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gii chi tit thi th i hc - m 702

Cu 1. p n DCc cht kt ta, khng tan khng to c dung dch

Cu 2. p n BGi PA l s proton c trong A, PB l s proton trong B, theo phn t AB3 c: PA + 3PB = 40 PA = 40 3PBV A chu k 3 nn: 11 PA 18 11 40 3PB 18Khi 40 3PB 11 PB 9,7Khi 40 3PB 18 PB 7,3T trn: 7,3 PB 9,7PA, PB ch c th nhn cc gi tr hp l: PA = 16; PB = 8Vy A l lu hunh v B l oxi

Cu 3. p n C

HF phn li: HF H+ + F-

Dung dch HF c pH = 2; Ka = 6,6.10

-4

ngha l [H

+

] = 10

-2

v Ka =

+ -. F

HF

, trong [HF] = C 10-2 C (v 10-2 thng qu nh so vi C nn c th b qua)

V th: [H+]2 = Ka.C C =

2 2+ -2

-4

a

H 10= = 0,152 (M)

K 6,6.10

Cu 4. p n B

Cu 5. p n CCng thc tng qut ca ru n chc l: CnH2n+2-2aO (a: s lin kt pi)

ca ru a chc CnH2n+2-2aOmSo snh cc cng thc cho trong vi cc cng thc tng qut trn cho thy:

CnH2nOH hay CnH2n+1O sai v s nguyn t H l s l

CnH2n+2(OH)2 hay CnH2n+4O2 sai v d s nguyn t HCH3-CH(OH)2 khng bn (khng tn ti), v 2 nhm OH nh vo cng mt nguyn t C

Cu 6. p n B2+Cl CH3-CH2Cl NaOH CH3-CH2-OH 2

O

men CH3COOH

CH3-CH3 O2 Mn2+ot

xt CH2=CH2 4

dd KMnO CH2OH-CH2OH 2 4H SO CH3-CHOCu 7. p n D

Dng Cu(OH)2 lm thuc thVi ru n propanol: Khng c hin tng g xy ra, v thuc th khng c phn ng vi

ru

Vi glixerin: Cho dung dch mu xanh lam, vCH2-OH CH2-O O-CH2

CuCH-OH + Cu(OH)2 CH-O O-CH + 2H2O

H HCH2-OH CH2-OH CH2-OH

(mu xanh)

Vi glucoz:

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Cho dung dch mu xanh lam nhit thng, khi un nng cho kt ta nu, v

CHO CHO

CHO CH-O O-CHCu

(CHOH)4 + Cu(OH)2 CH-O O-CHot

Cu2O + H H ( nu)

CH2OH (CHOH)2 (CHOH)2

CH2OH CH2OH(xanh lam)

Vi axit propionic: Cho dung dch mu xanh rt nht, vCu(OH)2 + 2CH3CH2COOH (CH3CH2COO)2Cu + 2H2O

(xanh rt nht)Cu 8. p n B

Gluxit c phn ng trng bc nn l mt anozK hiu cng thc ca X l R-CHO (a mol)Phn ng trng bc:

R-CHO + Ag2O 3dd NH R-COOH + 2Ag

a mol 2a mol

Theo trn v : 2a =1,728

= 0,046108

a = 0,008

Ta cng c: a =1,44

= 0,008R + 29

R = 151

V th nn: (CH2O)n = 151 + 29 = 30n n =180

= 630

Suy ra cng thc phn t ca X l C6H12O6Cu 9. p n C

Cc cht: Na2CO3; K3PO4; Ca(OH)2 v NaOH u c th lm mt tnh cng ca nc, v:Na2CO3 + Ca(HCO3)2 CaCO3 + 2NaHCO32K3PO4 + 3Ca(HCO3)2 Ca3(PO4)2 + 6KHCO3Ca(OH)2 + Ca(HCO3)2 2CaCO3 + 2H2O2NaOH + Ca(HCO3)2 CaCO3 + Na2CO3 + 2H2O

Cu 10. p n B

Cc cht H2SO4 c, P2O5 khng lm kh c NH3, v chng phn ng vi NH32NH3 + H2SO4 (NH4)2SO46NH3 + P2O5 + 3H2O 2(NH4)3PO4

Cc dung dch c Ba(OH)2, Ca(OH)2 khng phn ng vi NH3, nhng khng hp th H2OCn li, ch CaO mi lm kh c NH3, v CaO c tnh baz nn khng phn ng vi baz

m ch hp th nc: CaO + H2O Ca(OH)2Cu 11. p n D

V d, in phn dung dch mui Cu(NO3)2 s thu c dung dch axit, v:

Cu(NO3)2 + H2Odpdd Cu + 2HNO3 + 2

1O

2

Tri li, khi in phn mui NaCl, s thu c dung dch baz v:

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2NaCl + 2H2Opdd

c vch ngn NaOH + H2 + Cl2;

Cu 12. p n Dt cng thc ru no mch h X l CnH2n+2Om

Phn ng t chy: CnH2n+2Om +ot

2 2 2

3n + 1 - mO nCO + (n + 1)H O

2

Theo ta c:

3n + 1 - m 6 + m

= 3,5 n =2 3 Nghim hp l khi m = 3 n = 3Vy cng thc phn t ca ru X l C3H8O3

Cu 13. p n C

Phn ng x phng ha este: RCOOR'+ NaOH RCOONa + R'OH Theo trn v :

neste = nNaOH = nmui = 1.0,9 = 0,9 (mol) Meste =66,6

= 74 (g)0,9

Gi CnH2nO2 l cng thc phn t ca A, B theo kt qu trn ta c:14n + 32 = 74 n = 3

Vy cng thc phn t ca A, B l C3H6O2Cng thc cu to ca A, B:A: H-C-O-CH2-CH3; B: CH3-C-OCH3

O O Phn ng x phng ha ca A, B:

HCOOC2H5 + NaOH HCOONa + C2H5OHCH3COOCH3 + NaOH CH3COONa + CH3OH

t2 5 3 3HCOOC H CH COOCH

n = x; n = y

Theo trn v :x + y = 0,9

68x + 82y = 65,4

Gii h phng trnh c:x = 0,6 2 5HCOOC H

= 74 . 0,6 = 44,4 (gam)

y = 0,3 3 3CH COOCH

= 74 . 0,3 = 22,2 (gam)

Cu 14. p n CPhn ng ha tan oxit st

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

2SO

2,24n = = 0,2 (mol)

22,4;

2 4 3Fe (SO )

120n = = 0,3 (mol)

400

Theo trn ta c: (3x 2y).0,3 = 0,1.x 0,8.x = 0,6y x 0,6 3

= =y 0,8 4

Vy cng thc ca oxit st l Fe3O4Cu 15. p n C

Gi M l kim loi ha tr IIIM + 4HNO3 M(NO3)3 + NO + 2H2O

mol 0,05 0,2 0,0510M + 36HNO3 10M(NO3)3 + 3N2 + 18H2O

mol: 0,25 0,9 0,075t a l s mol kh NO c trong hn hp

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b l s mol kh N2 trong hn hp

Theo ta c: a + b =2,8

= 0,12522,4

(1)

Khi lng trung bnh ca hn hp kh = 4.d = 4.7,2 = 28,8

30a + 28b

= 28,8a + b

30a + 28b = 28,8.0,125 (2)

Gii h phng trnh (1), (2) c: a = 0,05 mol; b = 0,075 mol Tng s mol kim loi M l: nM = 0,05 + 0,25 = 0,3 (mol)

Vy M =m 8,1

= = 27 (gam)n 0,3

Kim loi M l Al

S mol HNO3 phn ng: 0,2 + 0,9 = 1,1 (mol)S mol HNO3 ban u: 2,5.0,5 = 1,25(mol)S mol HNO3 d: 1,25 1,1 = 0,15 (mol)

3dd HNO ban um : 2500.1,25 = 3125 (gam)

mdd sau phn ng =3dd HNO

m + mAl (mNO +2N)

= 3125 + 8,1 (30.0,05 + 28.0,075)= 3129,5 (gam)

Vy nng dd HNO3 d = 0,15.63.100% = 0,3%3129,5

Cu 16. p n DCc phn ng cho u xy ra:

C6H5ONa + CO2 + H2O C6H5OH + NaHCO3

Cl2 + 2KI I2 + 2KCl;

SiO2 + 2Cot Si + 2CO;

Cu 17. p n CGi Z l s proton, ng thi cng l s electron, N l notron, theo ta c:

2Z + N = 21 (1)

Suy ra: Z = 10,5 -N

2 Z 10 (2)

Ta li bit, do Z 10 nn 1 N

Z 1,2 N 1,2Z

Thay gi tr ca N vo (1) c:2Z + 1,2Z 21 hay Z 6,5. Kt hp gi tr ny vi (2) c: 6 Z 10Theo ta cng c: A = 21 ZKt hp hai biu thc trn lp bng:

Z 6 7 8 9 10

A 15 14 13 12 11N 9 7 5 3 1Nghim duy nht ng l nguyn t nit: 14

7 N

Cu 18. p n D

Theo pH = 3 [H+] = 10-3 mol/ltHA H

+ + A-S mol ban u ca HA l 0,1S mol HA in li l 0,001

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Vy in li =0,001

= 0,01 = 1%0,1

Theo phng trnh phn ng phn li trn, khi cho thm HCl, cn bng in li ca HA schuyn sang tri, nn in li gim

Cu 19. p n CCu 20. p n C

Cu 21. p n BCu 22. p n B

CH2=CH2 + H2Oo

2 4H SO , t CH3-CH2-OH

CH3-CH2-OHoCuO, t CH3-CHO + H2O + Cu

CH3-CHO 2 3Ag O / NH CH3-COOH + 2Ag CH2=CH2 + Br2 (dd) CH2Br-CH2Br

CH2Br-CH2Br + 2NaOHot CH2OH-CH2OH + 2NaBr

CH2-OH CH3-COO-CH3

+ 2CH3 COOHo

2 4H SO , t + 2H2O

CH2-OH CH3-COO-CH3(X)

Cu 23. p n DAxit axetic c nhm th ht electron c tnh axit mnh hn axit khng c nhm th (v nhm htelectron lm cho lin kt O H trong phn t axit c nhm - O C , d t nn d to H+)

O-HTrong cc halogen, halogen no c m in mnh hn, tnh ht electron cng mnh, lm cho axitc tnh axit cng mnhCu 24. p n D

Cu 25. p n BDng thm qu tm:

Dung dch lm qu tm chuyn sang mu xanh, l Ba(OH)2Dung dch lm qu tm chuyn sang mu hng, l HCl, H2SO4 v NH4HSO4Dung dch khng lm chuyn mu qu l BaCl2, NaClLy Ba(OH)2 cho vo 3 dung dch HCl, H2SO4, NH4HSO4. Cht no to ra kt ta trng l

H2SO4; cht no to ra kh mi khai v kt ta trng l NH4HSO4; v dung dch no khngc hin tng g l HCl, v

Ba(OH)2 + H2SO4 BaSO4 + 2H2O

Ba(OH)2 + NH4HSO4 NH3 + BaSO4 + 2H2O

Ly dung dch H2SO4 nhn bit dung dch BaCl2 v NaCl: cht no cho kt ta trng lBaCl2, v

H2SO4 + BaCl2 BaSO4 + 2HClCu 26. p n B

Da vo nguyn tc lm mm nc, c th dng cc ha cht sau kh cng tm thi v cng vnh cu ca nc: Na2CO3, Na3PO4, C17H35COONa, v:

2- 2+ 2+

3 3 3CO + (Ca , Mg ) CaCO + MgCO 3- 2+ 2+

4 3 4 2 3 4 2PO + (Ca , Mg ) Ca (PO ) + Mg (PO )

- 2+ 2+17 35 17 35 17 35 22C H COO + (Ca ,Mg ) C H COO Ca + (C H COO) Mg

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Cu 27. p n DPhn ng trung ha axit:

RCOOH + NaOH RCOONa + H2O (1)mol: 0,19 0,19.1Phn ng to este:

RCOOH + C2H5OH RCOOC2H5 + H2O (2)

S mol axit theo phn ng (1) l 0,19 molS mol ancol C2H5OH theo l 9,2/46 = 0,2 mol, v th lng este thu c tnh theo lng

axit:Theo phn ng (2):1 mol axit 1 mol este, nh vy 1 mol este so vi 1 mol axit tng 28 gamVy 0,19 mol axit lng este tng hn lng axit l: 28 . 0,19 = 5,32 gamSuy ra, lng este thu c theo l thuyt l: 10 gam + 5,32 gam = 15,32 gam

V lng este thu c thc t l: 15,32 .90

= 13,788 gam100

Cu 28. p n C

Theo : MY = 44.2 = 88 gam

nY =0,44

= 0,005 (mol)88

Khi t chy Y to ra CO2 v H2O c s mol bng nhau nn phn t Y c 1 ni i.Gi cng thc ca Y l CnH2nOm:

CnH2nOm + 23n - m

O2

ot nCO2 + nH2O

Khi lng bnh tng a gam chnh l khi lng ca CO2 v H2O, nn theo phn ng t chytrn, ta c:

44.0,005n + 18.0,005n = aHay 62.0,005n = a

Mt khc: 14n + 16m = 88m ch c th = 2 n = 4Vy cng thc phn t ca Y l C4H8O2. a = 62.0,005.4 = 1,24 gam

Cu 29. p n CTheo cc phn ng ca kim loi kim v kim th vi nc th sn phm thu c lun c:

-2HOH

n = 2n = 0,12.2 = 0,24 (mol)

V th: H+ + OH- H2Omol: 0,24 0,24

Do :2 4H SO

n = 0,12 mol 0,12

V = = 0,06 (lt) = 60 ml

2

Cu 30. p n Ct cng thc ca oxit st l FexOy v lng oxit st trong y gam l n molKhi ha tan trong H2SO4:

2FexOy + (6x 2y)H2SO4 2Fe2(SO4)3 + (3x 2y)SO2 + (3x 2y)H2O

n mol(3x - 2y)

n2

mol

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Phn ng kh: FexOy + yCOot xFe + yH2O

mol n xn

Ha tan st: 2Fe + 6H2SO4 ot Fe2(SO4)3 + 3SO2 + 6H2O

mol nx 1,5nx

Theo trn v ta c: 1,5nx =9(3x - 2y)n

2

24x = 18y x 18 3

= =

y 24 4

Vy cng thc ca oxit st l Fe3O4Cu 31. p n C

Cu 32. p n CGi P l s proton, ng thi cng l s electron

N l s notron, theo ta c: 2P + N = 155 (1)Cng theo : 2P N = 33 (2)T (1) v (2) 4P = 188 P = 47T (2) N = 155 94 = 61Vy s khi ca A = 47 + 61 = 108

Cu 33. p n D

t lng SO2 trong A l x mol, theo th lng khng kh l 5x mol, trong c x mol O2v 4x mol N2

2SO2 + O2oxt, t

2SO3

Theo : AB

M= 0,93

M; m mA = mB (v bo ton khi lng), nn:

nB = 0,93.nA = 0,93.6x = 5,58xS mol O2 d phn ng = 6x 5,58x = 0,42xS mol SO2 d phn ng = 2.0,42x = 0,84x

Vy hiu sut phn ng: h =0,84x

.100% = 84%x

Cu 34. p n DS mol ion H+ trong 10 ml ban u cng chnh l s mol H+ trong 250 ml dung dch sau khi

pha long c pH = 3

T pH = 3 [H+] = 10-3M +-3

-4

H

10 .250n = = 2,5.10

1000(mol)

Vy nng ca HCl trc khi pha long l:-4

-22,5.10 .1000 = 2,5.10 (M)10

Cu 35. p n D NH2Cu 36. p n C Br Br

3o

2 4

HNO ,1mol

H SO , t C6H5NO2

Fe + HCl dd C6H5NH3ClNaOH C6H5NH2 2

Br

BrC6H6 (1 mol) OHBr Br

2Cl 1 mol

bt Fe C6H5Cl

dd NaOH d C6H5ONadd HCl C6H5OH 2

Br

BrCu 37. p n B

Dng thuc th l Cu(OH)2, ln lt cho vo tng cht:Cht no khng ha tan Cu(OH)2 l etyl axetat

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Cht no to ra dung dch xanh lam l C2H4(OH)2, v

CH2-OH CH2-O O-CH22 + Cu(OH)2 Cu + 2H2OCH2-OH CH2-O O-CH2

H H (xanh lam)

Cht no to ra kt ta son l CH3CHO:CH3CHO + 2Cu(OH)2

ot CH3COOH + Cu2O () + 2H2OCu 38. p n C

Phn ng t chy A:

CnHmO2o

2O , t nCO2 + 2m

O2

mol: x nxx

2

Theo ta c: nx +x

2= 3x

Hay: 2n + m = 6Lp bng:n 1 2m 4 2Ch c gi tr n = 2, m = 2 l ph hp vi . Vy A l HOOC-CHOCu 39. p n D

Cu 40. p n C

Phn ng ha tan hp kim vo nc:2Na + 2H2O 2NaOH + H2

Ba + 2H2O Ba(OH)2 + H2 Qua cc phn ng trn cho thy s mol OH- thu c lun bng 2 ln s mol H2 nn theo ta

c:

-2HOH

6,72n = 2n = 2. = 0,6 (mol)

22,4

1

10dung dch X c 0,06 mol OH-

Phn ng trung ha:H+ + OH- H2O

mol: 0,06 0,06

Vy th tch dung dch HCl l0,06

= 0,061

lt = 60 ml

Cu 41. p n DGi a, b, c ln lt l s mol Fe, Fe2O3, Fe3O4 v d l s mol Fe st cha phn ng, ta c:

Fe + 21 O2

FeO

Mol: a a

4Fe + 3O2 2Fe2O3

Mol: b b

2

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3Fe + 2O2 Fe3O4

Mol: c c

3

3FeO + 10HNO3 3Fe(NO3)3 + NO +5H2OMol: a

a

3

Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O

3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + 14H2O

Mol: c3

c

9

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O

Mol: d dTheo cc phn ng trn v ta c:

b c72a + 160 + 232 + 56d = 12 (1)

2 3

a c 2,24+ + d = = 0,1 (2)3 9 22,4

T (1) 216a + 24b + 232c + 168d = 36 (3)(2) 3a + c + 9d = 0,9

Hay 24a + 8c + 72d = 7,2 (4)T (1) + (4) 240(a + b + c + d) = 43,2Hay a + b + c + d = 0,18Vy mFe = 56(a + b + c + d) = 56.0,18 = 10,08 (gam)

Cu 42. p n Dt cng thc ca axit l CnH2n+2-2mO2m (v axit no a chc nn m 2)

Phn ng t chy axit:CnH2n+2-2mO2m 2o

O

t nCO2 + (n + 1 m)H2O

Theo trn v ta c:7,3

.n = 0,314n + 30m + 2

(1)

7,3.(n + 1 - m) = 0,25

14n + 30m + 2(2)

T (1) v (2) cc nghim hp l ch c th l m = 2; n = 6V mch cacbon khng phn nhnh nn cng thc cu to ca axit phi l

HOOC-(CH2)4-COOHCu 43. p n C

Phn ng craking n butan (a mol)

C4H10 CH4 + C3H6Mol: x x x

C4H10 C2H4 + C2H6Mol: y y yHn hp thu c: CH4 = x mol; C3H6 = x mol

C2H6 = y mol; C2H4 = y molC4H10 d = a - (x + y)

Phn ng vi brom:

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C3H6 + Br2 C3H6Br2

Mol: x xC2H4 + Br2 C2H4Br

Mol: y y

Theo trn v ta c:160x + 160y = 25,60

42x + 28y = 5,32

Gii h trn c: x = 0,06 mol; y = 0,10 molNh vy, kh cn li: CH4 = 0,06 mol; C3H6 = 0,01 mol; C4H10 d = a 0,16

Theo c:(16.0,06) + (42.0,10) + 58(a - 0,16)

= 1,962516a

a = 0,2

Vy hiu sut ca phn ng crackinh nbutan l

h =x + y 0,16

= = 0,80 = 80 %0,2 0,2

Cu 44. p n C

Gi M l kim loi: MCl2 M(NO3)2Theo trn: 1 mol MCl2 bin thnh 1 mol M(NO3)2, khi lng tng 53 gam

Vy khi c khi lng tng 2,65 gam s c 2,65 = 0,05 (mol)53

MCl2 bin i

Suy ra: M + 71 =104

0,05 M = 137. Vy kim loi M l Ba

Cu 45. p n C

nHCl =400

.2 = 0,8 (mol)1000

Khi cho hn hp vo dung dch HCl, c:Fe3O4 + 8HCl 2FeCl3 + FeCl2 + 4H2O

Mol: 0,1 0,8 0,2 0,1

Cu + 2FeCl3 CuCl2 + 2FeCl2Mol: 0,1 0,2 0,1 0,2Theo trn nCu phn ng l 0,1 mol x gam cht rn B chnh l Cu d, c s mol:

0,2 0,1 = 0,1 (mol)Vy x gam = 64 . 0,1 = 6,4 (gam)

Dung dch A c:2 2FeCl CuCl

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

Phn ng ca dung dch vi NaOH toFeCl2 + 2NaOH Fe(OH)2 + 2NaCl

Mol: 0,3 0,3CuCl2 + 2NaOH Cu(OH)2 + 2NaCl

Mol: 0,1

0,12Fe(OH)2 + 2

1O

2+ H2O 2Fe(OH)3

Mol: 0,3 0,3

2Fe(OH)3ot Fe2O3 + 3H2O

Mol: 0,3 0,15Cu(OH)2

ot CuO + H2OMol: 0,1 0,1Theo trn, cht rn C c y gam gm:

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2 3Fe On = 0,15 mol v nCuO = 0,1 mol

Vy y = 0,15.160 + 0,1.80 = 32 (gam)Cu 46. p n D

Cc cht phn ng c vi nhau:2NaOH + CuCl2 Cu(OH)2 + 2NaCl2NaOH + CO2 Na2CO3 + H2O

Al + NaOH + H2O NaClO2 + 23

H2

NaOH + NH4Cl NaCl + NH3 + H2O

Fe2O3 + 2Alot 2Fe + Al2O3

3CuCl2 + 2Al 2AlCl3 + 3CuCu 47. p n B

Khi thm dung dch H2SO4, nng H+ tng ln, theo nguyn l L Satli cn bng schuyn dch sang tri

Ngc li, khi thm dung dch NaOH cn bng s chuyn dch sang phi, v khi nng H+ gim, theo nguyn l trn

V khi thm dung dch Na2CO3 cng lm cho nng H+ gim, do cng theo nguyn ltrn nn cn bng chuyn dch sang phiCu 48. p n C

Nh dung dch NaOH vo c 3 l:Cht no to ra mu xanh lam l CuSO4, v:

CuSO4 + 2NaOH Cu(OH)2 + Na2SO4Cht no to ra kt ta mu xanh nht, trong khng kh m chuyn thnh mu nu l FeSO4,

v:FeSO4 + 2NaOH Fe(OH)2 + Na2SO44Fe(OH)2 + O2 + 2H2O 4Fe(OH)3

Cht no to ra kt ta mu xanh ru, tan c trong NaOH d l Cr2(SO4)3, v:Cr2(SO4)3 + 6NaOH 2Cr(OH)3 + 3Na2SO4Cr(OH)3 + NaOH NaCrO2 + 2H2O

Cu 49. p n C

Cu 50. p n C