TRỌNG TÂM KIẾN THỨC HÓA HỌC 12 HÓA VÔ CƠ & HÓA HỮU CƠ - ĐỖ XUÂN HƯNG

8/21/2019 TRNG TM KIN THC HA HC 12 HA V C & HA HU C - XUN HNG

1/661

XUN H NG \

TR NG TMKIN TH C HA H C

12HA V C

oa NH NG V N TH NG G P TRONGCC K THI TUY N SINH I H C, CAO NG VT T NGHI P THPT

ca PH NG PHP TR C NGHI M

NH XU T B N I H C Q u c GIA H N I

WWW.FACEBOOK.COM/DAYKEM.QUYNHON

WWW.FACEBOOK.COM/BOIDUONGHOAHOCQUYNHON

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WWW.DAYKEMQUYNHON.UCOZ.COM

ng gp PDF bi GV. Nguyn Thanh T


2/661

NH XUT BN I HC QUC Gm H NI

16 Hng Chui - Hai B Trng - H Ni

in th oai: B in tp -Ch' bn : (04) 39714896:

Hnh chnh: (04) 39714899: Tng bin tp: (04) 39714897

Fay. (04^ 39714899

Ch u trch nhi m xu t b n

Gim d c: PHNG Q u c B O

T ng bin t p: PH M TH TRM

Bin t p: TH H NG - THU H NG

K thu t vi tnh: THANH VINH

Trnh by ba: XUN TH A

i tc lin k t xu t b n:

NH SCH H NG N

SCH LIN K T

TR NG TM KI N TH C HA H C 12 - HA V c

M s: 1L- 125H2012

In 2.000 cun, kh 17x 24cm ti Cng ty c phn Vn ha Vn Lang.S xut bn : 415 -2012/CX B /07-65 /HQ G HN , ngy 10 /4 /2012.

Quyt nh xut bn s: 133LK-TN/Q-NXBHQGHN

In xong v np lu chiu qu II nm 2012.



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3/661

L I NOI AU

Theo cu trc ,thi ca B Gio dc v o to K thi t't nghipTrung hc ph thng v tuyn sinh i hc hin nay, cc bn hc sinhmun t kt qu cao th r't cn cng c' kin thc sut qu trnh hc tp

ca mnh, ch yu l chng trnh lp 12.Nhm gip cc em n li kin thc c bn, nm vng kin thc trng

tm v bit phng php lm cc dng bi tp trc nghim. Chng ti binson cun sch "Tr ng tm kh th c Ha h c 12"

Sch g m 2 t p:Tp 1: gm 5 chuyn (phn Hu c)

Chuyn 1: Este - Lipit.

Chuyn 2: Cacbohirat (Gluxit).Chuyn 3: Amin - Aminoaxit - Protein.Chuyn 4: Polime.Chuyn 5: Tng hp h u c.

Tp 2: gm 6chuyn (phn V c)Chuyn 1: i cng v kim loiChuyn 2: Kim loi kim - Kim loi kim th - Nhm.Chuyn 3: Crom - st - ng.

Chuyn 4: Nhn bit - Chun .Chuyn 5: Mi trng.Chuyn 5: Tng hp v c.

Trong m i chuyn g m c:Tm tt kin thc c bn.Cc dng ton c bn (c gii chi tit)Bi tp trc nghim c hng dn gii.Bi tp trc nghim t luyn tp (c p n)

Chng ti hy vng cun sch s gip ch cho cc em trong qu trnh hctp, n luyn, rn luyn k nng gii nhanh bi ton ha hc v t tin trongcc k thi.

Trong khi bin son chng ti khng trnh khi nhng thiu st, r'tmong nhn c s gp chn thnh t qu bn c, ln ti bn sc hon thin hn.

Mi chi tit xin lin h:[email protected].

Tc gi



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mailto:[email protected]:[email protected]


4/661

A I C OIHG V KIM LO I

A. TM T T KI N TH C C B N

I. KIM LO I - H P KIM

1. Kim lo ia) V tr c a kim lo i trong b ng tu n hon

- Nhm IA (tr H) v IIA. Cc kim loi l nhng nguyn t s.- Nhm IIIA (tr B) l nhng nguyn t nhm p.- Mt phn ca nhm IVA, VA, VIA y l nhng nguyn t p.

- Cc nguyn t nhm B (IB -> VIIIB) v nhng nguyn t h lantan, actini.b) Tnh ch t v t l c a kim lo i

- Tnh cht vt l chung ca kim loi: tnh do, tnh dn in (Ag > Cu >Au > AI > Fe ...), dn nhit, nh kim do cc electron t do trong kim lo igy ra.

- Tnh cht vt l ring ca kim loi: khi lng ring (nh nht: Li, nngnht: Os), nhit nng chy (thp nht: Hg, eao nht: W), tnh cng (mmnht: Na, K,..; cng nht: w, Cr ...) ph thuc vo bn ca lin kt kimloi, khi lng nguyn t, kiu mng tinh th , ... ca kim loi.c) Tinh ch t h a hoc c a kim lo i

Tnh cht ha hc chung ca kim loi l tnh kh: M - Mn+ + ne Ph n ng v i phi Mm:halogen (K >Au u phn ng), 2 (Au, Pt,

Ag khng phn ng), s (Kim loi + s *mui sunua). Ph n ng v i axit

- Phn ng vi dung dch axit H2SO4long, HC1:

2M + 2nH+ - 2Mn+ + 11H2(n l mc oxi ha thp ca M)

- Phn ng vi H2SO4c, HNO3: kim loi boxi ha ln mc oxi ha

cao nht ng thi s +6v N+5bkh xung .mc thp hn. Ph n ng v i dung d ch mu i:

- Kim loi hot ng mnh y kim loi hot ng km hn ra khidung dch mui: m + nBm+ mAn+ + nB (mui ca A, B tan trongnc; A c tnh kh mnh hn B; Bm+ c tnh oxi ha mnh hn An+).

Phn ng v i n c:

- Nhit phng: M + H20 M(OH) + - H 2T (M: IA, Ca, Ba, Sr)

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5/661

AI + H2O ------> A1(0 H)3 + H2 1. Phn ng b mt.

- Nhit cao: Mg + H2O a > MgO + H2t

Fe + H20 >57{)Uc > FeO + H23Fe + 4H20 - Fe30 4 + 4H2

Phn ng vi dung dch kim (vi M: Be, Al, Cr, Zn, Pb...):

M + nH20 + (4 - n)NaOH - Na(4 - n)[M(OH)4] + - H 2

(cht kh) (cht oxi ha)

2. H p kima) nh ngha:Hp kim l vt liu kim loi c cha thm mt hay nhiu

nguyn t khc.

b) Tnh ch t:

- Tnh cht ha hc tng t tnh cht ca cc cht tham gia to thnhhp kim.

- Tnh cht vt l v tnh cht c hc khc nhiu vi tnh cht cc n cht.

II. DY I N HA C A KIM LO I

Cp oxi ha - kh:

V d : Fe + Cu2+------> Fe2+ + Cu.

Cip: Fe% e ; Cu% u gi ' c#p OTX h

1 . Pin i n haS cu to pin in ha:

Vit tt: (-) Zn2+/Zn| ICu2+/Cu (+)

Phn ng xy ra trong pin: Zn + Cu2+ ------> Zn2+ + Cu

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6/661

Nh n xt- Trong qa trnh phng in, [Cu2+] gim dn, [Zn2+] tng dn.

- Phn ng oxi ha - kh trong pin in ha sinh ra dng in mt chiu.

- Sut in ng E ca pin in ha ph thuc vo: bn cht cp oxi ha

- Kh ca kim loi, nng ca cc dung dch mui v nhit .- Sut in ng ca pin E = E. - E7. Trong cp Mn+/M no c E

ln hn l cc dng.

2. Dy i n ha chu n c a kim lo i v ngha

Dy in ha chun ca kim loi:----------------------- Tnh oxi ha c a ion kim lo i Mn+ tng-------------------

1+ IT+ r> 2+ 2+ K1 + . . 2+ . .3 + Zn2+ + Cu

cht kh mnh + cht oxh mnh * cht oxh yu + cht kh yu

=> Bi-c th t xy ra phn ng oxi ha kh.

Th d : Cho Zn d vo dung dch gm Cu(NC>3)2, AgN 3, Fe(N 3)3,NaNO. Th t phn ng xy ra nh sau:

Zn + 2Ag+ ------> Zn2+ + 2Ag (1)

Zn + 2Fe3+ - Zn2+ + 2Fe2+ (2)

Zn + Cu2+ ----- >Zn2++ Cu (3)

Zn + Fe2+ ----- > Zn2+ + Fe (4)

- Kim loi trong cp oxi h -k h c th in cc nh hn

E = 0,00 V y c 2ra khi dung dch axit.2%



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7/661

Mg + 2H+ ----- Mg2+ + H2T

- Xc nh sut in ng chun ca pin in ha. Epin = E - E >0,00 V.

3. S in phn Trong qu trnh in phn:

- cc m (catot) xy .ra s kh v cht no c tnh oxi ha mnh hn(cht c th in cc chun ln hn) s bkh trc:

Ion kim loi c tnh oxi ha yu hn Al3+:

Cht oxi ha Mn+ + ne -> M

lon kim loi c tnh oxi ha mnh hn Al3+:2H20 + 2e - 20H~ + H2

- cc d ng (anot) xy ra s oxi ha v cht no c tnh kh mnhhn (cht c th in cc chun nh hn) s bkh trc:

gc axit khng c oxi: 2X' (X: Cl, Br) - 2e >X2

Cht kh

gc axit c oxi (NO ' , s o 2H20 - 4e >4H* + 0 2

Ch : - S khc bi t v tn v d u cc i n c c trong pin i n ha.

Ph n ng oxi ha kh ch c th x y ra trong thi t b i n phn khic dng i n m t chi u bn ngoi cung c p. Khc v i pin i n ha.

4. ng d ng c a s i n phn '

- iu ch kim loi.- iu ch mt s phi kim nh: H2, O2, F2, CI2.- iu ch mt s hp cht, nh: KM11O4, NaOH, H2O2, nc Gia ven,..

- Tinh ch mt s kim loi, nh: Cu, Pb, Zn, Fe, Ag, A u ,. . .- M in: bo v kim loi chng n mn v to v p cho vt.

III. CC PH NG PHP I U CH KIM LO I

Phng php thy luyn: dng iu ch nhng kim loi c tnh khyu nh Cu, Hg, g, A u,. . .

Phng php nhit luyn: dng iu ch nhng kim loi c tnh khyu v trung bnh.

Phng php in phn (dng trong cng nghip):



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8/661

- i n phn nng ch y (mui, baz, oxit) iu ch nhng kimloi c tnh kh mnh, nh: K, Na, Ca, Ba, Al.

- i n phn dung d ch cht in li (dung dch mui) iu chnhng kim ioi c tnh kh yu v trung binh.

1 M

Khi lng cht thot ra in cc: m . = A 96500 n- n: s e trao i- I: cng dng in- t: thi gian dng in i qua (s: giy)

IV. N M N KIM LO I1. Khi ni m

S n mn kim loi l s ph hy kim loi hoc hp kim do tc dng

ca cc cht trong mi trng. Hu qu l: M ------> Mn+ + ne.2. Phn loi: c hai loi chnh : n mn ha hc v n mn in ha.

An mn ha hc n mn in haNguynnhn

- Do kim loi tc dng vi chtkh hoc hi nc nhit cao.

- Do km loi hoc hp kimtip xc vi dung dch chtin li v to ra cc cp pinin ha.

Th - Vt liu bng gang thp, ccb phn ca thit b l t hocnhng . thit b thng xuynphi tip xc vi hi nc v khoxi,... nhit cao:

3Fe + 4H20 Fe30 4+ 4H2

3Fe + 20 , ------> Fe,0 ,2 3 4

- S n mn Fe ln Cu trongkhng kh m:

(-)Fe|02, C02, S02, H20|Cu(+) (dung dch in li)

- cc dng:2H+ + 2 e -+ H20 2+ 2H20 + 4 e 4 0 H "

- cc m: Fe Fe2+ + 2eRi:Fe2+ + 20H~ - Fe(OH)24Fe(OH)2+ 0 2+ (2n -4)H 2b

>2Fe203 .nH20(gi st)

iukin

- Xy ra trong mi trng

khng c cHt in li, hoc kimloi nguyn cht tip xc vimi trng cht in li.

- C cc in cc khc nhau:

kim loi - kim loi; kim loi- phi kim; hoc kim loi -hp cht ha hc.- Cc in cc phi tip xc

9


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ng gp PDF bi GV Nguyn Thanh T


9/661

trc tip, hoc gin tip vinhau.

4

Cc in cc cng tip xcvi dung dch cht in li.

cim - Bn cht l phn ng oxi ha -kh.

- Khng pht sinh dng in.

- Nhit cng cao tc nmn cng ln.

- Bn cht l phn ng oxiha - kh, e di chuyn t cc(-) sang cc (+), pht sinh dngin theo chiu ngc li.- Vn tc n mn cngnhanh nu nng cht inli ln v 2 in cc cng xanhau trong dy in ha.

3. Bo v kim loi- Cch li kim loi khi mi trng nh sn, m,...

- To hp kim bn ha hc vi mi trng: nh hp kim (Fe - Cr - Ni).

- Bo v in ha: gn mt mnh kim loi hot ng hn (vt hi sinh)vo kim loi cn v.

B. CC D NG TON c B N

, D ng 1: Xc nh tn kim lo i.

Bi tp tm tn kim loi thng gp c gii trong cc dng sau y:- Tnh trc tip khi lng mol ca kim loi

- Tm gii hn khi lng mol ca kim loi v da vo gi thit (cc

tnh cht ca kim loi) tm ra tn ca kim loi.

* L u :Thng s dng cc phng php nhu: Phng php trung bnh,nh lut bo ton khi lng, nh lut bo ton eecon... tim kim loi.

V d 1: e ho tan hon ton 6,4 gam hn hp gm kim loi R (chc hotrII) v oxit ca n cn va 400ml dung ch HC1 IM. Kim loi R l:

A. Ba B. Ca c . Be D. Mg;(Trch thi tuy n sinh Cao ng nm 201 )

H ng d n gi iR + 2HC1 RCh + H2

RO + 2HC1 RC12 + H20

Ta c : 1HCI= 0 , 4 . 1 = 0 , 4 mol => I1(R RO) = 0 , 2 mol

=> M = = 320,2

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10/661

=> M 1 6< M M = 24(Mg)

=> p n D.V d 2: Cho 8,5 gam hai kim loi ha trI ng st nhau trong bng tun

hon ha hc vo nc thu c 3,36 lt kh bay ln. Tm tn ca hai kimloi ?A. Li v Na B. Na v K c . K v Rb D. Rb v Cs

H ng d n gi i

Gi A l cng thc tng qut ca hai kim loi cn tm, ta c PTHH:

2 + 2H20 -> 2 OH + H2

T a c :n =-5i2. = 0,15(mol)h 2 2 2 j 4

Theo PTHH : n- = 2.nH= 2.0,15 = 0,3 (mol)A r*2

Vy: M - = = 28,33A 0,3

Vy hai kim loi cn tm chnh l Na v K.=> p n B.

V d 3: Nhng mt l kim loi M (chc ho trII trong hp cht) c khilng 50 gam vo 200ml dung dch AgN 3 IM cho n khi phn ngxy ra hon ton. Lc dung dch, em c cn thu c 18,8 gam mui

khan. Kim loi M lA. Mg B. Zn c . Cu D. Fe

(Trch thi tuy n sinh Cao ng kh i A,B nm 2009)H ng d n gi i

Ta c : nANOj = 0,2.1 = 0,2(mol) => m AgNO = 170.0,2=34(g).

Ptp: M + 2AgN 3 M(N 3)2 + 2AgTa c: 2mol AgN03 lmol M(N 3)2 th AM gim = (2.108 - M)(g)

Vy:0,2mol AgNOs 0,lmol M(N 3)2th Am gim = 3418,8 =15,2(g)0,2.(216 - M) = 15,2 X2 => M = 64 (Cu)

=^>p n cV d 4: Nhng mt thanh graphit c ph mt lp kim loi ha tr (II)

vo dung dch C11SO4 d. Sau phn ng khi lng ca thanh graphitgim i 0,24 gam. Cng thanh graphit ny nu c nhng vo dungdch AgN 3 th khi phn ng xong thy khi lng thanh graphit tng

ln 0,52 gam. Kim loi ha tr(II) l kim loi no sau y?A.Pb. B.Cd. c . Al. D.Sn.

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11/661

H ng d n gi i

t kim loi ha tr(II) l M vi s gam l X (gam).

M + CuS0 4 d -----> MSO4 + CuC 1 mol M - 1 mol Cu, khi lng gim (M - 64) gamVy : Xmol M -> Xmol Cu khi lng gim. 0,24 gam

0,24= > X = M - 6 4Mt khc: M + 2AgN 3 -----> M(N 3)2 + 2Ag

C 1 mol M - 2 moi Ag khi lng tng (216 - M) gamVy : Xmol M -> 2x mol Ag khi lng tng 0,52 gam

0,52=>x = ----

216-M

_ 0 24 0 52Ta c: = -J. M = 112 (kim loi Cd).M - 6 4 2 16 -M => p n B.

V d 5: Ho tan hon ton 2,9 gam hn hp gm kim loi M v oxit ca nvo nc, thu c 500 ml dung dch 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. Na

(trch tuy n sinh i h c kh i B nm 20091

H ng d n gi i

M : a molM2On : b mol

2M + 2nH20 2M(OH) + nH21a a na/2M20 + nH20 -* 2M(OH)n

b 2b

Ta c: n = = 0,01 na = 0,02 (1)H2 2

V: a+2b = 0,5.0,04 = 0,02 => na + 2nb = 0,02nnb = 0 ,01n - 0,01 (2 )

Mt khc: Ma+(2M+16n)b = 2,9 => Ma + 2Mb + 16nb = 2,9 (3)Th (2) vo (3) ta c:

Ma + 2Mb + 16(0,Oln - 0,01) = 2,9 => Ma + 2Mb + 016n = 3,06=> M(a + 2b) + 0,16n = 3,06 => 0,02M + 0,16n = 3,06=> M + 8n = 153

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12/661

Lp bng:n 1 2 -jM 145 137 129

=> M l bari (Ba)

=> p n B.

V d 6 : Cho 1,9 gam hn hp mui cacbonal v hirocacbonat ca kimloi kim M tc dng ht vi dung dch HCI (d), sinh ra 0,448 lt kh (ktc). Kim loi M lA. Na. B. K. c . Rb. I). Li.

(Trch tuyn sinh D i h c khi B nm 2008)H ng d n gi i

T _ 0 4 4 8 _ n m Ila c: n = ----- = 0 ,0 2 moi

co 22,4Gi cng thc ca 2 mui MHCO3v M2CO3.MHCO3 + HC1 H> MCI + C 02t + 11,0M2CO3 +2HC1 -> 2MC1 + C 0 2t + H2Q

Theo phng trnh phn ng c :

n =n =0,02(mol) => Mhh = - ^ - = 95hh c o 2 0 02

Theo tnh cht ca M , ta c: Mmhco < Mhh =95 < Mmc0

=> Mm+ 61 < 95 < 2Mm+ 60 => 17 ,5< M m p n A.V d 7: in phn dung dch mui MCI,, vi in cc Ir. Khi catot thu

c 16g kim loi M th anot thu c 5,6 lt kh (ktc). M l:A. Cu B. Zn c . Fe D. Ag.

H ng d n gi i

Ta c: nr., = = 0,25 moic>2 2 2 ,4

Phng trnh in phn: MC1 -> M + c

0.25 motn

=> .Mm = 16 => Mm= 32nn

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13/661

=> Cp nghim n = 2 v M = 64 (Cu) l ph hp.:=> p n A.

V d 8 : X l kim loi thuc phn nhm chnh nhm II (hay nhm IIA).Cho 1,7 gam hn hp gm kim loi X v Zn tc dng vi lng d dung

dch HC1, sinh ra 0,672 lt kh H2(ktc). Mt khc, khi cho 1,9 gam X tcdng vi lng d dung dch H2SO4 long, th th tch kh hiro sinh racha n 1,12 lt (ktc). Kim loi X l:A. Ba B. Ca c . Sr D.Mg.

(Trch thi tuy n sinh Cao ng kh i A,B nm 2008)H ng d n gi i

Gi R l kim loi tng ng thay cho X v Zn.

* 1,7g hn hp X, Zn tc dng vi HC1 d:R + 2HC1 RC12 + H2T

0,03 ^ ^ = 0 , 0 3 m o l22,4

^ M r ==56,67 m Mzn= 65 > M r= 56,670,03

=> Mx< M r = 56,67 (1)* 1,9g X tc dng vi H2SO4long, d:

X + H2SO4 -> XSO4 + H2

Ta c: nu = =0,05mol22,4 :

Theo bi th tch kh hiro sinh ra cha n 1,12 lt => nH< 0,05 mol

Theo phng trnh phn ng:

n = nH => n < 0,05mol => Mx > ^ =38 (2)X H2X 0 > 0 5

T (1) v (2) => 38 < Mx < 56,67 (*)

=> X l Ca v Mca= 40 tho mn iu kin (*).

=> p n B.

V d 9: Ho tan hon ton 2,45 gam hn hp X gm hai kim loi kim thvo 200ml dung dch HC1 1,25M, thu c dung dch Y cha cc chttan c nng mol bng nhau. Hai kim loi trong X lA. Mg v Ca B. Be v Mg c . Mg v Sr D. Be v Ca

(Trch thi tuy n sinh i h c kh i B nm 2010)


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14/661

H ng d n gi i

Ta c: nna = 0,2.1,25 = 0,25 molGi s 2 kim loi kim th l A v B.Gi R l kim loi chung cho 2 kim loi kim th , B.

R + 2HC1 -+ RC12 + H2

XX

2

Gi s HC1 ht => nR = =0,125 mol2

=* Mr = = 19,6 ^ Ma + Mb = 19,6.2 = 39,2 (l) ^ lo i .

Vy d axit.

Y c nng mol/1 ca cc cht bng nhau nn:.nA = ne = nnci (d)=> 2x + 2x + X = 0,25 (vi X l s mol ca A)

=> X = 0,05 => M = = 24,5 Ma + Mb= 24,5.2 = 490,1

Cp nghim ph hp l Be (9) v Ca (40)=> p n c .

V d 10: Cho 7,1 gam hn hp gm mt kim loi kim X v mt kim loikim th Y tc dng ht vi lng d dung dch HC1 long, thu c 5,6

lt kh (ktc). Kim loi X, Y lA. Natri v magie. B. Liti v beri.c . Kali v canxi. D. Kali v bari.

(Trch thi tuy n sinh i h c kh i A nm 2010)H ng d n gi i

Ta c: n = 0,25 mol2

Gi R l kim loi chung cho kim loi kim X v kim loi kim th Y

R + nHCl RCln + " H it (n l ha tr ca R)

0,25n n

=> Mr. = 7,1 => Mr= 14,2nn

- Vi n = 1 => Mr = 14,2n = 2=>Mr = 28,4

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15/661

- Nhng 1 < n < 2 => 14,2 < Mr< 28,4=> 2 kim loi l Na (23) v Mg (24)=> p n A.

V d 11: Ha tan hon ton 8gam hn hp gm kim loi trII v oxit can trong 250ml dung dch HC1 va . Kim loi ha trII l:

A. Mg B. Ca c . Zn D. Fe.H ng d n gi i

Ta c: nHC1= 0,25.2=0,5(mol)

Gi R l kim loi ha trII => oxit ca n l RO.

R +2HC1 -> RC2 + H2

RO +2HC1 ->RC12 + H20Theo phng trnh phn ng c:

nhh=nH,..=.0,5=0,25(mol) => Mhh = =32"h 2 2 0,25

Theo tnh cht ca M , ta c: Mr Mr < 32 < Mr+ l => 16 < Mr < 32 => Chc Mg tha mn=> p n A.

V d 12: Cho 3,024 gam mt kim loi M tan ht trong dung dch HNO 3long, thu c 940,8ml kh NxOy (sn phm kh duy nht, ktc) c tkhi i vi H2bng 22. Kh NxOy v kim loi M lA. NO v Mg. B. N2O v AI c . N2O v Fe. D. NO2v AI.

(Trch th tuy n sinh i h c kh i A nm 2009)H ng d n gi i

Ta c: 0 /H = 22 => MN0 = 44 => Nx0yl N 20.

0 Q408Mt khc: nM = = 0,042 moln2o 22^4

Qu nh nhn e:2N + 8e -> N2o

0,336 0,042Qu trnh nhng e:

M > M + ne

W ( ' 0,336n

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16/661

Ta c: .Mm = 3,024 => Mm = 9nn

=> Cp nghim ph hp l n = 3 v Mm= 27 AI.^ p n B.

V d 13: Mt dung dch c cha cc ion: X mol M3+, 0,2 mol Mg2+, 0,3 molCu2+, 0,6 mol S042 0,4 mol NC>3~. C cn dung dch ny thu c 116,8gam hn hp cc mui khan. M l:A Cr B. Fe c. AI D. Zn.

H ng d n gi ip dng nh lut bo ton in tch, ta c:3x + 0,2.2 + 0,3.2 = 0,6.2 + 0,4 => X = 0,2 mol

=> 116,8 = 0,2.Mm + 0,2.24 + 0,3.64 + 0,6.96 + 0,4.62=> Mm = 52 => M l Cr =5- p n A.V d 14: t chy hon ton 7,2 gam kim loi M (c ho II khng i

trong hp cht) trong hn hp kh C2v O2. Sau phn ng thu c 23,0gam cht rn v th tch hn hp kh phn ng. l 5,6 lt ( ktc). Kimloi M l

A. Be B. Cu c . Ca D. Mg(Trch thi tuy n sinh Cao ng nm 2009)

H ng d n gi ip dng L bo ton khi lng ta c:

mhhkh= 23 -7 ,2 = 15,8g

Ta c: mmt,i= mM,++mM +m cu2++m. + m

Ta c: nhhkh - = 0,25 mol22,4

Gi s moi oxi v clo ln lt l X mol; y mol.Ta c h: J X + y = 0,25 rX = 0,05

1 __ _ _ J _ _{l3 2 x + 71y= 15,8 ^ "[ y = 0,2{ PTTH JWV PHQu trnh nhn e: MCl2 + 2 e 2C1"

0,2 0,4

0 2. + 4e 202"0,05 0,2Tng s mol e nhn = 0,2 + 0,4 = 0,6 mol

Qu trnh nhng e:M >Mn++ ne

0 ,6/n 0 ,6

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17/661

=> Mm = - = 12n => Cp nghim ph hp l n = 2; Mm = 24 (Mg)0,6

=> p n D.

a. D ng 2: Kim lo i tc d ng v i dung dch axit.

Dng bi tp ny c bn loi bi ton nh sau:Lo i : Mt kim loi tc dng vi mt axit.- Vit ng PTHH- Nu kim loi tc dng vi axit c tnh oxi ha (HNO3, H2SO4c) to

ra hn hp kh khc nhau th nn vit nhng PTHH phn bit (mi PTHHto mt kh). Neu cn ghp cc PTHH li vi nhau cn ch n t l giacc kh to thnh (thng thng, bi s cho t l ny, hoc vi vi phptnh n gin s tnh ra).

- Nu l kim loi tan c trong nc nhit thng (kim loi kim,Ca, Ba), cn ch '+ Nu axit d: chc phn ng gia axit v kim loi+ Nu axit thiu: ngoi phn ng gia axit v kim loi(xy ra trc),

cn c phn ng gia kim loi d vi nc to dung dch kim.Lo i 2:Hn hp hai kim oi tc dng vi mt axit.

- Vit ng PTHH+ Kim loi no c tnh kh mnh hn s un tin phn ng c.+ Nhng kim loi ng sau H trong dy in ha th khng phn

ng vi cc axit c tnh oxi ha yu nh (HC1, H2SO4long, H3PO4...)+ Vi cc axit c tnh oxi ha (HNO3, H2SO4 c) cn phn bit

c cc kh to thnh nhiu trng hp khc nhau. Ch : Al, Fe, Cr, Mnbth ng ha trong HNO3, H2SO4c, ngui.

- Cng thc tnh khi lng mui trong dung dch:

mmui= y 'm + y m .IUUUI / caliou / J anion

hay rtlmu i - mhh hai kimloi anj1)n

Lo i 3:Mt kim loi tc dng vi hn hp hai axit.Trng1hp i: Hn hp 2 axit khng mang tnh oxi ha (H+ ng vai

tr l cht oxi ha). Khi :

2 n H+ = n H C l+ ^ n H2S 0 4 + ^ n H 3P 0 4

- Vit PTHH di dng ion, sau gii thng thng.2X + 2nH+ 2X+ + nH2

Ltru :n l s oxi ha thp ca XTrng h~p 2: Hn hp 2 axit gm 1 axit khng mang tnh oxi ha v

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18/661

1 axit mang tnh oxi ha. (Thng thng l H2SO4long/HCl v HNO3) thtrong hn hp ny H+ ng vai tr mi trng cn N 3~ ng vai tr chtoxi ha.

- Tm s mol ca H+ v NC>3~

S n H+ = n H C I+ ^ n H2S 0 4 + ^ n H -P 0 4 + n H N 0 3

- Vit PTHH dng ion.- Xc nh cht phn ng ht t PTHH bng cch so snh cc t s mol

v h s tlng trong cc PTHH. T tnh ton bi ton bng cht phnng ht.

L u : NC>3~ trong mi trng axit c tnh oxi ha mnh nh HNO3V d :Cho Ag vo hn hp dung dch cha Cu(N 3)2v HC1 thy Ag

tan v H+ c trong HC1 gp NO3- trong Cu(N 3)2c tc dng nh HNO3. V

vy n oxi ha c Ag.- Trong mi trng trung tih (H2O): khng mang tnh oxi ha.

- Trong mi trng baz (OH-) : bAl, Zn kh n NH3Lo i 4:Hn hp nhiu kim loi tc dng vi hn hp hai axit.Nh n xt: Vidng bi tp ny, phi bin lun nhiu trng hp. ngin ha, thng thng ta s dng phng php bo ton electron trong

phn ng oxi ha - kh hay phng php ion - electron gii.- Dng bi tp ny rt kh vit PTHH to mui. V vy, tm khi

lng ca mui to thnh, ta s dng cng thc sau;n i m u i ~ n i h n h p k im lo i n i an i o n t o m u i

V d 1: Cho 13,5 gam hn hp cc kim loi Al, Cr, Fe tc dng vi lngd dung dch H2SO4 long nng (trong iu kin khng c khng kh),thu c dung dch X v 7,84 lt kh H2 ( ktc). C cn dung dch X(trong iu kin khng c khng kh) c m gam mui khan. Gi trcam lA.42,6. B.45,5. c.48,8. D.4 7 ,1.

(Trch thi tuy n sinh Cao ng nm 2008)H ng d n gi i

T ' _ 7 84 1Ta c: nu = _ - = 0,35 mol22,4 ,

p dng nh lut bo ton khi lng:

ITChh + m|_J SQ llm n i

=> mmui= 3,5 + 0,35.98 - 0,35.2 = 47,1 gam. p n D.

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19/661

V d 2: Cho 6,72 gam Fe vo dung dch cha 0,3 mol H2SO4 c, nng(gi thit SO2l sn phm kh duy nht). Sau khi phn ng xy ra honton, thu cA. 0,03 moi Fe2(S 4)3v 0,06 moi FeS 4B. 0,05 mol Fe2(SC>4)3v 0,02 mol Fe d.c . 0,02mol Fe2(S04)3v 0,08mol FeS4.

D. 0,12 mol FeS04.(Trch thi tuy n sinh i h c kh i B nm 2007)

H ng d n gi i

Ta c: npe = 0,12 mol

- Ptp: 2Fe + 6H2SO4c Fe2(S 04)3 + 3S02+ 6H200,1 0,3 0,05

Ta thy: ^ M => nFe d = 0,12 - 0,1 = 0,02 mol

2 6V Fe d nn tip tc xy ra phn ng:Fe + Fe2(S4)3 >3FeS04

0,02 0,02 0,06

Ta thy: < M => nF (so ^ d = 0,05 - 0,02 = 0,03 mol

Vy dung dch sau phn ng gm: 0,06 mol FeS 4v 0,03 mol Fe2(S04)3=> p n A.

V d 3: Cho 3,68 gam hn hp gm AI v Zn tc dng vi mt lng va dung dch H2SO410% thu c 2,24 lt kh H2( ktc). Khi lngdung dch thu c sau phn ng lA. 01,48gam. B. 101,68 gam. c . 97,80 gam. D. 88,20 gam.

(Trch thi tuy n sinh i h c kh i A nm 2009)

H ng d n gii

~ 2 24Ta c: n = n = -2~ =0,1 mol

2 4 2 2 2 4

9 8x100^ mH2so4 = 0 l x 9 8 = 9 8 gam => mddH2so4= = 98 gam


m h n h p KL + m U( H so nid sau phn ng

lUdd sau ph n ng rnh nh pKL ddH so

= 3,68 + 98 - ,1x2 = 101,48 gam=> p n B.

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20/661

V d 4: Ha tan ht 7,74g hn hp bt Mg, AI bng 500ml dung dch hnhp HC1 IM v H2SO4 0,28M thu c dung dch X v 8,736 lt H2 (ktc). C cn dung dch X thu c lng mui khan l:A. 38,93 gam B. 103,85 gam c . 25,95 gam D. 77,86 gam.

(Trch thi tuy n sinh Cao ng kh i A,B nm 2008)

H ng d n gi i_ , _ 8,736 _Ta c: nu - ' =0,39m ol

H2 22,4

nHCi = 0,5.1 = 0,5 mol

nH = 0,28.0,5 = 0,14 mol2 4


I f i h h m [ tC l SO _ m r nu i "* r n H

=> mmui- 7,74 + 0,5.36,5 + 0,14.98 - 0,39.2 = 38,93 gam.=>pnA.

V d 5: Cho 3,2 gam bt Cu tc dng vi 100ml dung dch hn hp gmHNO3 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 trca V lA. 0,746. B. 0,448. c . 1,792. D. 0,672.


Ta c : nr = =0, 05 molCu 64

r nHNOs : 0,08 mol ^ J nH+ : 0,08 +0 ,02.2 = 0,12 molnH2SC>4: 0,02 mol \ nN 3~ : 0,08 mol

Phn ng: 2N03 + 3Cu + 8H+ -> 3Cu2+ + 2NO + 4H200,08 0,05 0,12 _ TT+1X ._ _ , T+Ta thy: -7-> > => H ht, NO c tnh theo H .

2 3 8

1 1Theo phng trnh phn ng: nN0= .n += .0,12 = 0,03 mol

=> VNO= 0,03.22,4 = 0,672 ltp n D.

V d 6 : Ho tan hon ton 8,862 gam hn hp gm AI v Mg vo dungdch HNO3long, thu c dung dch X v 3,136 lt ( ktc) hn hp Ygm hai kh khng mu, trong c mt kh ho nu trong khng kh.

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21/661

Khi lng ca Y l 5,18 gam. Cho dung dch NaOH (d) vo X v unnng, khng c kh mi khai thot ra. Phn trm khi lng ca AI tronghn hp ban u lA. 12,80% B. 15,25% c . 10,52% D. 19,53%

(Trch thi tuy n sinh Cao ng nm 2009)H ng d n gi i2kh khng mu => 2 kh chc th l: N2, N2O, NO.Trong c mt kh ha nu trong khng kh kh l NO.

Ta c: n = = 0,14 mol => My = = 37hhY 22,4 0,14

T a thy : M n o = 30 < M y = 37 < Mkh cn li => kh c n li l N 20 (M = 4 4)

Gi IINO - Xmol; s mol N2O = y molJ-X + y = 0,14 rx = 0,07

l-30x + 44y = 5,18 ^ i - y = 0, 07Gi nAi = a mol; riMg = b mol.Cho dung dch NaOH (d) vo X v un nng, khng c kh mi khai

thot ra => trong dung dch X khng c mui NH4NO3Qu trnh nhng e:

Mg - 2e Mg2+a 2a

AI - 3e -> Al3+b 3b

Qu trnh nhn e:

N + 3e -> NO0,21 0,07

2N + 8e > N2o 0,56 0,07

p dng nh lut bo ton e, ta c : 2a + 3b = 0,21 + 0,56 = 0,77 (1)Mtkhac: 27a + 24b = 8,862 (2)T (1), (2) => a = 0,042; b = 0,322

0 042 27

Vy: %mAi = -.100% = 12,8%8,862= > p n A.

V d 7: Ho tan hon ton 1,23 gam hn hp X gm Cu v I vo dung .dch HNO3 c, nng thu c 1,344 lt kh N 02 (sn phm kh duy

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22/661

nht, ktc) v dung dch Y. Sc t t kh NH3 (d) vo dung dch Y,sau khi phn ng xy ra hon ton thu c m gam ktta.Phn trm vkhi lng c Cu trong hn hp X v gi trca m ln lt lA. 21,95% v 0,78 B. 78,05% v 0,78c . 78,05% v 2,25 D. 21,95% v 2,25

Trch thi tuy n sinh i h c kh i B nm 2009 H ng d n gi i

1 3 4 4 'Ta c: nMn = =0,06 mol; goi ncu = X mol; nAi = y mol

N2. 22,4

Qu trnh nhng e:Cu - 2e -> Cu2+X 2x

AI - 3e -> Al3+y 3y

Qu tenh nhn e:+5 +4

N + le -> N 0 2

0,06 0,06p dng nh lut b ton e ta c h:

r2 x + 3y = 0,06 rx = 0,015

i 64x + 27y= 1,23 ^ Ly = 0,01

=> %Cu = 1 5' 64 100% = 78,05%1,23

Cu(OH)2an c trong dung dch NH3d v to phc, do kt ta thuc l Al(OH)3.

P to kt ta: Al3+ + 3NH3 + 3H20 -> Al(OH)3 i + 3NH4+0 ,0 0,01

=> m4A,(0H)3= 0,01.78 = 0,78g=> p n B.

V d 8 : Cho 2,16 gam Mg tc dng vi dung dch HNO3 (d). Sau khiphn ng xy ra hon ton thu c 0,896 lt kh NO ( ktc) v dungdch X. Khi lng mui khan thu c khi lm bay hi dung dch X lA. 8,88 gam. B. 13,92 gam. c . 6,52 gam. D. 13,32 gam.

(Trch thi tuy n sinh i h c kh i B nm 2008)

H ng d n gi iTa c: nMg = 0,09 mol v nNo = 0,04 mol

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23/661

l qu trnh to mui

- Qu trnh nhng electron:

Mg - 2e ------> Mg

0,09 0,18

- Qu trnh nhn electron:

+5 +2N + 3 e ----- -> N

0,12 0,04

Ta thy: ne nhng = 0,18 mol > lie nhn =0,12 mol=> Cn thm qu trnh nhn elecon na

NH4NO3v ne nh n = 0,18 - 0,12 = 0,06 mol.+5 +3

N + 8e ------> N

0,06 0,075Vy:mnniI= m Mg(NOi)2 + mNHiNOi

= 0,09.148 + 0,075.80 = 13,92 (g)

=> p n BV d 9: Ha tan 14,8g hn hp gm Fe v Cu vo lng d dung dch hn

hp HNO3 v H2SO4 c, nng. Sau phn ng thu c 10,08 lt NO2(ktc) v 2,24 lt SO2(ktc). Khi lng Fe trong hn hp ban u l:

. 5,6 B. 8,4 c . 18 D. 18,2.H ng d n gi i

Gi a, b ln lt l s moca Fe, Cu56a + 64b = 14,8 (1)

- Qu trnh nhng electron:

Fe - 3e +3

> Fe

a 3a

Cu 2 e+2

Cu

b 2b

'y nhng (3a + 2b) moi

- Qu trnh nhn electron:

N + le ------> N

0,45 0,45

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24/661

+6 +4s + 2 e ------ s

0,2 0,1

=> X " nhn = O45 + O2= ^ 5mo1

- p dng nh lut bo ton electron:3a + 2b = 0,65 (2)

T (1), (2) => a = 0,15 v b = 0,1=5- mpe = 8,4 g

=> p n B.

V d 10: un nng m gam hn hp Cu v Fe c tl khi lng tng ng7 : 3 vi mt lng dung dch HNO3. Khi cc phn ng kt thc, thu

c 0,75m gam cht rn, dung dch X v 5,6 lt hn hp kh (ktc) gmNO v NO2(khng c sn phm kh k;hc ca N+5). Bit lng HNO3phn ng l 44,1 gam. Gi trca m lA. 44,8. B. 40,5. c . 33,6. D. 50,4.

(Trch .thi tuy n sinh i h c kh i A nm 2011)H ng d n gi i

- Khi lng Fe = 0,3m gam v khi lng Cu = 0,7m gam

Sau phn ng cn 0,75m gam => Fe chphn ng 0,25m gam; Fe d vysau phn ng chthu c mui Fe2+

Ta c: n = 0,7 ; nNO+ nN0 = 0,25, s mol ca Fe(N03)2= - ^ | m2 56

S p :

Fe + HNO3 Fe(N3 )2 + NO + NO^

0 ,7 5 ^ 5 .

56 56p dng LBT nguyn t N ta c:

0,7 = 2. - 5 + 0,25 =>m =50,4 (g)56

=> p n D.V d 11: Ha tan hon ton 13,00 gam Zn ong dung dch HNO3long, d

thu c dung dch X v 0,448 lt kh N2(ktc). Khi lng mui trong

dung dch X lA. 18,90 gam B. 37,80 gam c . 39,80 gam D. 28,35 gam

(Trch thi tuy n sinh Cao ng nm 2011)

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25/661

H ng dn gi i

Hr - 1 3 - n o . 0,448 _ n no ,Ta c: nzn = = 0,2 mol v nN, = - = 0,02 mol65 22,4

He nhng 2.ln 0,4 mol > le nhn 10. 0,2 moi

=> Phn ng c to thnh NH4NO3.

= 0,025 mol

+5 -3

(v khi to thnh NH4NO3: N + 8 e -> N )

Khi lng mui toong dung dch X l = 180.0,2 + 80.0,025 = 39,80 gam

=> p n c .L u : bi khng ni thu c kh X duy nh t nn c th c mui

NH4NO3to thnh.V d 12: Cho m gam hn hp bt X gm ba kim loi Zn, Cr, Sn c s moi

bng nhau tc dng ht vi lng d dung dch HC1 long, nng thuc dung dch Y v kh H2. C cn dung dch Y thu c 8,98 gammui khan. Nu cho m gam hn hp X tc dng hon ton vi O2(d) to hn hp 3 oxit th th tch kh 0 2(ktc) phn ng lA. 2,016 lit. B. 0,672 lt. C. 1,344 lt. D. 1,008 lt.

- Ba kim loi Zn, Cr, Sn khi tc dng vi dung dch HC1 long, nng uboxi ha thnh s oxi ha +2 nn: Gi R l kim loi chung cho Zn, Cr, Snkhi tc dng vi dung dch HC1.

Ta c: n f = ---- --------- = 0,06 molR2 236

( f + 7,)

Theo ptp: riR= n = 0 06 moi => s mol mi kim loi l 0,02 mol- P vi oxi:

2Zn + O2 - 2ZnO0,02 0,01


- Do s mol 3 kim loi bng nhau nn: M r=

-P : R + 2HC1 -+ RC12 + H2T0,06 0,06

65+52 + 119 2363

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26/661

2Cr + - 0 2 Cr20 32

0,02 0,015Sn + 0 2 Sn 2

0,02 0,02=> E a - = 0,045 mol => V = 1,008 lit

2

's. pnD.V d 13: Ha tan hon ton 14,6 gam hn hp X gm AI v Sa bng dung

dch HC1 (d), thu c 5,6 lt Hi ( ktc). Th tch kh 2( ktc) cn phn ng hon ton vi 14,6 gam hn hp X lA. 3,92 lt. B. 1,68lt c . 2,80 lt D. 4,48 lt

(Trch thi tuy n snh i h c kh i A nm 2009)H ng d n gi i

Ta c: n = 0,25 mol2

Gi IAI = X moi; nsn = y raol :=> 27x + 119y = 14,6 (1)Khi cho I, Sn phn ng vi dd HC1 th Al, Sn l cht kh, HG1 l cht

oxi ha:Qu trinh nhng e:

A - 3 e A13+

X 3xSn - 2e -* Sn2+

y 2yQu trnh nhn e:

2H+ + 2e -> H20,5 0,25

p dng L bo ton e, ta c: 3x + 2y = 0,5 (2)T (1) v (2) => X = y = 0,1 moiKhi cho Al, Sn phn ng vi O2th Al, Sn l cht kh, O2l cht oxi ha:

-- 3e >+3

AI

0,1 0,30 +4

Sn - 4 e Sn

0,1 0,4

^ nhng inol

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27/661

Qu trnh nhn e:

2 + 4e -> 200,175 0,7

=> Vco =0,175.22,4 =3,92 (lt)

= > p n A.L u :Sn c hai ha trl II v IV, khi tc dng vi cht OXH yu nh

HCI, H2SO4 long to hp cht thic (II); cn khi tc dng vi cht OXHmnh nh O2, CI2THNO3... to hp cht thic (IV).V d 14: Cho m gam bt Fe vo 800ml dung dch hn hp gm Cu(N 3)2

0,2M v H2SO40,25M. Sau khi cc phn ng xy ra hon ton, thu c0,6m gam hn hp bt kim loi v V lt kh NO (sn phm kh duy nht ktc). Gi trca 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.(Trch thi tuy n sinh H kh i B nm 2009)

H ng d n gi i

Ta c: nr. ,Nr, , = 0,16 mol=> n = 0,16m ol; n _ = 0,32 molCu(NO . ) 2 Cu NO3

nHso = 0,2mo => n += 0,4 mol

Fe + 4H+ + NO3' - Fe + NO + 2H200,1 0,4 0,1 0,1 0,1

Vn o ==2,24 (lt)Fe + Cu2+^ F e 2++Cu0,16 0,16 0,16

Fe d nn tip tc xy ra phn ng:Fe + 2Fe3; -> 3Fe2+

0,05 0,1 0,15

^ lFe phn ng 0 ,0 5 0 , 16 0 ,3 1 m o i

0,6m (g) hn hp bt kim loi sau phn ng gm Fe d v Cu sinh ra:

= > m - n i F e p + m cu = 0 ,6 m = > m - 0 ,3 1 .5 6 + 6 4 .0 ,1 6 = 0 ,6 m

m = 17,8 (g)Cch khc:

0,6m (g) hn hp bt kim loi sau phn ng gm Fe d v Cu sinh ra. DoFe d nn mui st sinh ra l mui Fe +

Qu trnh nhn e:N 0 3' + 4H+ + 3e NO + 2H200,4 0,3 0,1

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28/661

Cu2+ + 2e ^ Cu0,16 0,32 0,16

=> Vno = 0,1.22,4 = 2,24 litQu trnh nhng e:

Fe - 2e - Fe2+

0,31 0,62Ta C: mpc bd - nit'c p + mcu = 0,6m

=> m - 0,31-56 + 64.0,16 = 0,6m ==>111= 17,8 (g)=> p n B.

V d 15: Ha tan hon ton 12,42 gam AI bnti dung dch HNO.} long(d), thu c ung dch X v 1.344 lt ( ktc) hn hp kh Y gm haikh l N20 v N2. T khi ca hn hp kh Y so vi kh 12 l 18. C cndung dch X, thu c m gam cht rn khan. Gi trca m lA. 97,98. B. 106,38. c. 38,34. D. 34,08.

(Trch thi tuyn sinh D i h c kh i A nm 2009)H ng d n gii

Ta c: nAi = 0,46 mol; M hhkhi = 18.2 = 36 ; nhh = 0.06 molGi s mpl ca N2O v N2ln lt l Xmol v y moi.

Ta c h :

N + 8e > N H 4 N O 3(1,38-0,54) 0,105

Vy: mn, i = mAI(NOi)i + mNHjNO = 0,46.213 + 0.105.80 = 106,38 (g)

=> p'nB.

Qu trnh nhng e: AI - 3e >Al+0,46 1,38

+5 +1

Qu trnh nhn e : 2N + 8e ----- >N2O

0,24 0,03

2N + lOe-----

N20,3 0,03

=i> Tng s moi e nhn = 0,54 (mol)Ta thy: nc nhng > ne nhn => cn phn ng to NH4NO3

+5 -3

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29/661

D ng 3: Gi ton kim lo i tc d ng v i dung ch mu i.Nhn dng: Cho kim loi N (ng trc) phn ng vi dung dch mui M"

(ng sau). Tnh khi lng N, hoc tng gim khi lng sau phn ngH ng gi i:

mMm+ -> nMm+ + mNm n ra

am am

n n

am= - -.N -a M

a(mN-nM)

n n

*_ I u am a(Mn-m N)

A m = a M - .N = - :-------- -n n- Trong a l s mol kim loi M tham gia phn ng, gi s ton b

lng kim loi gii phng ra bm vo thanh kim loi M.

V d 1: Cho 2.7 gam hn hp bt X gm Fe v Zn tc dng vi dung dchCuS04. Sau mt thi gian, thu c dung dch Y v 2,84 gam cht rn z.Cho ton b z vo dung dch H2SO4 (long, d). Sau khi cc phn ngkt thc th khi lng cht rn gim 0,28 gam v dung dch thu c ch

cha mt mui duy nht. Phn trm khi lng ca Fe trong X l:A. 58,52% B. 51,85% c. 48,15% D. 41,48%(Trch thi tuy n sinh i h c kh i A nm 2011)

H ng d n gi i

* Cch 1:

Zn c tnh kh mnh hn Fe, Zn s phn ng vi dung dch CuS 4trc.

Theo : hn hp rn zphn ng vi dung dch H2SO4long thu c 1

mui duy nht => trong hn hp rn z c Cu v Fe d. Vy riiFed = 0,28 (g)v mcu - 2,84 - 0,28 = 2,56 (g).Ta c: Khi lng hn hp X phn ng vi Cu2+ = 2,7 - 0,28 = 2,42 (g).Gi nz = X mol; liFe pr = y moi.

Zn + C11SO4 Cu + ZnS 4X X

Fe + C11SO4 - Cu + FeS04y y

Ta c h:f65x + 56y = 2,42 Jx = 0,02L64x + 64y = 2,56 ^ ty = 0,02

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30/661

=> mFe bi. = 0,02.56 + 0,28 = 1,4 (g)

=> %Fe= .100% =51,85%2.7

=> p n B.

* Cch 2:V z + H2SO4chcha mt mui >Zn ht , Fe d

56x + 65y + 01,28 = 2,7 jy = 0 ,0 2

[64 y + 64x + 0,28 = 2,84 |x = 0,02

%Fe = .+>28 ,100 = 51,85%2.7

= p n B.

V d 2: Cho m gam bt Zn vo 500ml dung dch Fe2(SC>4)30,24M. Sau khicc phn ng xy ra hon ton, khi lng dung dch tng thm 9,6 gamso vi khi lng dung dch ban u. Gi trca m lA. 32,50 B. 20,80 c . 29,25 D. 48,75

(T'rch thi tuy n sinh i h c kh i B nm 2011)H ng d n gi i

Ta c: np, ... . = 0,12 mol => n_ = 0,24 mol

0,12 .0,24 0,24Zn + Fe2+ Zn2+ + FeX X

T a c o ; Aldd (ng niZn phn na Fe sinh ra

=> 9,6 = (0,12 + x).65 - 56.X X= 0,2Vy: mZo = (0,12 + 0,2).65 = 20,8 (g)

= p n B.

V d 3: Cho 19,3 gam hn hp bt gm Zn v Cu c tl mol tng ng l1 : 2 vo dung dch cha 0,2 mol Fe2(S 4)3. Sau khi cc phn ng xyra hon ton, thu c m gam kim loi. Gi trca m l

A.6,40. B. 16,53. c . 12,00. D. 12,80.


Gi X l s mol ca Zn th s mol ca Cu l 2x => 65x + 64.2x = 19,3 (1)

X= 0,1 => nz = 0,1; ncu = 0,2; nFe3+= 0,4.31



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31/661

Zn kh mnh hn Cu nn s phn ng vi Fc3' trc.Cc phn ng: Zn + 2Fe3->2Fe2+ + Zn2+

0,1 0,2

=> n , cn = 0,4 - 0,2 = 0,2 mol nn tip tc xy ra phn ng:

Cu + 2Fe3+-2Fe2+ + Cu2+

0,1 0,2

Nn s mol Cu d l 0,1 => m = 6,4 (g)

=> p n A.V d 4: Nhng mt l kim loi M (ha tr II) nng 56 gam vo 200ml dd

AgN 3 IM cho n khi phn ng xy ra hon ton. Lc dung dch,em c cn c 18,8g mui khan kim loi M l:

A. Mg B. Zn c . Cu IX F(Trch thi tuy n sinh D i h c kh i B nm 2009)

H ng d n gi i

Ta c: nANOj = 0,2 moi

M + 2AgN03 M(N03)2+ 2Ag0,2 0,1

M + 62.2 = = 188 => M = 640 ,1

=>p n c.V d 5: Nhng mt thanh st nng lOOg vo lOOml dung dch hn hpgm Cu(N 3)2 0,2M v AgN 3 0,2M. Sau mt thi gian ly thanh kimloi ra, ra sch cn c 101,72 gam (gi thit cc kim loi t thnhu bm ht vo thanh st). Khi lng st phn ng l:A.2,16 B. 1,40 c .0 ,8 4 D. 1,72.

(Trch th tuy n sinh i h c kh i B nm 2009)H ng d n gi i

n ASN O , = 0 2 m 0 ; n C u ,N O .,)2 = 0 2 m o !

+ Gi s AgNC>3phn ng ht; Cu(N03)2cha phn ngFe + 2 AgN3-> Fe(NOj)2+ 2 Ag0,01 0,02 0,02

mthanhkim lo i = 1 0 0 - 0 , 0 1 . 5 6 + 0 , 0 2 . 1 0 8 = 1 0 1 , 6 * 1 0 1 , 7 2 = > v l

+ Gi s AgN 3v C u (N3)2 u phn ng ht I

Fe + Cu(N03)2> Fe(N03)2 + CuX X X

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32/661

mthanhkimioi = 100 - (0,01 + 0,02).56 + 0,02.108 + 0,02.64= 101,76* 101,72 => v l.

=> AgN 3phn ng ht: Cu(N 3)2phn ng 1phn.Ta c: 0,02.108 + 64x - 56x - 0,0156= 1,72

X= 0,015 mol => iripe phn ng = (0,01 + 0,015).56 = 1,4 gp n B.

V d 6 : Cho m(g) hn hp bt Zn v Fe vo lng d dung dch CuS 4.Sau khi kt thc cc phn ng, lc b dung dch thu c m(g) cht rn.Thnh phn % theo khi lng ca Zn ong hn hp ban u l:A. 90,27% B. 82,20% c . 85,30% D. 12,67%.

(Trch thi tuy n sinh i h c kh i A nm 2007)H ng d n gii

Gi a,b ln lt l s mol ca Zn v Fe.

Zn + CuS04 ------> Z11SO4 + Cu (1)a a

Fe + C11SO4 ----- > FeS 4 + Cu (2)b b

Phn ng (1) lm gim khi lng hn hp kim loi, phn ng (2) lmtng khi lng kim loi.

M theo bi ban u c m(g). Sau phn ng c m(g) cht rn.

Vy: Am / = Am \ => (65 - 64)a = (64 - 56)b => = 'd 8

=> %Zn = ----- -100% - 90,27% =? p n A.65.8 + 56.1

V d 7: in phn (vi in cc tr) 200ml dung dch Q 1SO4nng Xmol/1,sau mt thi gian thu c dung dch Y vn cn mu xanh, c khi lnggim 8g so vi dung dch ban u. Cho I6 ,8g bt Fe vo Y, sau khi cc

phn ng xy ra hon ton, thu c 12,4g kim loi. Gi trca X lA.2,25 B. 1,5 c . 1,25 D. 3,25(Trch thi tuy n sinh i h c kh i B nm 2010)

H ng d n gi i

T a co : npe ban u Qj3 niolPhng trnh in phn dung dch C11SO4:

C11SO4 + H20 ^ Cu + H2SO4 + - O 2 (1)

1a a a a

2

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33/661

Gi ncuso4p = a mo => 64a+ 32. j a - 8=> a = 0,1 mol

_ Dung dch Y gm: 0,1 mol H2S04v Cu2+ d

P: Fe + H2SO4 > FeS04 + H2T0,1 0,1

= > niFe cn 16,8 0,1.56 11,2 (g)P: Fe + Cu2+ - Fe2+ + Cu (2)

b b b

ONaC! = 0 ,2 . 0 , 7 = 0 , 1 4 m o l

AgNOj + NaCl - AgCU + NaN030,14 0,14

=> L ngAgNOj phn ng vi M l: 0 ,2 -0 14 = 0,06 molM + nAg+-> Mn+ + nAg

^ 0,06 0,06 0,06n

Ta c: 0,05.108 - = 5,94nx T-l

=>6,84n - 0,05M = 5,94n M = 9n

Khi n = 3 th M = 27, vy M l AI => p n c.

Gi nc2+p = bmol => A ra = 6 4 b -5 6 b = 12 ,4-11,2

Theo p (1),(2) ta c: nCuS04b = 0,1 + 0,15 = 0,25 molb - 0,15 moi

=> [CUSO4] = ^ = 0,125 M

p n c.

A. Mg B. Zn c . AI

i vo dung dchloi ra cn li thyta hon toan bi

- A 1 D. PbH ng d n gi i

n



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34/661

H ng d n gi i

Phng trnh phn ng:

Fe + Cu2+ -> Fe2+ + Cu1 1 1 1 (mol)56a 64a (g)

Tac: Am T= 64a-5 6a = 1,2=3>a = 0,15mol

_ _ 0,15=> CM,r. , = = IM

M (C u S(J4 ) ( ) i 5

=> p n A.V d 10: Ngm mt vt bng Cu c khi lng 15g trong 340g dung dch

AgN 3 6 %. Sau mt thi gian ly vt ra thy khi lng AgN 3 trong

dung dch gim 25%. Khi lng ca vt sau phn ng l:A. 3,24g. B. 2,28g c . 17,28g. D. 24,12g.H ng d n gi i

T a c: m . Nn = = 20 ,4 (g) => n . = 0 ,12 m olAgNO? J 00 AgN03 9

Khi lng AgN03gim 25% chnh l lng AgNOs phn ng.

Vy: nAgNOj phn ng =0,12.25% = 0,03 mol

Phn ng: Cu +2AgN

3 - Cu(N

3)2 + 2Ag0,015 0,03 0,03

mVt = 15 + (0,03.108 - 0,015.64) = 17,28 (g)

p n c .

V d 11: Tin hnh hai th nghim sau:

-T h nghim 1: Cho m gambt Fe (d) vo Vi lt dung dch Cu(N 3)2 IM.- Th nghim 2: Ch m gam bt Fe (d) vo V2lt dung dch AgNC>30,1M.Sau khi cc phn ng xy ra hon ton, khi lng cht rn thu c

hai th nghim u bng nhau. Gi trca V 1 so vi V2l:A . V i = V 2 B.V! = 10V2 c . Vi = 5V2 ! D . V | = 2 V 2.

(Trch thi tuy n sinh i h c kh i B nm 2008)H ng d n gi i

Th nghi m 1 : nCu(N03)2= V, mol

Fe + Cu(N03)2 -> Fe(N03)2 + CuVimol Vimol Vimol

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35/661

Fe d nn Cu(N 3)2 ht => ncu= riFc = nCu(NOOi = !V| mol

=> tng khi lng: (64 - 56).vJ = 8VI mol.

Th nghim 2: nA,N0 = 0,1.V2 raol

Fe + 2 AgN 3 -> Fe(N 3)2 + 2Ag0,05V2mol 0 ,l.V2 mol 0 ,l.V'2 mol

Fe d nn AgNC>3 ht => nAa= n ANOi = 0,1.V2 moi

V n . = n . Mn = -0 ,l .V ,= 0,05.V,moIFc 2 AgNO} 2

=> tng khi lng: 108.0,1 V2 - 56.0,0 5V2= 8V2mol.Theo bi: sau phn ng khi lng cht rn thu c bng nhau.=> tng khi lng hai th nghim cng bng nhau.

Hay: 8V1= 8V2 => V, = v 2=> p n A.V d 12: Nhng thanh kim loi R ha trII vo dung dch C11SO4. Sau mt

thi gian ly thanh kim loi ra thy khi lng gim 0,05%. Mt khcnhng thanh kim loi trn vo dung dch Cu(N 3)2, sau mt thi gianthy khi lng tng 7,1%. Bit rng s mol R tham gia 2 trng hpnh nhau. R l:A. Cd B. Zn c . Fe D. Sn.

H ng d n gi i .Gi a l khi lng ban u ca thnh kim loi R.

R + CuSC>4 ----- > RSO4 + CuX X

=> (M -64).x = ^ a (1)R 100

R + Pb(N 3)2 ---- > R(N03)2 + Pb

=> (2 0 7 -MJ.X = a (2)R 100

Ly (2): (1) =>Mr= 65 (Zn)=>p n B.

3. D ng 4 : Bi ton v tnh kh ca CO, H?.Bi ton t ng qut: n V lt kh c o hoc V lt H2d qua ng ng rri|gam mt hn hp hoc mt oxit baz nhit cao. Sau khi phn ng ktthc thu c m2 gam hn hp hoc mt cht rn y dn kh sinh ra vo



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36/661

nc vi trong d th thu c m3gam kt ta..- Xc nh cng thc phn t ca oxit ban u (nu d liu cho mt oxit baz)- Xc nh mi nu bit m2v m3. V ngc li ca cc thng s.- Xc nh V, m nu bit mi v m2. V ngc li ca cc thng s.

dng bi tp ny, thng thng ta s dng phng php bo ton

nguyn t.Ta c PTHH : c o + [ 0 ] C02

hoc H2 + [ 0 ] -> H20

Vy s mol oxi nguyn t c trong hn hp oxit bng s mol ca CO2

Vy : m2= mi - nco *16

Tng t, rri-0-= I1| - m2 => nco = ri|-0j => m3

L u : CO, H2chkh c cc oxit ca kim loi ng sau AI trong dyhot ng ha hc.______________________________________________

V d 1: Dn kh c o d qua ng ng bt mt oxit st (FxOy) nhit cao. Sau khi phn ng kt thc thu c 0,84 gam st v dn kh sinh ravo nc vi trong d th thu c 2 gam kt ta. Xc nh cng thcphn t ca FexOy?Fe30 4 B. FeO c . Fe20 3 D. Fe20

H ng d n gi i

T a c : n caco3= ^ = 0 ,0 2 (m o l)

nFe= =0,015 (mol)

PTHH: FexOy + yCO -> xFe + yC02

C 02 + Ca(OH)2 -> CaC03+ H20

0,02

0,02Ta c: no (trong oxit) rico phn tig nco 0,02 (moi)

Mtkhc: nF = = 0,015mol => t l = = Fc 56 y 0 ,0 2 4

Vy CTPT ca oxit l Fe304=> p n A.V d 2: Kh 16 gam hn hp cc oxit kim loi : Fe, Fe2 3, Fe3 4, CuO

v PbO bng kh c o nhit cao, khi lng cht rn thu c l 11,2gam. Tnh th tch kh c o tham gia phn ng (ktc).A. 22,41 B. 4,481 c . 5,6 1D.6,721

37



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37/661

H ng d n gi iCch 1:Ta c: moxit = mKt. + mo

=> mo= m0xit - IKL = 16 - 11,2 = 4,8 (g):=> no = 4,8 : 16 = 0,3(mol)

PTHH: CO + [ o ] C 0 2

=> nco = no = 0,3 (mol)Vy: Vco = 0,3.22,4 = 6,72 lt

Cch 2:p dng LBT khi lng:

Ta c: nco = nco = Xmol

m ox,t + m c o = m ch i rn + m c o

=> 28x - 44x = 11,2- 16

=> X= 0,3-Vy: Vco = 0,3.22,4 = 6,72 lt=> p n D.

V d 3: Cho V lt hn hp kh ( ktc) gm c o v H2phn ng vi mtlng d hn hp rn gm CuO v Fe3 4nung nng. Sau khi cc phn ngxy ra hon ton, khi lng hn hp rn gim 0,32g. Gi trca V l:

A. 0,1121t B. 0,560 lt c . 0,448 lt I). 0,224 lt(Trch thi tuy n sinh D i h c kh i A nm 2008)

H ng d n gi i

Ta c: tigim = rrij-Q-j = 0,32g

= > X n c o+ H = n o = 0 ,0 2 m o ! => V = 0 ,0 2 X 2 2 , 4 = 0 , 4 4 8 lt

=> p n c .

V d 4: Cho lung kh c o (d) i qua 9,1 gam hn hp gm CuO vAI2O3nung nng n khi phn ng hon ton, thu c 8,3 gam cht rn.Khi lng CuO c trong hn hp ban u l:A. 0,8 gam B. 8,3 gam c . 2,0 gam D. 4,0 gam

(Trch thi tuy n sinh D i h c kh i A nm 2009)

H ng d n gi i

V AI2O3 khng bkh bi cc cht kh thng thng nh c o nn cccht sau phn ng s l Cu v AI2O3

Gi s mol CuO v AI2O3trorig hn hp u ln lt l a v b. Ta c h

^ 80a + 102b = 9,1 fa = 0,05phng trnh sau: c=> p n D.

V d 5: Thi mt lung kh c o d qua ng s ng m gam hn hp gmCuO, Fe2O j, FeO, A120 3nung nng thu c 2,5 gam cht rn. Ton b

kh thot ra sc vo nc vi trong d thy c 15 gam kt ta trng. Khilng ca hn hp oxit kim loi ban u l:A. 7,4 gam B. 4,9 gam c . 9,8 gam D. 23 gam

H ng d n gi iCc phng trnh ha hc:

M xOy + y C O xM + yC O

Ca(OH)2 * C 02 - CaC03+ H20T c o : IXloxit n^km loi m moxit = 2,5 + 0,15 . 16 = 4,9 (gam) => p n B.

V d 6: Dn t t V lt kh c o (ktc) i qua mt ng s ng lng dhn hp rn gm CuO, Fe2 3( nhit cao). Sau khi cc phn ng xyra hon ton, th c kh X. Dn ton b kh X trn vo lng ddung dch Ca(OH)2th to thnh 4 gam kt ta.V c gi trl:A 1,120 B. 0,896 C. 0,448 D. 0,224.

(Trch thi tuy n sinh Cao ng nm 2008)

H ng d n gi i

_ 4Ta c: n = =0,04 moicaco, 100

C 2 + Ca(OH)2 -> CaCOsi + H200,04 . 0,04

Trong phn ng kh CuO, Fe2 3bng c o , ta lun c:

nco = nco = 0,04 molVco = 0 ,04.22,4 = 0,896 lt

= > p n B.V d 7: Cho 4,48 lt kh c o ( ktc) t t i qua ng s nung nng ng 8

gam mt oxit st n khi phn ng xy ra hon ton. Kh thu e sauphn ng c tkhi s vi hiro bng 20. Cng thc ca oxit st v phntrm th tch ca kh C 2trong hn hp kh sau phn ng lA. FeO; 75%. B. FejOs; 75%.

C.Fe20 3;65%. D. Fe30 4; 75%.(Trch th tuy n sinh Cao ng nm 2007)

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39/661

H ng d n gi i

T 1 4 ,4 8 n ,Ta c: ncobanu = -r -T = 0,2 mol

22,4

M hh khi = 20.2 = 40 < Mco = 44 => hn hp kh sau phn ng c c o d

Gi ncop = Xmol => ncod = 0,2 - XTrong phn ng kh oxit st bng kh c o , ta luri c:

fk) (trong oxit) flco ~ ^co liol

, 2 8(0 ,2- x)+ 44 x _ A 1c __ 1Ta c: r ---------- =40 => X= 0,15 mol(0,2-x) + X

=> % v = xl00% = 75%c o, 0 2

_ 8-0,1 5.16 _ n . ,Ho (trong oxit) X 0 ,1 5 m o l IlFe (trong oxit) r UjlTOOl

56

Gi cng thc oxit st l FexOy

=> - = ^ = -5 :1 = - => OxitFe20 3y nG 0,15 3

=> p n B.V d 8 : Thi t t V lt hn hp kh (ktc) gm 0 0 v H2i qua mt ng

ng 16,8 gam hn hp 3 oxit: CuO, Fe30 4 , AI2O3nung nng, phn nghon ton. Sau phn ng thu c m gam cht rn v mt hn hp kh vhi nng hn khi lng ca hn hp V l 0,32 gam. Tnh V v m.A. 0,224 lt v 14,48 gam. B. 0,448 lt v 18,46 gam.c . 0 ,112 lt v 12,28 gam. D. 0,448 lt v 16,48 gam.

H ng d n gi iThc cht phn ng kh cc oxit trn l:

CO + o C02

H2 + o * h 2o.Khi lng hn hp kh t thnh nng hn hn hp kh ban u chnh l

khi lng ca nguyn t oxi trong cc oxit tham gia phn ng. Do vy:

0 32mo = 0,32 gam. => n0= - = 0,02 mol

16

= > ( n + n H2 ) = > 0 2 m O 1 -

p dng nh lut bo ton khi lng ta c:

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40/661

nioxit Ittcht rn 0 ,3 2

=> 16,8 = m + 0,32 => m = 16,48 gam.

=> V c o +H2) -0,0 2x 22 ,4 = 0,448lt.

p n D.

V d 9: Cho hi nc i qua than nng , thu c 15,68 lt hn hp khX (ktc) gm CO, C 2v H2. Cho ton b X tc dng ht vi CuO (d)nung nng, thu c hn hp cht rn Y. Ha tan ton bY bngdungdch HNO3 (long, d) c 8,96 lt NO (sn phm kh duynht, ktc). Phn trm th tch kh c o trong X l:A. 18,42% B. 28,57% c . 14,28% D. 57,15%

(Trch thi tuy n sinh i h c kh i B nm 201 ) H ng d n gi i

h 20 + c c o + h 2

X X X

2H20 + C * > C 02+2H2

2y y 2yTa c : nx = 0,7 mol :=> 2x + 3y = 0,7 (1)* hhX {CO, H2} + CuO Cu + HNO3 0,4 mol NO.

CO +CuO > CO2 + CuH2 + CuO -> H2 0 + Cu

3Cu + 8HNO3 - 3Cu(N03)2 + 2NO + 4H200,6 0,4

Trng phn ng kh oxi kim loi bi CO, H2

Ta c: n(C0H J= n0 (trong cuO) ^ncu = 0 ,6 m o l =>2x+ 2y = 0,6 (2)

T (l), (2) =>x = 0,2; y = 0,l

Vy : %Vco = . 100% = 28,57%

=> p n B.

JSkD NG 5; Gi i ton i n phnNhn dng: in phn cht X sau mt thi gian t vi cng dng

in l I. Tnh khi lng cht thu c in cc hoc pH ca dung dch

sau in phn.H ng gi i:

- Vit phng trnh in phn

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41/661

- Xt th t in phn, cht no in phn ht trc, cht no ht sau t d on pH ca mi trng

- Khi lng cht thot ra in cc:

A: Khi lng nguyn t (g/mol)

A.I.tm =

n.F

I: cng dng in (A)t: Thi gian in phn (s)

F: Hng s Faraday; F = 96500C

n: S' e trao i

V d 1: in phn 500ml dung dch C11SO40,2M (in cc tr) cho nkhi catot thu c 3,2 gam kim loi th th tch kh (ktc) thu c

anot :A. 3,36 lt B. 1,12 lt c . 0,56 lt D. 2,24 lt


- Ta c: ncu = 0,05 mol < nc so = 0,1 mol => C11SO4d, catot chc

Cu2+ bin phn, anot nc bin phn.- Ti catot: Cu2+ + 2e > Cu

0,1 0,05- Ti anot: 2 H2O - 4e > O2 + 4H+

0,1 0,025

=> nQ = 0,025 => v 0 = 0,56 lt

=>p n c .V d 2: in phn (in cc tr) dng-dch X cha 0,2 mol C11SO4 v 0,12

mol NaCl bng dng in c cng 2A. Th tch kh (ktc) thot ra anot sau 965 giy in phn lA. 2,240 lt. B. 2,912 lt. c . 1,792 lt. D. 1,344 lt.


, _ It _ 2.9650 _Tac .n = - 7 = -------= 0,2mol

c F 96500

Anot (+): c r , S042, H20

2CF - 2e Cl20,12 0,12 0,06

= > He c n = 0 , 2 - 0 , 1 2 = 0 , 0 8 m o l

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42/661

2H20 - 4 e -> 0 2+ 4H+0,08 0,02

=> Vkh: = VC| + VQ = 22,4.(0,06 + 0,02) = 1,792 lt => p n c .

V d 3: in phn c mng ngn 500ml dung dch cha hn hp gm

CuC2 IM v NaCl 0,5M (in cc tr, H = 100%) vi cng I = 5Atrong 3860 (giy). Dng dch thu.c sau in phn c kh nng ha tanm (g) Al. Gi trln nht ca m l:A.4,05 B. 2,7 c . 1,35 I). 5,4.


_ , . A.I.t _ 35 ,5.5 .38 60T c: mn = = ------ = 7,1 (g)

C,J nF 96500

7 1 .=> nr . = - f - = 0,1 mol

CI2 71

CuCl2 -> Cu + CI20,05 0,05

=> CuCl2 phn ng ht => NaCl in phn 1 phn

2NaCl + 2H20 > 2NaOI + H2+ Cl20,1 0,05

A + NaOH + H20 NaA102+ - H21

0,1 0,1 .

mAi = 0 ,l X 27 = 2 ,7 (g )

=> p n B.V d 4: in phn dung dch gm 7,45 gam KCi v 28,2 gam Cu(N 3)2

(in cc tr, mng ngn xp) n khi khi lng dung dch gim i

10,75 gam th ngng in phn (gi thit lng nc bay hikhng ngk). Tt c cc cht tan trong dung dch sau in phn lA. KN03v KOH. B. KNOj, KC1 v KOH.c . KNO v Cu(N03)2. D. KNO3, HNO3v Cu(N03)2.


Ta c: nKu = 0,1 m ol; nCu( ' = 0,15 mol

2KC1 + Cu(N03)2 - Cu + 2 KNO3 + Cl20,1 0,05 0,05 0,05

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43/661

KC1 ht, s moi Cu(N 3)2cn = 0,15 - 0,05 = 0,1

Cu(N03)2 + H20 -> Cu + 2HN 3 + - 0 2

1X X -7 X

2.

Khi lng dung dch gim = khi lng ca Cu kt ta + khi lngca C2v O2bay ra

= (0 , 0 5 + x ) 6 4 H 0 , 0 5 .7 1 + - X . 3 2 = 1 0 , 7 5 => X = 0 , 0 5

=>Cu(N 3)2vn cn d =>dung dch sau p ch KNO3; HNO3v Cu(N 3)2.* Cch khc:

Ti anot: 2CF, NO3 H2O Ti catot: Cu2+, K+, H2O

2C1 - 2 -* Cl2 Cu2+ + 2 e > Cu0,1 0,1 0,05 0,15 0,3

Ta thy: ne ( Cu(N03)2cn d => lo i A, B.Gi s anot H2 khng bin phn

=> Ar= mCI + mcu = 0,05.71 + 0,05.64 = 6,75 (g) < 10,75 (g)

=> H2O c bin phn catot lo i c.

=> p n D.V d 5: Ha tan 13,68 gam mui MSO4vo nc c dung dch X. in

phn X (vi in cc tr, cng dng in khng i) trong thi giant giy, c y gam kim loi M duy nht catot v 0,035 mol kh anot.Cn nu thi gian in phn l 2t giy th tng s mol kh thu c chai in cc l 0,1245 mol. Gi trca y lA. 4,480. B. 3,920. c 1,680. D. 4,788.

- (Trch thi tu n sinh i h c kh i A nm 2011)

H ng d n gi i in phn trong thi gian t giy thu c 0,035 mol kh vy 2t giy ta s

thu c 0,035.2=0,07 mol kh, nhng thc t ta thu c 0,1245 mol kh,s chnh lch s mol l do in phn nc to kh H2

=> n H = 0 , 1 2 4 5 - 0 , 0 7 = 0,0545

H20 - H2 + - O 22

0,0545 0,02725

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44/661

=> s mol 0 2 to ra do mui in phn = 0,07 - 0,02725 = 0 04275

=> Mmui = ' - = 160 => M + 96 = 160 => M = 640,0855

=> Khi lng Cu tnh theo t giy l mcu = 2.0.035.64 = 4,480 gam

=> p n D.V d 6 : in phn 200g dung dch AgN 3trong thi gian l 16 pht 5giy

vi I = 5A. e kt ta ht ion Ag+ cn li trong dung dch sau in phncn dng 25ml dung dch NaCl 0,4M. Vy c%ca dung dch AgN 3ban u l:A. 5,1% B.5,4% c.4,8% D. 10,8%

H ng d n gi i i t = 16 pht 5s = 16.60 + 5 =965 (s)Khi lng Ag thu c catot

=> p n A.V d 7: in phn dung dch cha 0,02 mol FeS 4v 0,06 mol HC1 vi

I = 1,34A trong 2 gi (in cc tr, mng ngn). B qua s ha tan cakh clo trong H2O, coi hiu sut in phn 100%. Khi lng kim loi

thot r catot v th tch kh (ktc) thot ra anot l:A. 11,2 g v 8,96 lt B. 1,12 g v 0,896 ltc . 5,6 g v 4,48 lt D. 0,56 g v 0,448 lt.

0,0855

MSO4 + H20 M + H 2SO 4 + - O 22

0,04275

A l t 108.5.965= 5,4 (g)

AgN03 + H20

0,05 mol

Ag + HNO3 + - O 2

0,05 molTa c: nNaci = 0,025.0,4 = 0,01 mol

^AgNCV^NaC^ 0 0111101

^ 2 nAgNQ, - 0,05 + 0,01 = 0,06 mol

c%= 7 0 . 1 0 0 = 5,1%

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45/661

H ng d n gi i

It 34 2S mol electron trao i: n = ^- = = 0,1 mol

c F 26,8Th t in phn catot: Th t in phn anot:

2C1" - 2e -> Cl20,06 0,032H20 - 4 e 0 2 + 4H+

2H+ + 2e H2

0,06 0,06 0,06Fe2+ + 2e > Fe

0,02 0,04 0,02 0,04 0,01=> mFe = 0,02.56 = 1,12 (g) Vkhano, = (0,03 + 0,01).22,4 = 0,896 lt=> p n B.

V d 8 : i n phn dung dch ch a a mol CuS 4v b moi NaCl (v i i ncc tr, c mng ngn xp). dung dch sau in phn lm

phenolphtalein chuyn sang mu hng th iu kin ca a v b l (bit ionS042- khng bin phn trong dung dch)A. b > 2a. B. b = 2a. c . b < 2a. D. 2b = a.


Phng trnh in phn:

C11SO4 + H20 ----- * Cu + H2SO4 + V2 O2 (1)

a a2NaCl + 2H20 ------> 2NaOH + Cl2 + H2 (2)

b b

H2SO4 + 2NaOH ----- Na2S04 + 2H20 (3)a 2 a

Dung dch sau in phn lm phenolphtalein ha hng chng t sau phnng (2): NaOH d => b > 2a.

* Cch khc:Phng trnh in phn:

C11SO4 + 2NaCl ? Cu + Cl2 + Na2S04 (1)

Dung dch sau in phn lm phenolphtalein ha hng chng t sauphn ng (1) NaCl cn d v tip tc bin phn cho mi trng baz

=> b> 2a

=> p n .V d 9: in phn dung dch C11CI2vi in cc , sau mt thi gian thu

c 0,32 gam Cu catt v mt lng kh X ant. Hp th hon ton

a 2a

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46/661

lng kh X trn vo 200ml dung ch NaOH ( nhit thng). Sauphn ng, nng NaOH cn li l 0.05M (gi thit th tch dung dchkhng thay i). Nng ban u ca dung dch NaOH lA. 0,15M. B .0,2 c . 0,1 D. 0,05

(Trch thi tuy n sinh i h c kh i A nm 2007)

H ng d n gi iTa c: ncu = 0,005 mol

Phng nh in phn: C11CI2 Cu + CI20,005 0,005

Cl2 + 2NaOH - NaCl + NaClO + H200,005 0,01

Sau phn ng trn nng NaOH cn li 0,05M, ngha l NaOH d.=> nNaOHd = 0,0 5.0 ,2 = 0,01 m oi

=> nNa H b u = 0 ,01 + 0,01 = 0 ,0 2 m ol

:=> [N aOH] = = 0,1M

=>p n c .

V 10: in phn 100ml dung dch A cha ng thi HC10,1M v NaCl0,2M vi in cc tr c mng ngn xp ti khi anot thot ra 0,224 ltkh (ktc) th ngng in phn. Dung dch sau khi in phn c pH (coi

th tch dung dch thay i khng ng k) l:A. 6 B. 7 c . 12 D. 13.H ng d n gi i

T ac:n Hci = 0,01 moinNaci = 0,02 rnol

_ 0 ,224 _ n n i .lkli anot ~ ~ 0 ,01 m ol

22,4

Th t in phn anot: 2 c r - 2e -> CI2

0,02 0,01

Th t in phn ctot: 2H+ + 2e -> H20,01 0,01

2H20 + 2e -> H2 + 20HT0,01 0,01

=>[OH "] = = 0,1M => pOH = 1 => pH =13

=^ p n D.

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47/661

V d 11: Tin hnh in phn (c mng ngn xp) 500ml dung dch chahn hp HC1 0,02M v NaCl 0,2M. Sau khi anot bay ra 0,448 lt kh (ktc) th ngng in phn. Th tch dung dch HNO30,1M cn trungha dung dch thu c sau in phn l:A. 200ml B. 300ml c . 250ml D.400ml.

H ng d n gi i

Ta c: nnci =0,01moln N a c i = 0 , 0 1 m o l

rikhi anot 0 ,0 2 m o i

Th t in phn anot:

2CP - 2e - Cl20,04 0,04 0,02

=> lon c r cha bin phn ht v s mol electron trao i ne = 0,04 mol.Th t in phn catot:

2H+ + 2e -> H20,01 .0,01

2H20 +2 e ->H 2 + 20H~0,03 0,03

Phn ng trung ha gia dung dch HNO3v dung dch sau in phn:I f + Oil -> 1-I2O0,03 0,03

=> VHN0 = =0,3 (1) = 300ml => p n B.

V d 12: in phn dung dch cha m (g) hn hp 2 mui C11SO4v NaClvi cng dng in I = 5A cho n khi 2 in cc H2O cng in

phn th dng li. Dung dch sau in phn ha tap va l,6g CuO v anot ca bnh in phn c 448ml kh bay ra (ktc). Gi trca m l:A.5,97g B.4,8g c.4 ,95g D. 3,875g

H ng d n gi i

Ta c: n = = 0,02 molC u 80

0 448Hkhi allot ~ 2 2 ~ ~ ~ 0 , 0 2 m o i

Phng trnh in phn:

C11SO4 + 2NaC l----- > Cu + Cl2 H- Na2S04 (1)0,01 0,02 0.01



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48/661

Dung dch sau in phn ha tan c CuO => sau (1) CuS 4cn d vtip tc bin phn.

C 11SO 4 + H 20 ------- Cu + H 2 SO 4 + - O 2 (2)

0,02 0,02 0,01

CuO + H2SO4 ----- > CuS04 + H200,02 0,02

Theo (2): nD = 0,01 mol => n = 0,02 - 0,01 = 0,01 mol

Theo (1) v (2) => nCuS0 = 0,01 + 0,02 = 0,03 mol

nNact = moi=>m =0,03.160 + 0,02.58,5 = 5,97 (g)

=> p n A.Cu 13: in phn nng chy AI2O3vi anot than ch (hiu sut in phn

100%) thu c m kg Al catot v 67,2 m3( ktc) hn hp kh X c tkhi so vi hiro bng 16. Ly 2,24 lt ( ktc) hn hp kh X sc vodung dch nc vi trong (d) thu c 2 gam kt ta. Gi trca m lA. 54,0 B. 75,6 c . 67,5 . 108,0


2Ta c: n,.,, = nf. = = 0,02mol2 CaC03 100

Trong 2,24 lt hn hp X c nco = 0,02 mol

Vy: Trong 67,2m3 hn hp X c nco = 0,6 kmol

Gi s trong hh X ngoi CO2chc CO:

Ta c: nhh = 3 kmol => nco = 3 - 0,6 = 2,4 kmol

7T 2,4.28 + 0,6.44 _ ,1 A x=> Mx = ---------_----------= 31,2 > Mx cho = 16.2 = 323

=> Trong hh X ngoi CO2v c o (x kmol) cn c O2d (y kmol)Ta c: X+ y = 2,4 (1)

IU' 28x + 32y + 0,6.44 _Mt khc: Mx = ------------Z------------ = 323

=> 28x + 32y = 69,6 (2) = > x = l , 8 ; y = 0,6.Cc phn ng xy ra anot:

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2C + 2 2C00,9 1,8

c + 0 2 - C0 20,6 0,6

=> n0 p = 0,9 + 0,6 = 1,5 kmol

^ n o ban u n o pir + n o dir 1 >5 + 0 ,6 2,1 kinol

Phng trnh phn ng in phn nng chy AI2O3 :2AI2O3 -> 4A1 + 302

2,8 2,1

=> mAi= 2,8.27 = 75,6 (kg) => p n B.L u :Trong qu trnh in phn nu cc in cc khng Dhi l in

cc tr th c th xy ra phn ng gia cht lm in cc vi sn phm caqu trnh in phn sinh ra in cc .V d 14: in phn 400ml dung dch 2 mui KC1 v C11CI2vi in cc

tr v mng ngn cho n khi anot thot ra 3,36 lt kh (ktc) th ngngin phn. trung ha dung dch sau in phn cn 100ml dung dchHNO3 0,6M. Dung dch sau trung ha tcdng vi AgN 3 d sinh ra.2,87 gam kt ta trng. Nng mol ca mi mui trong dung dch trcin phn l:A. [CuCh] = 0,3M; [KC1] = 0,02M B. [CuCl2] = 0,25M; [KC1] = 3M c . [CuCl2] = 2,5M; [KC1] = 0,3M D. [CuCl2] = 0 ,3M; [KC1] = 0,2M.

H ng d n gi i

Ta c: nichanot = ^ T = 0,15 moi22,4

2 87n, ... = = 0 , 0 2 moi

AsCI 1 4 3 ,5

nHNOj = 0 ,6 .0,1 = 0,06 mol

t: KC1: Xmo Cu2+; y molCuC2: y mol ^ L c r : (x + 2y) mol

Dung dch sau in phn phi trung ha bng dung dch HNO3, chng t catot ion Cu2+ in phn ht v H2O bin phn.

Cu2+ + 2e -> Cu

y 2y2H20 + 2e -> H2 + 20H-

0,06 0,06

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50/661

Phn ng trung ha: H+ + OH- - H2O! 0,06 0,06

Dung dch sau khi trung ha tc dng vi AgNOj sinh ra kt ta trngchng t anot ion cr cha bin phn ht. Vy kh thot ra anot chlkh CI2.

2C1" - 2 e C20,3 0,3 0,15

cr + Ag+ -* AgCl0,02 0,02

=> n _ bu = 0,3 + 0,02 = 0,32 mol

Hay X + 2y = 0,32 (1)Da vo bn phn ng catot ta c: ne = 2y + 0,06 = 0,3 => y = 0,12 mol

T ( l ) => X = 0,08 mol

= > [ m ] = ^ = 0,2M ; [CuCl2] = =0,3M

=> p n D.

^ D ng 6 . XC NH THE I N cc CHOAN,SU T I N NG CHU N C A PIN I N HA

+ Xt pin Me - N

+ Chiu phn ng: Vit cp oxi ha - kh c th in cc nh bntri, cp oxi ha - kh c th in cc chun ln hn bn phi ri vitph ng trnh ph n ng oxi ha - kh theo quy t c (X.

V d 1: Cho gi trth in cc chun ca mt s G p oxi ha - kh:

Cp oxi ha/kh

m 2+// m x 2+// x z%

E (V) -2,37 -0,76 -0,13 +0,34

Phn ng no sau y xy ra?

a . x + z 2+-> x 2++ z b . x + m 2+-> x 2++ m

c . z + Y2+ - z2++Y D.Z + M2+->Z2++M(Trch thi tuy n sinh Cao ng nm 2011)

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51/661

H ng d n gi i

- Phn ng xy ra l: X + z2+>x2++ zChiu xy ra phn ng oxi ha kh l chiu: Cht oxi ha mnh tc dng

vi cht kh mnh to ra cht oxi ha yu v cht kh yu.

=> p n A.

V d 2: Cho bit E" ...... = -2,37V; E 2+, = -0,76V; E,- . . = -0,13 V;M y / M g Z n / Z n P b ^ / P b

E 2+ = +0,34V. Pin in ha c sut in ng chun bng 1,6 IVti = +U,J4V. rin ain na co :Cu /Cu r

c cu to bi hai cp oxi ha-kh.A. Pb2+/Pb v Cu2+/Cu B. Zn2+/Zn v Pb2+/Pbc . Zn2+/Zn v Cu2+/Cu D. Mg2+/Mg v Zn2+/Zn


Ta c: E , - E 2 + = -0,76 - (-2,73) = 1,61 VZn /Zil Mg /My v ;

= > p n D.V d 3: Cho sut in ng chun ca cc pin in ha: Zn-Cu l 1,1 V;

Cu-Ag l 0,46 V. Bit th in cc chun E( = +0,8V . Th in ccAg /Ag

chun E v E c gi tri ln lt lZn /Zn Cu /Cu

A. +1,56 V v +0,64 V B. - 1,46 V v - 0,34 V

c . - 0,76 V v + 0,34 V D. - 1,56 V v +0,64 V


Ta c: E pinCu-Ag E 2 => 0,46 =0,8 E 26 . Ag /Ag Cu /Cu Cu /Cu

=> E , = 0,8 - 0,46 = 0,34 (V)Cu /Cu v '

Mt khc: EpinZn-c = E - E" 2+ => 1,1 = 0,34 - Ev Cu / c Zn /Zn Zn 2 +/Zn

^ E 2+f = 0,34 - -0,76 (V)Zn /Zn v '

p n c .

V d 4: Cho cc th in cc chun: E = -1,66V; E = -0 ,76V ;AI3+/AI Zn /Zn

Eb2+/Pb = -0 ,13V ;E" 2+c = +0,34V. Trong cc pin sau y, pin no c

sut in ng churi ln nht?A. Pin Zn - Pb B .PinP b-C u C. P inA l-Zn D .PinZ n-C u

(Trch th tuy n sinh i h c kh i B nm 2009)

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H ng d n gi i

+P in Z n - Pb : Epi Zn - Pb = E ph2+/pb - E z2+/z = - 0 ,13 - ( -0 ,76) = 0 ,63V

+ P in P b - C u: Epin Pb-Cu = E cu2*/c- E PbJ+/Pb ~ = 34 ~ ( - 13) = 0,47V

+ P in A

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TRỌNG TÂM KIẾN THỨC HÓA HỌC 12 HÓA VÔ CƠ & HÓA HỮU CƠ - ĐỖ XUÂN HƯNG