36 (gam)

100

Gi x l s gam mui KCl thm vo, theo ta c:36 + x 12

= x = 20,45 gam450 + x 100

Cu 32. p n DPhn ng trung ha axit:

R-COOH + NaOH RCOONaMol 0,05 0,05

S mol NaOH trung ha = 0,05 naxit = 0,05 Khi lng phn t axit = 3,6 : 0,05 = 72 R = 72 45 = 27

Vy R l CH2=CH-Suy ra axit l CH2=CH-COOHCu 33. p n D

Phn ng t chy: n 2n+2 z 2 2 2(3n + 1 - z)

C H O + O nCO + (n + 1)H O2

Theo trn v :3n + 1 - z

= 3,52

Hay 3n + 1 z = 7 vi iu kin: 1 z nCh gi tr n = 3, z = 3 l hp lVy cng thc phn t ca ru l C3H8O3

Cu 34. p n D

S mol kim MOH = 0,15.0,1 = 0,015 (mol)Phn ng trung ha v ha tan H2SO4 ca dung dch A:

R-COOH + MOH R-COOM + H2O

Mol 0,015 0,015(R-COO)2Ba + H2SO4 2RCOOH + BaSO4

Mol 0,02 4,66

= 0,02233

Theo trn v :

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CM(R-COOH) =0,015

= 0,3 (M)0,05

; CM((R-COO)2Ba) =0,02

= 0,4 (M)0,05

Cng theo trn v , ta c:0,015 (R + 44 + M) + 0,02 (2R + 225) = 6,33 R =

1,17 - 0,015M

0,055

Ch c gi tr hp l khi M = 23 R = 15 ng vi gc CH3

Vy cng thc cu to ca axit: CH3-COOHCu 35. p n D

Cc phn ng vi, nAl =10,8

= 0,4 (mol)27

3 4Fe O

34,8n = = 0,15 (mol)

232;

2H

10,752n = = 0,48 (mol)

22,4v x l s mol Al tham gia phn

ng8Al + 3Fe3O4 9Fe + 4Al2O3 (1)

Mol x 3

x8

9

x8

1

x2

Fe + H2SO4 FeSO4 + 2H (2)

Mol9

x8

9

x8

9

x8

2Al + 3H2SO4 Al2(SO4)3 + 3H2 (3)

Mol 0,3 x 3

(0,4 - x)2

3

(0,4 - x)2

Fe3O4 + 4H2SO4 FeSO4 + Fe2(SO4)3 + 4H2O (4)

Mol 0,15 -3

x8

3

4(0,15 - x)8

Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O (5)

Mol 1 x2

3. 1 x2

Tnh hiu sut phn ng theo Al (c th tnh theo Fe3O4):Theo cc phn ng (2), (3) v , ta c:

2H

9 3x + (0,4 - x) = n = 0,48 x = 0,32

8 2

Vy hiu sut h =0,32.100

= 80%0,4

Theo (2), (3), (4), (5):2 4

H SO

9 3 3 1n = x + (0,4 - x) + 4 0,15 - x + 3 x = 1,08 (mol)

8 2 8 2

Vy2 4dd H SO

V dng =1,08.98.100

= 464,21 (mol)20.1,14

Cu 36. p n D

Cu 37. p n CPhn ng cn bng:

46Al + 168HNO3 46Al(NO3)3 + 9N2 + 6N2O + 84H2OCu 38. p n C

Ly mt t mi cht rn cho vo tng ng nghim cha nc c kt qu:

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Hai cht tan trong nc l K2SO4 v NaNO3Hai cht khng tan l CaSO4 v BaCO3 Sc kh CO2 vo ng nghim cha cc cht khng tan, ng no thy cht rn tan

dn n ht, ng cha BaCO3, vBaCO3 + CO2 +H2O Ba(HCO3)2(tan)

ng c cht khng tan l CaSO4 Ly dung dch Ba(HCO3)2 to thnh trn cho vo hai ng nghim ng NaNO3

v K2SO4 ng no c kt ta trng l ng cha K2SO4, v:K2SO4 + Ba(HCO3)2 BaSO4 + 2KHCO3

ng khng c phn ng g l ng cha NaNO3Cu 39. p n C

Cu 40. p n CCu 41. p n C

Nhit si ca cc cht ph thuc vo:- Khi lng phn t: cht c khi lng phn t ln hn, nhit si s cao hn- Lin kt hiro gia cc phn t: Cht c lin kt hiro c nhit si cao hn cht

khng c lin kt hiro (Trong cc cht u c lin kt hiro th cht no c lin kt hiro mnh hns c nhit si cao hn nh axit v ru, axit c lin kt hiro mnh hn ru nn axit c nhit si ln hn ru)Cu 42. p n C

ng vi cng thc phn t C3H6O2:- Khng th l ancol 2 chc cha no c mt lin kt pi:

CH=CH-CH2-OHOH v ancol ny khng tn ti

- Cng khng th l xeton hay anehit no 2 chc nhCH=CHO, H-C-C-CH3CHO O O

V 2 cht khng ng ng cng thc phn t choCu 43. p n D

Axit c gc y electron nh R- lm gim tnh axit (HCOOH c tnh axit mnh hnCH3COOH, v gc CH3

- y electron). Tri li, axit c gc c nhm th ht electron (nh halogen slm tng tnh axit); gc cng nhiu nhm th ht electron, tnh axit cng mnh

Do axit CCl3-COOH mnh nhtCu 44. p n C

Xem cch gii tng t cu 10 9Cu 45. p n D

Gi A1 v A2 l s khi ca 2 ng v, theo :A1 = 35 + 44 = 79A2 = 35 + 46 = 81

Theo trn v , xc nh c khi lng nguyn t trung bnh ca nguyn t l:79.27 + 81.23A= = 79,92

27 + 23

Cu 46. p n D

RCOOH RCOO- + H+

mol b: 0,1Mol li: 0,1.0,05 0,005 0,005Mol Cb: 0,095 0,005 0,005

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Vy Ka =2

-4(0,005) = 2,6.100,095

Cu 47. p n D

Phn ng xy ra trong bnh: S + O2 SO2C + O2 CO2

Theo cc phng trnh phn ng trn, s mol kh trc (n1) v sau phn ng khng i (n2)Theo th th tch bnh v nhit cng khng iDo , t PV = nRT cho ta:

1 1

2 2

P n= = 1

P n(v n1 = n2) (P1 v P2 l p sut trc v sau phn ng)

P1 = P2 = 2 atmCu 48. p n B

2O

10,08n = = 0,45 (mol)

22,4

t CnH2n+3N l amin no n chcPhn ng t chy amin:

CnH

2n+3N +

2

6n + 3O

4

nCO

2+

2n + 3

2

H

2O +

2

1N

2

Mol:6,2

14n + 17

6,2 6n + 3.

14n + 17 4

Theo trn v ta c:6,2

14n + 17.

6n + 3

4=

10,08= 0,45

22,4 n = 1

Vy amin l CH3NH2Cu 49. p n D

T 2 olefin ng ng lin tip suy ra 2 anehit cng l ng ng lin tip.t cng thc phn t trung bnh ca hai anehit n chc, no: n 2n+1C H CHO c x molCc phn ng xy ra:

n 2n+1C H CHO + H2 oNi

t n 2n+1 2C H CH OH Mol x x

n 2n+1 2C H CH OH 2 4oH SO d

170 C +1 2(n+1)C H

Mol x x

+1 2(n+1)C H + 23(n + 1)

O2

ot ( n +1)CO2 + ( n +1)H2O

Mol x 3(n + 1)x

2( n +1)x

S mol O2 ban u =1,68

= 0,075 (mol)22,4

S mol kh sau phn ng (gm CO2, O2 d) = 0,055 (mol)Hay:

2On d +

2COn = 0,075 -

3(n + 1)x

2+ ( n +1)x = 0,055

( n +1)x = (0,075 0,055.2) = 0,04 (1)Khi lng anehit: (14 n + 30).x = 0,8 (2)T (1) v (2) ta c x = 0,015 (mol)

T (1) cho ta: n =0,04 0,04

- 1 = - 1 = 1,66x 0,015

1 < n = 1,66 < 2,3

Vy n = 1 CH3CHO; n = 2 C2H5CHO

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Cu 50. p n C

nHCl = 400 .2 = 0,8 (mol)1000

Cc phn ng xy ra:Fe3O4 + 8HCl 2FeCl3 + FeCl2 + 3H2O (1)

Mol 0,1 0,8 0,2 0,1Cu + 2FeCl3 CuCl2 + 2FeCl2 (2)

Mol 0,1 0,2 0,1 0,2Sau phn ng (2): nCu d = 0,2 0,1 = 0,1 mol, chnh l cht rn B.

a = mCu d = 0,1.64 = 6,4 (gam) Cng sau phn ng (2):

2FeCln = 0,1 + 0,2 = 0,3 (mol) ;

2CuCln = 0,1 (mol)

FeCl2 + 2NaOH Fe(OH)2 + 2NaCl (3)Mol 0,3 0,3

CuCl2 + 2NaOH Cu(OH)2 + 2NaCl (4)Mol 0,1 0,1

2Fe(OH)2 + 21

O

2

+ H2O 2Fe(OH)3 (5)

Mol 0,3 0,32Fe(OH)3

ot Fe2O3 + 3H2O (6)

Mol 0,3 0,15

Cu(OH)2ot CuO + H2O (7)

Mol 0,1 0,1T (3), (5), (6), (7) rt ra: b =

2 3Fe O CuOm + m = 0,15.160 + 0,1.80 = 32 (gam)