7/28/2019 Gii chi tit 392

1/9

1

gii chi tit thi th i hc - m 392

Cu 1. Lp electron M (lp th n = 3) c 3 phn lp electron l 3s, 3p v 3d. Do c s electronti a l 2 + 6 + 10 = 18 electron. Vy lp electron M bo ha khi c 18 electron. p n ng l BCu 2. Cc nguyn t cng chu k (theo th t t tri sang phi l Na, Mg, Al

Cc nguyn t cng nhm (theo th t t trn xung di) l Na, K

Trong cng chu k, t tri sang phi tnh kim loi ca cc nguyn t gim dnTrong mt nhm, t trn xung di tnh kim loi ca cc nguyn t tng dn. Vy dynguyn t ca cc nguyn t c tnh kh tng dn l: Al, Mg, Na, K

p n ng l BCu 3. phn cc ca lin kt cng ln khi hiu m in gia 2 nguyn t cng ln

- Cc nguyn t F, Cl, Br, I cng nhm VIIA (nhm halogen). Da vo quy lut bin i m in ca cc nguyn t trong nhm (t trn xung di gim dn) ta c th c lng chiu m in tng i ca chng phn cc ca lin kt

p n ng l ACu 4. +2y/x +3Xt s : FexOy + HNO3 Fe(NO3)3 + l phn ng oxi ha kh +2y/x +3

+3

x 2, y 3 (Fe2O3)Cc trng hp cn li u tha mn l phn ng oxi ha khx = 1, y = 0 Fe (kim loi st)x = 1, y = 1 FeO (st (II) oxit)x = 3, y = 4 Fe3O4 (st t oxit)Vy c 3 cp gi tr x,y (x = 1, y = 0, x = 1, y = 1,x = 3, y = 4) tha mnp n ng l C

Cu 5. Trong cc phn ng, HCl th hin tnh oxi ha khi n tc dng vi kim loi (tnh oxi ha caion H+). Do trong cc phn ng vi kim loi, HCl th hin tnh oxi ha:

2HCl + Fe FeCl2 + H2

cht oxi ha cht kh6HCl + 2Al 2AlCl3 + 3H2

cht oxi ha cht kh c 2 phn ng. p n ng l DCu 6. thu c kh O2 tinh khit t hn hp kh O2, Cl2, SO2, CO2, c th dn hn hp i quadung dch NaOH d, khi Cl2, SO2, CO2 s b gi li trong dung dch, cn kh O2 thot ra:

Cl2 + 2NaOH NaCl + NaClO + H2OSO2 + 2NaOH Na2SO3 + H2OCO2 + 2NaOH Na2CO3 + H2O

O2 khng tc dng v khng tan trong dung dch NaOH nn thot ra khi dung dch. p n ng l

CCu 7. Xt phn ng: aA + bB cC + dDTc phn ng tnh theo biu thc:

v = k. (CA)x. (CB)

y (1)- Khi tng gp i nng A (CB = const):

v' = k. (2CA)x.(CB)

y = 2v (2)- Khi gim nng B hai ln (CA = const):

v" = k. (CA)x. (CB/2)

y = v/8 (3)T (1), (2) 2x = 2 x = 1T (1), (3) (1/2)y = 1/8 = (1/2)3 y = 3

7/28/2019 Gii chi tit 392

2/9

2

p n ng l BCu 8. Theo thuyt Bronstet (thuyt proton):

Axit l nhng cht c kh nng cho protonBaz l nhng cht c kh nng nhn proton

phn ng axit baz c s nhng v nhn protonTh d: HCl + NaOH dung dich NaCl + H2O

Nhng electron nhn proton

Axit Bazp n ng l C

Cu 9. Theo bi ra: nCu =19,2

= 0,364

(mol)

Cc qu trnh xy ra:+2

Cu 2e Cu0,3 0,6 (mol)+5 +2

N + 3e N0,6 0,2 (mol)

4NO + 3O2 + 2H2O 4HNO30,2 (mol)

2O

3.0,2n = = 0,15 (mol)

4

2O (ktc)V = 0,15.22,6 = 3,36 (lt)

p n ng l A

Cu 10. Xt cn bng: CaCO3ot

CaO + CO2(k), H > 0 cn bng trn chuyn dch theo chiu thun cn tng nhit phn ng v gim nng

kh CO2. p n ng l BCu 11. Cp tn ca cng mt hp cht l

Ancol isobutylic v 2 metylpropan 1 ol

CH3-CH-CH2OHCH3

p n ng l CCu 12. Gi s crackinh 1 mol C4H10 (58 gam C4H10). Suy ra, s mol hn hp sau phn ng:

nhh (sau p/) = 58/36,25 = 1,6 (mol)(v tng khi lng cc cht sau phn ng bng khi lng C4H10 v bng 58 gam)Trong phn ng crackinh, ta thy c 1 mol C4H10 phn ng th sinh ra 2 mol ca hai

hirocacbon, do s mol C4H10 tham gia phn ng ng bng s mol hirocacbon tng sau phnng. Vy:

4 10C H (p/-)n = 1,6 1 = 0,6 mol

Hiu sut phn ng: H =0,6

.100% = 60%

1

p n ng l D

Cu 13. PTHH: CxHy2+O xCO2 + 2

yO

2

V nguyn t C, H c bo ton, nn:

X Y X Y X Y 2 2C (C H ) H (C H ) C H C (CO ) H (H O)m + m = m = m + m

m =11 9

.12 + .2.1 = 4,0 (gam)44 18

p n ng l A

7/28/2019 Gii chi tit 392

3/9

3

Cu 14. p n ng l BCu 15. Cc loi lin kt hiro c th c hnh thnh

C2H5OHOHC6H5; C2H5OH.OHC2H5C6H5OHOHC2H5; C6H5OH.OHC6H5

C th c 4 loi lin kt hirop n ng l DCu 16. Xt cc phng n:

A. ng: CH3CH2COOCH=CH2 + Br2 CH3CH2COOCHBr-CH2BrB. ng CH3CH2COOCH=CH2 + NaOH CH3COONa + CH3CHOC. ng nCH=CH2 T.H (-CH-CH2-)n

CH3CH2COO CH3CH2COOD. Sai CH3CH2COOCH=CH2 khng cng dy ng ng vi

CH2=CHCOOCH3p n ng l D

Cu 17. p n ng l CCu 18. Cc cht A, B, C l:

A. H2N-C3H6-COOH: nguyn t N c lai ha sp3

B. C4H9-NO2; nguyn t N c lai ha sp2

C. C3H5COONH4 nguyn t N c lai ha sp3

*Cc PTHH xy ra:H2N-C3H6-COOH + HCl ClH3N-C3H6-COOH2H2N-C3H6-COOH + Na2O 2H2N-C3H6-COONa + H2OC4H9-NO2 + 6[H] C4H9NH2 + 2H2O

B BC4H9NH2 + HCl C4H9NH3Cl

B BC4H9NH3Cl + NaOH C4H9NH2 + NaCl + H2O

B BC3H5COONH4 + NaOH C3H5COONa + NH3 + H2Op n ng l A*Ch : xc nh (d on) trng thi lai ha ca nguyn t cn da vo tng s lin kt (zichma) v cp electron cha chia (cp electron ha tr cha tham gia lin kt):

(n + n cp e ch-a chia) = A Nu A = 4 lai ha sp3; A = 3 lai ha sp2; A = 2 lai ha sp.

Cu 19. c im cu to ca monome tham gia phn ng trng hp l c lin kt bi (thng llin kt i). p n ng l ACu 20. Kh H2, nhit cao kh c cc oxit ca cc kim loi yu (CuO), kim loi trung bnh(sau Al, nh Fe2O3, ZnO):

H2 + CuO ot Cu + H2O

3H2(d) + Fe2O3ot 2Fe + 3H2O

H2 + ZnO ot Zn + H2O

MgO + H2ot khng phn ng (vn cn MgO)

Sau phn ng, trong cht rn c Cu, Fe, Zn, MgOp n ng l B

Cu 21. Theo bi ra:2 3Cr O

15,2n = = 0,1 (mol)

152

7/28/2019 Gii chi tit 392

4/9

4

Trong phn ng nhit nhm: mAl +2 3Cr O

m = mhn hp rn sau phn ng

mAl = 23,3 15,2 = 8,1 (gam) nAl = 0,3 (mol)PTP: 2Al + Cr2O3

ot 2Cr + Al2O3Ban u: 0,3 0,1 (mol)Phn ng: 0,2 0,1 0,2 (mol)Cn li: 0,1 (mol) 0,2 (mol)

Al + 3HCl

AlCl3 + 1,5H2

0,1 (mol) 0,15 (mol)Cr + 2HCl CrCl2 + H2 0,2 (mol) 0,2 (mol)

2H

n = 0,15 + 0,2 = 0,35 (mol) 2H

V = 0,35.22,4 = 7,84 (lt)

p n ng l CCu 22. Theo bi ra:

2 4H SOn = 0,5.0,1 = 0,05 (mol)

S phn ng xy ra: Oxit + H2SO4 mui sunfat + H2O- p sng nh lut bo ton nguyn t (cho nguyn t H), ta c:

2 2 4H O H SOn = n = 0,05 (mol)

- Theo nh lut bo ton khi lng, ta c:moxit +

2 4H SOm = mmui +

2H Om

mmui = moxit +2 4H SO

m -2H O

m = 2,81 + 0,05.98 0,05.18 = 6,81 (gam)

p n ng l DCu 23. Cch tt nht l dng dung dch NaHCO3 d:

HCl + NaHCO3 CO2 + NaCl + H2Op n ng l C

*Ch :

- Khng dng nc ct, tuy HCl tan tt trong nc, cn CO2 tan rt t trong nc nhng CO2s li cun HCl i theo

- Khng dng dung dch AgNO3, v:HCl + AgNO3 AgCl + HNO3CO2 s li cun HNO3 i theo

Cu 24. trng gng, trng rut phch, ngi ta s dng phn ng:CH2(OH)[CH(OH)]4CHO + Ag2O 3

NH 2Ag + CH2(OH)[CH(OH)4]COOHGlucoz

Bc sinh ra mt cch t t, u n v bm chc vo gng, rut phchp n ng l A

*Ch : Ngi ta khng s dng anehit trng gng, trng rut phch, v phn ng xy ranhanh, bc sinh ra khng u v tnh cht c hi ca anehit

Cu 25. Z = 17: 1s

2

2s

2

2p

6

3s

2

3p

5

([Ne] 3s

2

3p

5

)

c 7 electron ha tr

nguyn t phi kim. pn ng l BCu 26. p n ng l ACu 27. phn cc ca lin kt ph thuc vo hiu m in gia 2 nguyn t tham gia linkt

- Da vo s bin thin hiu m in ca cc nguyn t trong cng chu k, cng nhm hiu m in tng i trt t phn cc tng i ca lin kt

p n ng l CCu 28. Phn ng trao i lun khng phi l phn ng ha kh (v s oxi ha ca cc nguyn tkhng thay i)

7/28/2019 Gii chi tit 392

5/9

5

p n ng l DCu 29. Cc cht u phn ng c vi dung dch HCl, dung dch NaOH l:

- 2Al + 6HCl AlCl3 + 3H2- 2Al + 2NaOH + 6H2O 2NaAl(OH)4 + 3H2- Al2O3 + 6HCl 2AlCl3 + 3H2O- Al2O3 + 2NaOH + 3H2O 2NaAl(OH)4- Zn(OH)2 + 2HCl ZnCl2 + 2H2O- Zn(OH)2 + 2NaOH Na2[Zn(OH)4]- NaHS + HCl NaCl + H2S- NaHS + NaOH Na2S + H2O- (NH4)2CO3 + 2HCl 2NH4Cl + CO2 + H2O- (NH4)2CO3 + 2NaOH 2NH3 + Na2CO3 + H2O c 5 chtp n ng l A

Cu 30. Xy ra phn ng: 3O2(k) phng in 2O3(k)Ta thy 3 th tch kh O2 phn ng to ra 2 th tch kh O3 gim 1 th tch kh

Nh vy c gim 1 th tch kh th to ra 2 th tch kh O3. Do khi gim 9 lt kh th to ra x

lt kh O3 x =

9.218

1(lt)

p n ng l CCu 31. Gi s th tch ca h bng 1 lt

CH3COOH + C2H5OH CH3COOC2H5 + H2OBan u: 1 mol 1 molPhn ng: 2/3 mol 2/3 mol 2/3 mol 2/3 molCn bng: 1/3 mol 1/3 mol 2/3 mol 2/3 molHng s cn bng ca phn ng:

KC = 3 2 5 2

3 2 5

CH COOC H . H O 2/3.2/3= = 4CH COOH . C H 1/3.1/3

CH3COOH + C2H5OH CH3COOC2H5 + H2OBan u: 1 mol x molPhn ng: 0,9 mol 0,9 mol 0,9 mol 0,9 molCn bng: 0,1 mol (x 0,9)mol 0,9 mol 0,9 molV T = const nn KC = 4, do :

KC =0,9.0,9

= 40,1.(x 0,9)

0,81 = 0,4.x 0,36 x = 2,925 mol

V V = 1 lt 2 5C H OH

n = 2,925 (mol) . p n ng l D

*Ch :

+) Phn ng este ha: RCOOH + ROH RCOOR + H2O

Ni dung: KC =

2RCOOR' . H O

RCOOH . R'OH

Khi dung dch long th [H2O] coi l hng s, do :

KC =

RCOOR'

RCOOH . R'OH

7/28/2019 Gii chi tit 392

6/9

6

+) Khi cho axit cacboxylic v ancol nguyn cht (khng dng dung dch) phn ng vi nhau thnh:

+ Tnh nng cc cht (c th gi s th tch ca h bng 1 lt)+ [H2O] c mt trong biu thc hng s cn bng ca phn ng

Cu 32. Xt cc phng n:A. loi, v c phng trnh ion rt gn:

Ba2+ + 2OH- + 2H+ + SO42 BaSO4 + 2H2O

B. loi, v c phng trnh ion rt gnBa2+ + 2CH3COO

- + 2H+ + SO42 BaSO4 + 2CH3COOH

C. ngBa2+ + SO4

2 BaSO4 D. loi, v c phng trnh ion rt gn:

Ba2+ + 2OH- + Mg2+ + SO42 BaSO4 + Mg(OH)2

p n ng l CCu 33. Phn ln supephotphat n c thnh phn:

Ca(H2PO4)2 v CaSO4.2H2Op n ng l A

*Ch : Phn ln supephotphat n c sn xut qua 1 giai onCa3(PO4)2 + H2SO4 (c) 2Ca(H2PO4)2 + CaSO4

Cu 34. Theo bi ra:2CO

6,72n = = 0,3 (mol)

22,4; nNaOH = 0,8.1 = 0,8 (mol)

V2

NaOH

CO

n 0,8 8= = > 2

n 0,3 3 To ra Na2CO3; NaOH d

PTHH: CO2 + 2NaOH Na2CO3 + H2Omol 0,3 0,6 0,3

2 3 2Na CO COn = n = 0,3 (mol)

2 3Na CO= 0,3.106 = 31,8 (gam)

nNaOH (d) = 0,8 0,6 = 0,2 (mol) mNaOH = 0,2.40 = 8 (gam)

Vy, khi lng cht rn sau phn ng lmrn =2 3Na CO NaOH (du)

m + m = 31,8 + 8 = 39,8 (gam)

p n ng l CCu 35. Hp cht cha C, H, O

2 2CO H On = n

Hp cht c CTPT dng CnH2nOaXt cc phng n:A. Axit cacboxylic no, a chc:

CxH2x+2-y (COOH)y Cx+yH2x+2O2y CnH2n+2-2yO2yV y 2 CnH2n+2-2yO2y CnH2nOa Hp cht khng phi l axit cacboxylic no, a chc

B.Anehit no, h, n chc:CXH2x+1CHO CX+1H2x+2OCnH2nO

tha mnC. Este n chc to bi axit no, h v ancol no, h:

CXH2x+1COOH + CyH2y+1OH CXH2x+1COOCyH2y+1 + H2OCXH2x+1COOCyH2y+1 CX+Y+1H2x+2y+2O2 CnH2nO2

tha mnD. Ancol khng no cha mt ni i, mch h

CxH2x-y(OH)y CxH2xOy CnH2nOa CxH2xOy tha mnp n ng l A

7/28/2019 Gii chi tit 392

7/9

7

Cu 36. Xt cc phng n:

A, B ng: Ankan CnH2n+2 (n 1)C, ng: (cc loi lin kt n C-C v C-H)D, sai. V d H-O-H ch c lin kt n trong phn t nhng khng phi l ankan

p n ng l DCu 37. Gi s t chy hon ton x mol hirocacbon A thu c 2 mol CO2 v 1 mol H2O

Suy ra * mA =2 2

C(CO ) H(H O)+ m = 2.12 + 1.2.1 = 26 (gam)

*2O

n va t =2 2 2 2O (CO ) O (H O)

n + n = 2.1 + 1.0,5 = 2,5 (mol)

2O

n ( ly d) = 0,2.2,5 = 0,5 (mol)

2 2CO O

n + n (d) = 2 + 0,5 = 2,5 (mol) nA = 2,5/2,5 = 1 (mol)

Vy MA = AA

m 26= = 26

n 1(C2H2). p n ng l C

Cu 38. p n ng l CCu 39. Quy tc tch Zaixep: Trong phn ng tch nc ca ancol, nhm OH u tin tch ra cngvi H nguyn t C bc cao hn bn cnh to thnh lin kt i C=C

CH3-CH-CH-CH

3

o2 4H SO d,170 C CH

3-C=CH-CH

3+ H

2O

CH3OH CH33- metylbutan 2 ol 2 metylbut 2 enp n ng l B

Cu 40. Tnh lng Ag sinh ra:PTP: 3Ag + 4HNO3 3AgNO3 + NO + 2H2O

0,3 (mol) 0,1 (mol)nNO = 2,24/22,4 = 0,1 (mol)Theo PTP: nAg = 3.nNO = 3.0,1 = 0,3 (mol)t anehit n chc X l RCHO (R H):PTHH trng gng:

RCHO + Ag2Oo

3NH , t RCOOH + 2Ag

0,15 (mol) 0,3 (mol)

nRCHO = Ag1 0,3

.n = = 0,15 (mol)2 2

Suy ra: MRCHO =m 6,6

= = 44n 0,15

R + 29 = 44 R = 15 (CH3)

Vy X l CH3CHO (anehit axetic). p n ng l B*Ch : Nhiu bi ton cn c thm trng hp X l anehit fomic (HCHO)

Anehit n chc, cn chia 2 trng hp:+ TH1: HCHO 2 3

+Ag O/NH 4Ag

+ TH2: RCHO2 3+Ag O/NH

2Ag Cu 41. p n ng l DCu 42. p n ng l CCu 43. Polime khng c nhit nng chy xc nh, do polime l hn hp nhiu phn t c phnt khi khc nhau (tc l c h s trng hp n khc nhau)

p n ng l CCu 44. Ta c:

[Dng oxi ha] Fe2+ Cu2+ Fe3+ Ag+[Dng kh] Fe Cu Fe2+ Ag

7/28/2019 Gii chi tit 392

8/9

8

Do , chiu gim dn tnh oxi ha: Ag+ > Fe3+ > Cu2+ > Fe2+p n ng l B

Cu 45. Nguyn t ca cc kim loi trong nhm IA (kim loi kim) ging nhau v:- S electron ngoi cng ca nguyn t (s electron ha tr)- S oxi ha ca nguyn t trong hp cht +1- Kiu mng tinh th ca n cht (kiu mng tinh th lp phng tm khi) Khc nhau v cu hnh electron nguyn t. p n ng l D

Cu 46. Xt cc phng n:A. Khng ng: V ch c Cr2O3, Cr(OH)3 (crom c s oxi ha +3) c tnh lng tnh; cn

CrO, Cr(OH)2 c tnh baz (khng th hin tnh axit)Th d: Cr(OH)3 + 3HCl CrCl3 + 3H2O

Cr(OH)3 + NaOH Na[Cr(OH)4]B. ng. i vi cc hp cht ca crom:S oxi ha + 2 +3 +6Tnh cht Tnh kh Va c tnh kh, va Tnh oxi hac trng c tnh oxi ha mnhTh d: 2CrCl2 + Cl2 2CrCl3

K2Cr2O7 + 14HCl 3Cl2 + 2KCl + 2CrCl3 + 7H2OC. ng. i vi cc oxit, hiroxit ca crom:S oxi ha + 2 +3 +6Tnh cht Tnh baz tnh lng tnh Tnh axit

D. ng: CrO42 + 2H+ Cr2O7

2 + H2OCromat icromat

Khi thm kim (OH-) th H+ + OH- H2O [H+] gim cn bng trn chuyn sang tri, tc l mui icromat s chuyn thnh mui

cromat. p n ng l ACu 47. Hu ht cc hp cht ca natri tan nhiu trong nc v khng c mu, nn ngi ta khngdng phn ng ha hc, nhn bit ion Na+ m dng phng php vt l th mu ngn la nh

sau: Tm bt t mui rn hoc dung dch mui natri ln dy platin ri a dy vo ngn la nkh khng mu, ta s thy ngn la nhum mu vng ti. p n ng l DCu 48. Mt trong nhng tc nhn ch yu gy nn nao, au u, khi ung ru l do trong ru cln etanal (anehit axit) CH3CHO. p n ng l B

Cu 49. Theo bi ra nNO =0,56

= 0,025 (mol)22,4

0 3+

Ta c s : Fe 3e Fe

Fe0 2O + 2e O3 m

(mol)8

FeO, Fe3O4, Fe2O3, Fe (d)+5 +2N + 3e N

3 . 0,025(mol)

Fe(NO3)3

3m(mol)

56

Ta c:3m 3 m

= + 3.0,02556 8

3m = 21 7m + 4,2

10m = 25,2 m = 2,52. p n ng l C

Cu 50. Theo bi ra:2H

5,6n = = 0,25 (mol)

22,4

7/28/2019 Gii chi tit 392

9/9

9

Xy ra qu trnh: 2H+ + 2e H20,5 (mol) 0,25

+2HH

n = 2.n = 2.0,25 = 0,5 (mol) = nHCl +HClCl Hn = n = n = 0,5 (mol)

Khi lng mui khan: m = mkim loi + Cl = 20 + 0,5.35,5 = 37,75 (gam)

p n ng l B