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gii chi tit thi th i hc - m 492

Cu 1. Ni dung ng l C. p n ng l CCu 2. Z = 1, nguyn t hiro (H): Phi kim

Z = 2, nguyn t Heli (He): Kh himZ = 3, nguyn t Liti (Li): Kim loi kim (IA)

Z = 4, nguyn t Beri (Be): Kim loi (nhm IIA)Cc kim loi l Z v T. p n ng l B*Ch : Cc kim loi thng c 1, 2 hoc 3 electron lp ngoi cng (tr B, He v H). Ngoi ra,mt s kim loi c th c 4 electron (Thic Sn, Ch Pb), 5 electron (Bitmut Bi) hoc 6 electron(Poloni Po) lp electron ngoi cngCu 3.

- D kin thc nghim l thuc tnh (tnh cht) vt l ca cht, do n ph thuc vo tinhth cu to ca cht

- Tinh th mui n (NaCl) l tinh th ion, tinh th nc (H2O rn) c cu trc tinh thphn t

- Vy c th kt lun: tinh th ion bn hn tinh th phn t. Tht vy, tinh th ion c lc linkt gia cc ion trong tinh th l lc tnh in, lc ny ln tinh th bn. Cn cc phn ttrong tinh th phn t lin kt vi nhau bng tng tc yu gia cc phn t (lin kt hiro,tng tc Vanecvan) tinh th km bn

p n ng l CCu 4. Cc phn ng (1), (2) c phng trnh ion rt gn:

(1) Ag+ + Fe+2 Fe+3 + Ag(2) Mn + 2H+ Mn2+ + H2T (1) Ag+ c tnh oxi ha mnh hn Fe+3T (2) H+ c tnh oxi ha mnh hn Mn2+Trong dy in ha: Ag ng sau H Ag+ c tnh oxi ha mnh hn H+

Do tnh oxi ha tng dn theo trt t: Mn2+, H+, Fe+3, Ag+

p n ng l DCu 5. Ta c s phn ng: Kim loi + oxi oxit

-2

( O + 2e O )Do : mkim loi + moxi = moxit moxi = moxit - mkim loi = 3,33 2,13 = 1,2 gam nO = 1,2/16 = 0,075 (mol) = 2-On

Qu trnh ha tan oxit vo dung dch axit:-2

O + 2H+ H2O

+ 2-H On = 2.n = 2.0,075 = 0,15 (mol)

nHCl =

+Hn = 0,15 (mol)Vy: Vdd HCl =

0,15= 0,075

2lt = 75 (ml). p n ng l C

Cu 6. S :X (Cu, Fe) 2 4H SO c, nng dung dch Y + hn hp cc kim loi

V thu c hn hp cc kim loi cc kim loi Fe v Cu u d (axit H2SO4 ht). Vy ccphn ng xy ra:

Fe + 6H2SO4(c, nng) Fe2(SO4)3 + 3SO2 + 6H2OCu + 2H2SO4(c, nng) CuSO4 + SO2 + 2H2O

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Fe (d) + Fe2(SO4)3 3FeSO4Cu + Fe2(SO4)3 FeSO4 + CuSO4Fe (d) + CuSO4 FeSO4 + Cu

Vy trong dung dch Y ch cha FeSO4p n ng l C*Ch : Gii thch da vo dy in ha ca kim loi:

Fe2+ H+ Cu2+ Fe3+

Fe H Cu Fe2+Cu 7. ktc (0oC, 1atm); 1 mol bt k kh no cng chim th tch 22,4 lt nng kh =1/22,4 (mol/lt)

Gi s ban u c n1 mol NH3 ( 0oC), th tch ca bnh l V lt

Ta c: Co = 1n 1

=V 22,4

(mol/lt)

0oC (273oK) P1.V = n1.R.T1 546oC (819oK) P2.V = n2.R.T2

Theo bi ra: P2 = 3,3.P1T2 = 3.T1

n2/n1 = 3,3/3 = 1,1

n2 = 1,1.n1Suy ra: s mol kh tng 1,1.n1 n1 = 0,1.n1 (mol)

2NH3 N2 + 3H2Ban u: n1 mol 0 0Phn ng: 2x x 3xCn bng: n1 2x x 3x

n2 = n1 2x + x + 3x = n1 + 2xTa c: n2 = 1,1.n1 n1 + 2x = 1,1.n1 x = 0,05.n1

Do : [N2] = 1nx 0,05

= 0,05. = (M)V V 22,4

[H2] =3x 0,15

= (M)V 22,4

[NH3] = 1 1 1 1n - 2x n - 2.0,05.n n 0,9

= = 0,9. = (M)V V V 22,4

Vy KC =

33

2 2 -7

2 2

3

N . H 0,05/22,4.(0,15/22,4)= = 4,15.10

(0,9/22,4)NH

p n ng l ACu 8. Cc ion phn ng c vi ion CO3

2- khi:+ To ra cht khng tan (mui cacbonat khng tan)+ To ra cht kh (kh CO2)

Dy cc ion l:Ca2+ + CO32 CaCO3

Mg2+ + CO32 MgCO3

2HSO4- + CO3

2 CO2 + H2O + 2SO4

22H+ + CO3

2 CO2 + H2Op n ng l B

Cu 9. Cc PTHH xy ra:FeCl3 + 3NaOH Fe(OH)3 + 3NaClFe(OH)3 + NH3

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CuCl2 + 2NaOH Cu(OH)2 + 2NaClCu(OH)2 + 4NH3 [Cu(NH3)4]

2+ + 2OH-Tan

AlCl3 + 3NaOH Al(OH)3 + 3NaClAl(OH)3 + NaOH Na[Al(OH)4]

TanNa[Al(OH)4] + NH3 ZnCl2 + 2NaOH Zn(OH)2 + 2NaClZn(OH)2 + 2NaOH Na2[Zn(OH)4]

TanNa2[Zn(OH)4] + NH3

Vy ch c 1 ng nghim cui cng vn c kt ta. p n ng l ACu 10. Cc kim loi c tnh kh trung bnh (sau Al) v yu (sau H) th oxit ca n b CO khthnh kim loi tng ng nhit cao:

ZnO + CO ot Zn + CO2

CuO + CO ot Cu + CO2

FexOy + yCO ot xFe + yCO2

p n ng l DCu 11. Cc cht (cc ng phn) c CTPT: C2H2ClF

n =2.2 + 2 - 2 - 1 - 1

= 12

()

(V ch c 2 nguyn t C, cc nguyn t H, Cl, F ch c ha tr I nn hp cht khng th c cu tovng)

H Cl H H H FC = C C = C C = C

H F Cl F Cl H c 3 cht. p n ng l B

*Ch : +CH2=CFCl: khng c ng phn hnh hc+ CHF= CHCl: c ng phn hnh hc

Cu 12. Hirocacbon c cu toCH3

CH3-CH2-CH-CH2-C-CH3CH3-CH CH3

CH3Vit li:

CH3CH3 - CH - CH - CH2 - C - CH3

CH3 C2H5 CH3C tn gi: 4 etyl 2,2,5 trimetylhexan. p n ng l D*Ch :

Mch cacbon chnh ca ankan:+ Mch cacbon di nht:+ Cha nhiu nhnh nht

Cch nh s th t nguyn t cacbon trn mnh chnh:+ T pha gn nhnh nht+ T pha nhiu nhnh nht

(m bo sao cho tng ch s cc mch nhnh l nh nht)

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Th t gi tn cc mch nhnh: Theo th t vn A, B, CCu 13. Theo bi ra: nA = 1,12/22,4 = 0,05 (mol)

3 4 3 6 3 8C H C H C H= 40; m = 42; m = 44

3 6A C HM = 21.2 = 42 = M

Do c th gi thit coi hn hp A ch cha C3H6 (0,05 mol):C3H6 2

+O 3CO2 + 3H2O

mol 0,05 0,15 0,15 tng khi lng bnh ng nc vi trong d chnh bng tng khi lng CO2v H2O (sn

phm chy):m =

2 2CO H O+ m = 0,15.44 + 0,15.18 = 9,3 gam

p n ng l ACu 14. p n ng l BCu 15. t cng thc phn t ca ancol n chc CxHyO

Theo bi ra, ta c: 12.x + 1.y = 16.1.3,625 = 58Ta c bng: iu kin: x, y N*, y 2x + 2

x 1 2 3 4 5

y 46 34 22 10 -2KL Loi Loi Loi C4H10, tha LoiVy CTPT ca ancol ny l C4H10OC4H10O c cc ng phn ancol l

CH3CH2CH2CH2OH, CH3CH2CH(OH)CH3(CH3)2CHCH2OH (CH3)3C-OH (4 cht)

p n ng l BCu 16. t cng thc ca lipit l: (R1COO)2 (R2COO)C3H5

Theo bi ra: nglixerol = 46/92 = 0,5 (mol)PTHH thy phn:(R1COO)2(R2COO)C3H5 + 3H2O C3H5(OH)3 + 2R1COOH + R2COOH

mol 0,5 0,5Theo PTHH: nlipit = nglixerol = 0,5 (mol)

Mlipit =m 444

= = 888 (gam/mol)n 0,5

Suy ra: 2R1 + R2 + 3.44 + 41 = 888 2R1 + R2 = 715

Gi s R1 < R2 R2 > 7153

= 238,3

Suy ra R2 = 239 (C17H35) R1 = 238 (khng tha mn)

Gi s R1 > R2 R1 > 7153

= 238,3 R1 = 239 (C17H35); R2 = 237 (C17H33)

Vy hai axit bo l C17H35COOH v C17H33COOHp n ng l A*Ch : C17H35 = 239; C17H33 = 237; C17H31 = 235; C15H31 = 211Cu 17. p n ng l CCu 18. p n ng l CCu 19. Cao su buna S c tng hp t buta 1,3 ien (CH2=CH-CH=CH2)

V stiren (C6H5-CH=CH2) bng phn ng ng trng hp:

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CH2=CH-CH=CH2 + C6H5CH=CH2 oxt , t , p (-CH2-CH=CH-CH2-CH(C6H5)CH2-)

Buta 1,3 ien Stiren Poli (butaien stiren)(Cao su buna S)

p n ng l BCu 20. Ta c:

[Dng oxi ha] Fe2+ H+ Cu2+ Fe3+ Ag+[Dng kh] Fe H2 Cu Fe2+ Ag

Xt cc phng n sau:A. ng Fe2+/Fe + Cu2+/Cu (quy tc ):

Fe + Cu2+ Fe2+ + CuB. ng: Tnh oxi ha Fe2+ < H+ < Cu2+ < Ag+C. ng 2Fe3+ + Cu Cu2+ + 2Fe2+D. Sai. Fe2+ c tnh oxi ha yu hn Cu2+ nn Cu + Fe2+ khng xy rap n ng l D

Cu 21. Tnh kh ca cc kim loi (kim loi kim th) tng theo chiu tng ca bn knh nguynt. p n ng l B

*Ch : Tnh kh ca kim loi (kim loi kim th,) gim theo chiu tng:- Nng lng ion ha (I2)- Th in cc chun 2+0M /M(E ) - m in

Cu 22. Cc PTHH xy ra:Al2O3 + 6HNO3 2Al(NO3)3 + 3H2O

mol a 6aCuO + 2HNO3 Cu(NO3)2 + H2O

mol b 2bAg2O + 2HNO3 2AgNO3 + H2O

mol c 2c 2c HNO3 c ly va (6a + 2b + 2c) mol thu c Ag tinh khit, cn cho bt Cu vo dung dch Y:

Cu + 2AgNO3 2Ag + Cu(NO3)2mol c 2cVy thm c mol bt Cu vo Y. p n ng l B*Ch : thu c Ag tinh khit th khng th cho bt Al vo v

Al + 3AgNO3 3Ag + Al(NO3)32Al + 3Cu(NO3)2 3Cu + 2Al(NO3)3

Cu 23. Phng php kt ta ch c th thc hin c khi c s to thnh cht khng tan. M thct, tt c cc mui nitrat u tan tt trong nc, do khng th dng phng php kt ta nhn

bit ion NO3-

p n ng l A*Ch : - Phng php kt ta c dng nhn bit cc anion:

Cl- + Ag+ AgCl (trng)CO3

2- + Ca2+ CaCO3 (trng)SO4

2- + Ba2+ BaSO4 (trng)- C th nhn bit NO3

- bng Cu/H+3Cu + 2NO3

- + 8H+ 3Cu2+ + 2NO + 4H2OXanh

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2NO + O2 2NO2 (Nu)Cu 24. x l thy ngn ri vi, ngi ta s dng bt lu hunh

Hg (lng) + S k thng HgS p n ng l D

Cu 25. Phn lp nd ni chung, phn lp 3d ni ring, c s electron ti a l 10 (phn lp d c 5obitan nguyn t AO, mi AO c ti a 2 electron)

p n ng l CCu 26. Cch gii nhanh bi ny l va gii va da vo cc phng n (A, B, C, D) bin lun

- Vit cng thc ha hc ca cc cht phng n:A. KNO3 B. NaNO3 C. MgSO4 D. KClO3- Tm A (cation kim loi trong hp cht AB); Theo cc phng n, ta thy A c th K hoc Na

hoc MgTheo bi ra: n+ n- AB ABA BE + E = E = P

n+A

E + 32 = 50 n+A

E = 50 32 = 18 (K+)

Vy cation n+A l K+ (Z = 19) s electron lp v ion l 19 1 = 18 (electron)- Tm B (gc axit trong hp cht AB): Theo cc phng n, ta thy B l NO3-V n-B t 2 nguyn t cng chu k v 2 nhm lin nhau

p n ng l ACu 27. + Th hnh l cc dng tn ti khc nhau ca n cht ca cng mt nguyn t ha hc

+ Kim cng, than ch l 2 dng th hnh ca nguyn t cacbon; oxi v ozon l 2 dng thhnh ca nguyn t oxi; lu hunh n t v lu hunh t phng l 2 dng th hnh ca nguyn tlu hunh

+ Hiro 11( H) v teri

2

1( H) khng phi l 2 th hnh ca nguyn t hiro, m l 2 ng v canguyn t hirop n ng l C*Ch : Hin tng th hnh c th do:

+ Cng thc phn t n cht khc nhau (v s lng nguyn t v cu to phn t)V d: O

2/O

3; P

4(photpho trng); P

n(photpho )

+ Cu to tinh th khc nhauV d: kim cng/ than ch, lu hunh n t lu hunh t phng

+2 +2 -2 0 +2 -2 +3 +4

Cu 28. S phn ng: Cu2 FeS2 + O2 CuO + Fe2O3 + SO2Fe+2 Fe+3 + 1e2S-2 2S+4 + 2.6eCuFeS2 Fe

+3 + Cu+2 + 2S+4 + 13eVy 1 phn t CuFeS2 nhng 13 electron. p n ng l A

+1 +3 -2

*Ch : Nu coi CuFeS2 th

Cu+1

Cu+2

+ 1e2S-2 2S+4 +2.6eCuFeS2 Cu

+2 + 2S+4 + 13eu c kt qu nh nhau (nhng 13 electron)

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Cu 29. Theo bi ra: nNaOH = -OHn = 0,01.V (mol)

nHCl = +Hn = 0,03.V (mol)

H+ + OH- H2OBan u: 0,03.V 0,01.VPhn ng 0,01.V 0,01.VCn li 0,02.V 0 (mol)

[H+] (sau phn ng) = 0,02.V = 0,01 (M)V + V

pH = -lg 0,01 = -lg10-2 = 2p n ng l D

Cu 30. Theo bi ra:2H

1,344n = = 0,06 (mol)

22,4

2 4 2H SO Hn = n = 0,06 (mol)

S phn ng: Kim loi + H2SO4 (l) Mui sunfat + H2 Theo nh lut bo ton khi lng: mkim loi +

2 4H SOm = m +

2H

m = mkim loi +2 4H SO

m -2H

= 3,22 + 0,06.98 0,06.2 = 8,98 (gam)

p n ng l BCu 31. Hiu sut phn ng tnh theo ancol etylic (v axit dng d):

2 5C H OHn (ban u) = 1 mol

2 5C H OHn (phn ng) = 0,6 + x = 0,6 + 0,18 = 0,78 (mol)

Vy H =n (phn ng) 0,78

.100% = .100% = 78%n (ban u) 1

p n ng l BCu 32. Dung dch mui lm qu tm chuyn sang mu khi dung dch mui c mi trngaxit mui to t axit mnh v baz yu

l cc mui: Zn(NO3)2, NH4Cl, Al2(SO4)3

Th d: Khi ha tan Zn(NO3)2 vo H2OZn(NO3)2 Zn

2+ + 2NO3-

Zn2+ + H2O ZnOH+ + H+

ZnOH+ + H2O Zn(OH)2 + H+

dung dch c mi trng axit (H+) lm qu tm ha p n ng l C

Cu 33. phn bit Fe2O3 v Fe3O4 c th dng dung dch HNO3 long:Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO + 14H2O2NO + O2 2NO2 (nu)

p n ng l DCu 34. Da vo tnh cht khc nhau ca CO2 (c tnh oxi ha yu, khng c tnh kh) v SO2 (ctnh oxi ha, c tnh kh) phn bit:

CO2 + Br2 + H2O SO2 + Br2 + 2H2O H2SO4 + 2HBr

Nu khng muVy c th dng dung dch Br2 phn bit CO2 v SO2p n ng l B

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Cu 35. Xt cc phng n:A.Nu X l HOOC-COOH khng tha mn, v

HOOC-COOH + AgNO3/NH3 khng phn ngB.Nu X l OHC-CHO khng tha mn, v

OHC-CHO + H2O khng phn ngC.Nu X l OHC-COOH khng tha mn, v

OHC-COOH + AgNO3/NH3 khng phn ng

D. + X l CHCH tha mnCHCH + 2AgNO3 + 2NH3 AgCCAg + 2NH4NO3CHCH + H2O

oxt, t CH3CHO+ Y l OHC-CHO, tha mn:

OHC-CHO + 4AgNO3 + 6NH3 + 2H2O H4NOOC-COONH4 + 4Ag + 4NH4NO3+ Z l OHC-COOH tha mn:

OHC-COOH + 2AgNO3 + 4NH3 + H2O H4NOOC-COONH4 + 2Ag + 2NH4NO3OHC-COOH + NaOH OHC-COONa + H2O

+ T l HOOC-COOH tha mnHOOC-COOH + 2NaOH NaOOC-COONa + 2H2O

p n ng l DCu 36. PTHH:CH3-CH2-CH3 + Br2 khng phn ng (c kh thot ra)Propan (kh)

+ Br2 BrCH2-CH2-CH2Br

Xiclopropan nu nht khng mu nc brom nht mu dnVy mu ca dung dch nht dn v c kh thot ra. p n ng l C*Ch : Ankan khng phn ng vi nc brom

Cu 37. S phn ng:CnH2n 4

dd KMnO CnH2n+2O2

CnH2n+2O2 2+ O nCO2 + (n + 1)H2O

mol x nx (n + 1)xCO2 + Ca(OH)2(d) CaCO3 + H2O

mol nx nxTheo bi ra:

3CaCOn = 4,5/100 = 0,045 = n.x (1)

Khi lng bnh ng dung dch Ca(OH)2 tng chnh bng tng khi lng CO2 v H2O:mtng =

2 2CO H Om + m 3,06 = nx.44 + (n +1).x.18 = 62.nx + 18x (2)

T (1) v (2) ta c x = 0,015; n = 3Vy X l C3H8O2 . p n ng l DCu 38. p n ng l CCu 39. C 2 kh nng xy ra:

+ Hn hp X gm CH3OH v C2H5OH+ Hn hp X gm 2 ancol no, n chc, bc 1

R1CH2OH v R2CH2OH (RCH2OH)

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Xt trng hp 1:Cu (r)

CH3OH o+CuO, t H2O (h) 2

+Ag O 4Ag + CO2 + H2Ox 4x

HCHO (k)

Cu (r)

CH3CH2OH o+CuO, t H2O (h) 2

+Ag O 2Ag + CH3COOHy 2yCH3CHO (h)

Theo bi ra, ta c: 4x + 2y = nAg = 64,8/108 = 0,6 (mol) (1)30.x + 44.y + 18(x + y)

= 13,75.2 = 27,52.x + 2.y

7y 7x = 0 y x = 0 y = x (2)T (1, 2) x = y = 0,1 (mol)Suy ra: m = 0,1.32 + 0,1.46 = 7,8 (gam)p n ng l B

Xt trng hp 2Cu (r)

RCH2OH o+CuO, t H2O 2

+Ag O 2Ag + RCOOHz 2zRCHO (h)

Theo bi ra, ta c: 2z = nAg = 0,6 z = 0,3 (mol)18.z + (R + 29).z

= 27,52.z

18.z + R.z + 29.z = 55.z R = 8 R1 = H (M = 1) < 8 (loi)

R2 = CH3 (M = 15) > 8Trng hp ny khng tha mn*Ch : Hai ancol ng ng th phi cng bc ancolCu 40. Xt cc phng n:

A. Loi, v etilen CH2=CH2B. ng:

HCHO + Ag2O 3NH HCOOH + 2Ag

Hoc HCHO + 2Ag2O 3NH CO2 + H2O + 4Ag

2CH2=CH-CCH + Ag2O 3NH 2CH2=CH-CCAg + H2O

2CH3-CCH + Ag2O 3NH 2CH3-CCAg + H2O

C.Loi, v etilenD. Loi, v butin 2

p n ng l B*Ch : Phn bit:

Anehit + Ag2O/NH3 Ag (phn ng trng gng) Ankin 1 + Ag2O/NH3 kt ta mu vng ( R C CAg) Ankin 2 + Ag2O/NH3 khng phn ng

Cu 41. p n ng l CCu 42. p n ng l DCu 43. S phn ng gi nh: kC2H3Cl + Cl2 C2kH3k-1Clk+1 + HCl

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Theo bi ra, ta c:35,5(k + 1) 63,96 63,96

= = = 1,77524k + 3k - 1 (100 - 63,96) 36,04

35,5k + 35,5 = 47,925k 1,775 12,425k = 37,275 k = 3p n ng l B

*Ch : Phn ng clo ha PVC ch c th l phn ng th v phn t PVC l hp cht no

Cu 44. Trong cng nghip, phng php in phn nng chy c p dng iu ch cc kimloi mnh (trc Al v Al) nh Na, Ca, Al

NaCl dpnc Na + 21

Cl2

CaCl2dpnc Ca + 2Cl

2Al2O3dpnc

criolit4Al + 3O2

p n ng l CCu 45. Nc cng khi un si m mt tnh cng l nc cng c tnh cng tm thi

nc cng cha mui hirocacbonat ca canxi v magie (Ca(HCO3)2, Mg(HCO3)2)p n ng l B

Cu 46. S cc phn ng:CuCl2+KOH d Cu(OH)23

NH d [Cu(NH3)4] (OH)2 (dd)

ZnCl2+KOH d K2[Zn(OH)4] (dd)

FeCl3 +KOH d Fe(OH)3 3NH d Fe(OH)3

AlCl3 +KOH d K[Al(OH)4] (dd)

Vy ch thu c 1 kt ta (Fe(OH)3 )p n ng l C

*Ch : Cc PTP xy ra trong s :CuCl2 + 2KOH Cu(OH)2 + 2KCl

Cu(OH)2 + 4NH3

[Cu(NH3)4](OH)2ZnCl2 + 2KOH Zn(OH)2 + 2KClZn(OH)2 + 2KOH (d) K2[Zn(OH)4]FeCl3 + 3KOH Fe(OH)3 + 3KClAlCl3 + 3KOH Al(OH)3 + 3KClAl(OH)3 + KOH (d) K[Al(OH)4]

Cu 47. Mu sc ca mt s hiroxit kim loi:- Fe(OH)2: Mu trng hi xanh- Cu(OH)2: Mu xanh- Ni(OH)2: Mu xanh lc- Mg(OH)2: Mu trng

- Fe(OH)3: Mu nu - Cr(OH)3: Mu xanh- Al(OH)3: Kt ta keo, mu trng cp n ng l A

Cu 48. S chuyn ha: 2CH4 C2H3Cl2.18kg 62,5kg

m 250 kg

4CH

250.2.18= = 144 (kg)

62,5

4

33

CH

144.10n = = 8.10 (mol)

18

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V H = 50% 4CH

n (thc t) =3

38.10 .100 = 16.10 (mol)50

4CH

V (thc t, ktc) = 16.103.22,4 = 358,4.103 (lt) = 358,4 (m3)

Vkh thin nhin =358,4.100

= 448,080

(m3)

p n ng l D

Cu 49. Gi x, y ln lt l s mol Zn v Fe c trong hn hp ban u.Theo bi ra, ta c: 65x + 56y = m (1)PTHH: Zn + CuSO4 ZnSO4 + Cu

x x (mol)Fe + CuSO4 FeSO4 + Cu y y (mol)

Theo bi ra, ta c: 64x + 64y = m (2)T (1, 2) ta c: 65x + 56y = 64x + 64y x = 8y

Vy, % mZn =65.x.100% 65.x.100% 65.8y.100%

= = = 90,27%m 65x + 56y 65.8y + 56y

p n ng l C

Cu 50. Theo bi ra:2H

5,32n = = 0,2375 (mol)22,4

nHCl = 0,25.1 = 0,25 (mol)

2 4H SOn = 0,25.0,5 = 0,125 (mol)

+Hn (ban u) = 0,25.1 + 0,125.2 = 0,5 (mol)C qu trnh xy ra: 2H+ + 2e H2

2.0,2375 0,2375 (mol) +

Hn (phn ng) = 2.

2Hn = 2.0,2375 = 0,475 (mol)

+H

n (d) = 0,5 0,475 = 0,025 (mol)

[H+] = -10,025 = 0,1 = 10 (M)0,25

pH = -lg10-1 = 1p n ng l A