MỘT SỐ PHƯƠNG PHÁP GIẢI BÀI TOÁN HÓA HỌ1

8/7/2019 MT S PHNG PHP GII BI TON HA H1

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MT S PHNG PHP GII BI TON HA HC

GV: V PHN ( YN S- HONG MAI- HN).

C: 0436.453.591;D: 01236.575.369

I. Phng php ghp n s:

Trong ton hc h phng trnh vi s n nhiu hn s phng trnh gi l h v inh v

h thng cho v s nghim v kh gii c. Tuy nhin trong ha hc th nhng h nh

vy vn c th gii c nh nhng tnh cht ring ca ha hc v mt s th thut ca

ton hc.Phng php ghp n ss cho thy iu thng qua mt s v d sau:

V d 1: Ha tan hon ton 27,4 gam hn hp hai mui cacbonnat ca hai kim loi phn

nhm IA v IIA bng dung dch HCl ( va ) thu c dung dch A v 6,72 lt kh ( o

ktc).

1/ C cn dung dch A thu c bao nhiu gam mui khan?

2/ Xc nh tn ca hai kim loi bit khi lng nguyn t ca chng hn km

nhau 1 n v.Gii: Gi X, Y ln lt l tn v khi lng nguyn t ca hai kim loi.Hai mui

cacbonnat l X2CO3 ; YCO3 vi s mol tng ng x;y. Cc phn ng:

X2CO3 + 2HCl 2XCl + H2O+ CO2 (1)

Mol: x 2x x

YCO3 + 2HCl YCl2 + H2O + CO2 (2)

Mol: y y y

1 /Theo bi ra v theo cc phn ng ta c h:

=+=+++

3,0

4,2 7)6 0()6 02(

yx

YyXx.


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T h trn bng cch ghp n s ta c: 2xX + yY = 9,4.(*) Vy tng khi lng mui

clorua khan thu c l : m clorua = 2x( X + 35,5) + y ( Y + 71 ) = (2xX + yY) + 71 ( x + y

) = 30,7 (g)

2/ Ta c y = 0,3 x v X = Y 1 thay vo (*) c 2x ( Y 1) + ( 0,3 x ) Y =

9,4.

T y suy ra 44/3 < Y < 100/3 ( V 0


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C s ca phng php ny l nh lut bo ton khi lng (BTKL).

V d 3: Ha tan hon ton 3,22 g hn hp X gm Fe, Mg v Zn bng mt lng va

va dung dch H2SO4 long thu c 1,344 lt kh H2 ( ktc) v dung dch cha m

gam mui. Gi tr ca m l:

A. 8,98. B.9,52 C. 10,27 D. 7,25.

( Trch TSH-C A -2007)

Gii:Phng trnh chung: M + H2SO4MSO4 + H2

Ta c: molnn HSOH 06,04,22

344,1242

=== . p dng nh lut BTKL ta c:

m mui = mX+ m axit - 2Hm = 3,22+ 0,06.98 0,06.2 = 8,98 (g).

Chn p n A.

V d 4: t chy hon ton m gam hn hp X gm CH 4, C3H6 v C4H10 cn V lit O2

(ktc) thu c 4,4 gam CO2 v 2,52 gam H2O. Tnh m v V?

Gii:p dng nh lut BTKL v BTNT ta c:

m = mC+ mH= (4,4/44).12 + (2,52/ 18).2 = 1,48 (g)

OHCOOHOCOOO nnnnpun 2222222 2

1)( )()( +=+= = 0,1 + 0,07 = 0,17 (mol) .Vy V =

0,17.22,4 = 3,808(l)

V d 5: un nng hn hp hai ancol n chc, mch h vi H 2SO4 c, thu c hn

hp cc ete. Ly 7,2 gam mt trong cc ete em t chy hon ton, thu c 8,96 lit

kh CO2 ( ktc) v 7,2 gam H2O. Hai ancol l

A. CH3OH v CH2=CH-CH2-OH B. C2H5OH v CH2=CH-CH2-OH

C. CH3OH v C2H5OH D. C2H5OH v CH3OH

( Trch TSH A -2009)

Gii:Xt ete Y em t chy. p dng nh lut BTKL v BTNT ta c: m C ( Y) = mC

(CO2)=

(8,96/22,4).12= 4,8 (g); mH(Y) = mH (H2O) = (7,2/18).2 = 0,8 (g) ; mO = mY (mC + mH)=1,6 (g)

T nC : nH : nO = 0,4 : 0,8 : 0,1 = 4:8:1. CTGN ng thi cng l CTPTca Y l

C4H8O.

CTCT ca Y ch c th l CH3-O-CH2-CH=CH2 ( C 1 lin kt i trong gc ancol).


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Chn p n A.

III. Phng php tng gim khi lng:

C s ca phng php vn l nh lut BTKL.

Cho phn ng c dng: RX + xM RY + yN ( R l thnh phn khng i) th khi

lng ca cht to thnh sau phn ng RY tng hoc gim so vi cht phn ng RX l |

Y X | ( vi 1 mol RX hoc a mol cht no trong phn ng).

V d 6: em nung mt khi lng Cu (NO3)2 sau mt thi gian dng li lm ngui, ri

cn thy khi lng gim 0,54 gam. Khi lng mui Cu(NO3)2 b nhit phn l

A. 0,5 gam B. 0,49 gam C. 9,4 gam D. 0,94 gam

Gii: Cu(NO3)2 0t

CuO + O2+ 2 NO2

C 1 mol cht rn Cu(NO3)2 b nhit phn th khi lng cht rn gim 188 80 = 108

(g).

S mol Cu(NO3)2 b nhit phn l: 0,54/ 108 = 0,005 (mol). Khi lng: 0,005.188= 0,94

(g).

Chn p n D.

V d 7: Cht hu c X c cng thc phn t C5H8O2. Cho 5 gam X tc dng va ht vi

dung dch NaOH,thu c mt hp cht hu c khng lm mt mu nc brom v 3,4gam mt mui. Cng thc ca X l

A. CH3COOC(CH3)=CH2 B. HCOOC(CH3)=CHCH3

C. HCOOCH2CH=CHCH3 D. HCOOCH=CHCH2CH3

( Trch TSH A 2009)

Gii:

X c cng thc RCOOR. Phn ng:

RCOOR + NaOH RCOONa + ROH.

Ta c: nX = 5/ 100= 0,05 (mol).C 1 mol X phn ng khi lng mui thu c gim so

vi X l R 23. Khi lng gim l 0,05 (R -23) = 5 3,4 = 1,6 R = 32+23 = 55

R = 1.

ROH l ancol khng no , khng bn b phn hy thnh xeton ( V khng lm mt mu

nc brom). ( C th tnh 0,05( R + 67) = 3,4 R =1 R = 55)


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Chn p n B.

V d 8:(Xem VD 1/ 1) C 1mol kh CO2khi lng mui Clorua tng ln so vi

mui cacbonnat l 71 60 =11 (g). Khi lng tng ln l 0,3.11=3,3 (g). Vy khi

lng mui clorua khan thu c l: 27,4 + 3,3 = 30,7 (g).

Ch :. (1) ca v d ny cn gii c bng phng php BTKL . Khng th ni cch

gii no hay hn v cn lin quan ti (2).

. V d 2 cng gii c bng phng php tng gim khi lng v nhiu

phng php khc!

IV. Phng php khi lng mol trung bnh ( KLMTB):

Khi nim: Mt hn hp cc cht c th coi l mt cht c KLMPTTB ( M ) c xc

nh bng t s gia tng khi lng v tng s mol ca hn hp.

Cng thc:hh

hh

n

mM = =

k

kk

nnn

MnMnMn

+++

+++

...

...

21

2211

=k

kk

VVV

MVMVMV

+++

+++

...

...

21

2211( Vi hn hp gm k cht kh o cng

iu kin)

= x1M1 +x2M2 ++ xkMk ( xi l % theo s mol hoc th tch cc cht )

Ch : min{ Mi} < M< max {Mi}.

i vi cc hp cht hu c cn nh KLM ca vi cht u trong dy ng

ng, vi cht v c cn nh cc cht lin quan n nguyn t u nhm,

V d 9: X phng ha hon ton 1,99 gam hn hp hai este bng dung dch NaOH thu

c 2,05 gam mui ca mt axit cacboxylic v 0,94 gam hn hp hai ancol l ng

ng k tip nhau. Cng thc ca hai este l

A. HCOOCH3 v HCOOC2H5 B. C2H5COOCH3 v C2H5COOC2H5

C. CH3COOC2H5 v CH3COOC3H7 D. CH3COOCH3 v CH3COOC2H5

( Trch TSH A -2009)

Gii:


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Theo bi ra cng thc chung ca hai este l: RCOO 'R ( phn ng thy phn cho mui

ca mt axit cacboxylic).Phn ng:

RCOO 'R + NaOH RCOONa + 'R OH

p dng nh lut BTKL ta c m NaOH= 2,05 + 0,94 1,99 = 1 (g) ; n NaOH= 0,025 (mol)

Suy ra: 0,025( R + 67) = 2,05 v 0,025( 'R + 17) = 0,94 R = 15 v 'R = 20,6.

Chn p n D

V d 10: Hn hp kh X gm anken M v ankin N c cng s nguyn t cacbon trong

phn t. Hn hp X c khi lng 12,4 gam v th tch 6,72 lt ( ktc). S mol, cng

thc phn t ca M v N ln lt l

A. 0,1 mol C2H4 v 0,2 mol C2H2 B. 0,1 mol C3H6 v 0,2 mol C3H4

C. 0,2 mol C2H4 v 0,1 mol C2H2 D. 0,2 mol C3H6 v 0,1 mol C3H4

( Trch TSH A 2009 )

Gii:nhh = 6,72/22,4 = 0,3 (mol). M= 12,4/0,3 = 124/3=41,3 Loi A,C.

Chn p n D.

V d 11:Cho dung dch cha 6,03 gam hn hp gm hai mui NaX v NaY ( X;Y l hai

nguyn t c trong t nhin, hai chu k lin tip thuc nhm VIIA, s hiu nguyn t

ZX< ZY) vo dung dch AgNO3 (d), thu c 8,61 gam kt ta. Phn trm khi lng

ca NaX trong hn hp ban u l

A. 58,2% B. 41,8% C. 52,8% D. 47,2%( Trch TSH B 2009)

Gii:Phn ng: (NaX + NaY) + 2 AgNO3 (AgX + AgY)+ 2 NaNO3

p dng phng php tng gim khi lng ta c: nhh = ( 8,61 6,03) : ( 108- 23) =

2,58/85 =0,0304 (mol). KLMNTTB ca X;Y l (6,03.85/2,85) 23 = 180 23 =157.Do

X l I = 127 v Y l At = 210.

t nNaI = x (mol) ta c: 150x + (0,0304 x).233=6,03 x = 0,01256 (mol)

% NaI = 0,01256.150/6,03 = 31,2%. Khng c p n ng!

bi c sai khng? n s liu tnh ton qu l ( khng ph hp vi xu hng ra

hin nay). At l nguyn t iu ch bng phng php phng x, khng tn ti trong

t nhin. Trng hp ny b loi. Vy cn mt kh nng khc!Xem li tnh cht ca


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nguyn t u nhm VIIA l F th AgF tan. Nh vy kh nng th hai: X l F v Y l Cl.

Cht kt ta duy nht l AgCl; n NaCl =n AgCl = 8,61/ 143,5 = 0,06 (mol). % NaCl =

0,06.58,5/6,03 = 58,2 %;

Vy: %NaF = 41,8%.

Chn p n B.( ra ng!)

V d 12: Cho18 gam hn hp hai ancol gm mt ancol no n chc v mt ancol n

chc c mt lin kt i trong phn t vi s mol bng nhau tc dng ht vi Na d, thu

c 4,48 lt H2 ( ktc). CTCT ca hai ancol l

A. CH3CH2OH v CH2=CH-CH2OH B. CH3CH2CH2OH v CH2=CH-

CH2OH

C. CH3OH v CH2=CH-CH2OH D. CH3CH2CH2CH2OH v CH2=CH-

CH2OHGii:

t cng thc chung ca hai ancol l R OH. Phn ng:

2 R OH + 2 Na 2 R ONa + H2

Ta c: nhh ancol = 2nH2 = 2. 4,48/22,4 = 0,4 (mol). M ancol = 18/0,4= 45. Trong 2 ancol c

1 ancol c phn t khi nh hn 45. Ancol l CH3OH ( M = 32).

Chn p n C.

Ch : Cn th li p n bng php tnh 0,2( 32+ 58) = 18 (g) ( Tha mn)V. Phng php s nguyn t C, H trung binh ( KLMTB m rng):

V d 13: Hn hp X gm hai este no, n chc, mch h. t chy hon ton mt

lng X cn dng va 3,976 lt kh O2 ( ktc), thu c 6,38 gam CO2. Mt khc, X

tc dng vi dung dch NaOH thu c mt mui v hai ancol l ng ng k tip.

Cng thc phn t ca hai este trong X l

A. C2H4O2 v C5H10O2 B. C2H4O2 v C3H6O2

C. C3H4O2 v C4H6O2 D. C3H6O2 v C4H8O2

( Trch TSH B 2009)

Gii: Gi n l s nguyn t C trung bnh trong hn hp hai este. Cng thc tng

ng

nnHC

2 O2. Phn ng: nnHC

2 O2 +( 3n - 2)/2 O2 n CO2 +n H2O


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Ta c: n (O2) = 3,976/22,4 = 0,1775 ;n(CO2) = 6,38/44 = 0,145. Suy ra: 2n / (3 n - 2)

=0,145/0,1775 ; n = 29/8.Mt khc, X tc dng vi dd NaOH thu c mt mui v hai

ancol l ng ng k tip nn hai este cng l ng ng k tip.

Chn p n D.

V d 14: Cho hn hp X gm hai hp cht hu c no, n chc tc dng va vi 100

ml dung dch KOH 0,4M, thu c mt mui v 336 ml hi mt ancol ( ktc). Nu t

chy hon ton lng hn hp X trn, sau hp th ht sn phm chy vo bnh ng

dung dch Ca (OH)2 (d) th khi lng bnh tng 6,82 gam. Cng thc ca hai hp chthu c trong X l

A. CH3COOH v CH3COOC2H5 B. C2H5COOH v C2H5COOCH3

C. HCOOH v HCOOC2H5 D. HCOOH v HCOOC3H7

( Trch TSH B 2009)

Gii: nKOH= 0,04 (mol) > nancol = 0,015 (mol). Hn hp X gm mt axit cacboxilic no,

n chc v mt este no n chc. naxit= 0,025 (mol); neste = 0,015 (mol).

Gin

l s nguyn t C trung bnh trong hn hp X. Cng thc chung nnHC

2 O2.Phn ng:

nnHC

2 O2 + ( 3n -2)/2 O2 n CO2 + n H2O

Mol: 0,04 0,04n 0,04n

Ta c: 0,04n ( 44 + 18) = 6,82 ; n = 11/4.Gi x; y ln lt l s nguyn t C trong

phn t axit v este th: (0,025x + 0,015 y)/0,04 = 11/4 hay 5 x + 3y =22.T :

(x;y)=(2;4).

Chn p n A.

V d 15: Cho hn hp X gm hai ancol a chc, mch h, thuc cng dy ng ng.

t chy hon ton hn hp X, thu c CO2 v H2O c t l mol tng ng l 3:4. Hai

ancol l

A. C2H4(OH)2 v C3H6(OH)2 B. C2H5OH v C4H9OH

C. C2H4(OH)2 v C4H8(OH)2 D. C3H5(OH)3 v C4H7(OH)3


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( Trch TSH A 2009)

Gii: Gi n l s nguyn t C trung bnh trong hn hp hai ancol; k l s lin kt

trong phn t mi ancol ( kN). Cng thc phn t tng ng: CnH2 n +2-2kOz.

Phn ng:

CnH2 n +2-2kOz + 2O n CO2 + (n +1 k)H2O

Ta c: n/ (n +1 k) = suy ra: n = 3 3k. T : k=0; n = 3. Loi B v ancol

n chc.Loi A,D v khng ph hp vi n =3.

Chn p n C.

VI. Phng php s proton,ntron,electron trung bnh:

V d 16: Hp cht M c to thnh t cation X+ v anion Y 2-, Mi ion u do 5

nguyn t ca hai nguyn t to nn. Tng s proton trong X + l 11 v tng s electron

trong Y 2- l 50.Hai nguyn t trong Y 2- thuc cng mt phn nhm v thuc hai chu k lin tip.

Hy xc nh cng thc phn t v gi tn M.

Gii:Gi Z X l s proton trung bnh ca hai nguyn t to nn X+, ta c Z X = 11/5 =

2,2. Trong hai nguyn t to nn X+ phi c H hoc He. Nhng He l kh tr nn b loi.

Gi R l nguyn t th hai to ra X+, khi X+ l RnHm+ .

Theo thnh phn cu to ca X+ ta c:

=+

=+

1

5

mn Z

mn

R

suy ra: n( ZR 1) = 6

Ch c n =1, ZR = 7 ( R l N ) l ph hp. Cation X+ l NH4+.

Gi E Y l s electron trung bnh trong cc nguyn t ca anion Y2-.

Ta c: E Y = (50 2)/5 = 9,6.Trong Y2- c mt nguyn t c z < 9,6 , thuc chu k 2 v

nguyn t cn li thuc chu k 3. V u thuc chu k nh nn hai nguyn t cch nhau 8


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. Cng thc Y2- l AxBy2- vi:

=++

=+

4)8(

5

y ZZx

yxCh c x =1; y=4; Z = 8 l ph hp.A

l S cn B l O.

Anion Y2- l SO42-. Vy M l (NH4)2SO4 ( amoni sunfat).

VII. Phng php ha tr trung bnh v s nhm chc trung bnh:

V d 17: Cho 11 gam hn hp hai kim loi Fe, Al tan hon ton trong dung dch HCl

thu c 8,96 lt kh H2 ( ktc) v dung dch ca hai mui clorua. Tng khi lng hai

mui clorua (khan) l:


Gii:V phn ng hon ton nn ta thay hn hp Fe, Al bng kim loi tng ng M c ha tr trung bnh n . Phn ng: M+ nHCl M nCl + n/2 H2

Ta c:2H

n = 8,96/22,4 = 0,4 (mol) ; nHCl = 2.0,4 = 0,8 (mol). p dng LBTKL ta c:

m mui clorua = m kim loi + mHCl m H2 = 11 + 0,8.36,5 0,4.2 = 39,4 (g).

Chn p n B.

V d 18: Cho 3,68 gam hn hp gm Al v Zn tc dng vi mt lng va dung

dch H2SO4 10% thu c 2,24 lt kh H2 ( ktc). Khi lng dung dch thu c sau

phn ng l:A.101,48 gam B. 101,68 gam C. 97,80 gam D. 88,20 gam

( Trch TSH A 2009)

Gii:Tng t v d 17 c phn ng: 2M + aH2SO4 M 2(SO4)a + aH2 ( a l

ha tr trung bnh ca hai kim loi)

Ta c:242 HSOH

nn = = 2,24/22,4 = 0,1 (mol). p dng LBTKL ta c:

mddsau p = mkimloai + mddaxit mH2 = 3,68 + 0,1.98/10% - 0,1.2 = 101,48 (g).

Chn p n A.V d 19: Cho 27,2 gam hn hp gm hai ancol tc dng va vi Na thu c hn

hp hai mui v 6,72 lt kh H2 ( ktc). Tng khi lng hai mui thu c l:


Gii:Gi cng thc tng ng ca hai ancol l: R (OH)a ( a l s nhm chc trung

bnh ca hai ancol). Phn ng: R (OH)a + aNa R (ONa)a + a/2 H2


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Ta c: n H2 = 6,72/ 22,4 = 0,3 (mol); nNa = 2.0,3 = 0,6 (mol). p dng LBTKL ta c

mMui = mAncol+ mNa - mH2 = 27,2 + 0,6.23 - 0,3.2 = 40,4 (g).

Chn p n C.

V d 20: Trung ha m gam hn hp hai axit cacboxilic cn 300 ml dung dch NaOH1M

thu c 23,1 gam hn hp hai mui. Gi tr ca m l

A. 18,2 B, 20,1 C. 15,6 D. 16,5

Gii: Gi cng thc tng ng ca hai axit l R (COOH)a . Phn ng:

R (COOH)a + a NaOH R (COONa)a + a H2O

Ta c: n NaOH = nH2O = 0,3 (mol). p dng LBTKL ta c:

m = mMui + mH2O m NaOH = 23,1 + 0,3.18 0,3. 40 = 16,5 (g).Chn p n D.

VIII. Phng php s ng cho:

1/ Pha trn hai dung dch ca cng mt cht vi nng phn trm C1> C2 v khi lng

tng ng m1, m2 c dung dch c nng phn trm C, th c s ng cho:

m1: C1 C C2

C khi :CC

CC

m

m

=

1

2

2

1

m2: C2 C1 C

2/ Pha trn hai dung dch ca cng mt cht vi nng mol C 1, C2 ( C1>C2 ) v th tch

tng ng V1, V2, th vn c s ng cho nh trn vCC

CC

V

V

=

1

2

2

1.

3/ Vi hn hp gm hai cht ( hoc hai ng v) c KLM l M 1, M2 v s mol tng ng

n1,n2 nu coi l mt cht tng ng c KLMTB l M thMM

MM

n

n

=

1

2

2

1 . T suy


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ra % ( theo s mol ) tng ng ca mi cht l: % (Cht M 1) =21

2

MM

MM

; % (Cht M2)

=21

1

MM

MM

.

V d 21: Cn thm bao nhiu gam KCl vo 450 gam dung dch KCl 8% thu c

dung dch 12%.

Gii: Theo s ng cho ta c:12100

812

450

=

m. Vy: m = 20,45 (g).

V d 22: Cn thm bao nhiu lt nc ct vo 10 lt dung dch HCl c pH = 3 c

dung dch c pH = 4.

Gii: pH = 3 [ H+] = 10 -3 ; pH = 4 [ H+] = 10 -4 . Theo s ng cho ta c:

9

1

1010

1043

4

2

=

=

OH

HCll

V

V. Vy: VH2O = 9 VHCl= 90 lt.

V d 23: ng trong t nhin gm hai ng v 63Cu v 65Cu. Khi lng nguyn t

trung bnh l 63,54. T l % khi lng ca 63Cu trong CuCl2 l

A. 31,34% B. 34,18% C. 73,00% D. 31,48%

Gii: Gi x l thnh phn % ca ng v 63Cu . Theo s ng cho ta c:

x =6365

54,6365

= 73%. Vy % ( theo khi lng) ca 63Cu trong CuCl2 l:

18,3454,134

63.73= %.

Chn p n B.

V d 24: Cho 41,2 gam hn hp gm C2H5COOH v C2H5COOCH3 tc dng va

vi dung dch NaOH thu c 48,0 gam mui C2H5COONa. Thnh phn % theo s mol

v % theo khi lng ca C2H5COOH trong hn hp u ln lt l

A. 35,92% v 40,00% B. 40,00% v 35,92% C. 36,85%v 50,00% . D. 60,00% v

64,08%


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Gii:nhh = nmui = 48/96 = 0,5 (mol). Maxit = 74; MEste = 88; M= 41,2/0,5 = 82,4.

Vy %s mol ( axit) =7488

4,8288

.100% = 40%; % khi lng(axit)=2,41

74.5,0.40% =

35,92%

Chn p n B.

IX. Phng php bo ton electron ( Cho phn ng oxi hoa-kh):

V d 25: Ha tan hon ton 12 gam hn hp Fe, Cu ( t l mol 1 : 1) bng axit HNO 3,

thu c V lt ( ktc) hn hp kh X (gm NO v NO 2) v dung dich Y (ch cha hai

mui v axit d). T khi ca X i vi H2 bng 19. Gi tr ca V l:

A. 3,36 B. 2,24 C. 4,48 D. 5,60

( Trch TSH A 2007)

Gii: MX = 19.2 = 38. %V (NO) = %50%100.30463846

=

= %V (NO2) n NO = n NO2= x

(mol).

n Fe = nCu = 12/ 120 = 0,1 (mol). Cc qu trnh oxi ha kh:

Fe 3e Fe 3+ ; Cu 2e Cu 2+

0,1 0,3 0,1 0,2 Tng nhng e : 0,3 + 0,2

= 0,5 (e)

N+5 + 3e N+2 ( NO) ; N+5 + 1e N+4 ( NO2)

3x x x x Tng nhn e: 3x + x = 4x (e).p dng LBT electron ta c: 4x = 0,5 ; x = 0,125 (mol) n X= 2,0,125 = 0,25 (mol).

Vy :V = 0,25.22,4 = 5,60 (lt).

Chn p n D.

V d 26: Ha tan hon ton 12,42 gam Al bng dung dch HNO 3 long (d), thu c

dung dch X v 1,344 lt ( ktc) hn hp kh Y gm hai kh l N 2O v N2. T khi ca

hn hp Y so vi kh H2 l 18. C cn dung dch X, thu c m gam cht rn khan. Gi

tr ca m lA. 97,98. B. 106,38. C. 38,34. D. 34,08.

( Trch TSH A 2009)

Gii:nAl= 12,42/27 =0,46 (mol); M Y= 2.18 = 36; nY= 1,344/22,4 = 0,06 (mol).


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% V (N2O) =2844

2836

.100%=50% n N20 = n N2 = 0,03 (mol). Cc qu trnh oxi ha

kh:

N+5 + 4e N+ ( N2O) ; N+5 + 5e N0 (N2 ); Al 3e Al3+

0,24 0,06 0,3 0,06 0,46 1,38

Ta thy: 1,38 0,54 = 0,84. Do cn xy ra qu trnh: N+5 + 8e N-3

0,84 0,105.

Tng khi lng cht rn khan( gm Al(NO3)3 v NH4NO3 ) l:

m = 0,46. 213 + 0,105. 80 = 106,38 (g). Chn p n B.

Ch : Khi kim loai tc dng vi HNO3 rt long cn c th to ra c mui

NH4NO3.V d 27: Cho 61,2 gam hn hp X gm Cu v Fe3O4 tc dng vi dung dch HNO3

long, un nng v khuy u. Sau khi cc phn ng xy ra hon ton, thu c 3,36 lt

kh NO ( sn phm kh duy nht, ktc), dung dch Y v cn li 2,4 gam kim loi.. C

cn dung dch Y, thu c m gam mui khan. Gi tr ca m l

A. 151,5 B. 137,1. C. 97,5. D. 108,9.

( Trch TSH B 2009)

Gii:nNO = 3,36/22,4 = 0,15 (mol). Cc qu trnh oxihoa kh:

Cu 2e Cu2+ ; 3 Fe+8/3 - e 3Fe 3+ ; N5+ + 3e N2+ ; Fe3+ + e

Fe2+

x x2 x 3y y 3y 0,45 15,0 3 y 3y

3y

p dng LBT electron v LBTKL ta c h:

=++

+=+

2,6 14,22 3 26 4

34 5,02

yx

yyx.

Gii h c: x = 0,375; y = 0,15. Vy tng khi lng mui khan thu c ( gm

Cu(NO3)2 v Fe(NO3)2 ) l: m = 0,375.188+ 0,45.180 = 151,5 (g).

Chn p n A.

X. Phng php bo ton in tch:


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V d 28: Mt dung dch cha 0,02 mol Cu2+; 0,03 mol K+; x mol Cl v y mol SO42 - .

Tng khi lng cc mui tan c trong dung dch l 5,435 gam. Gi tr ca x,y ln lt l

A. 0,01 v 0,03 B. 0,02 v 0,05 C. 0,05 v 0,01 D. 0,03 v

0,02

( Trch TSH A 2007 )

Gii: p dng LBT in tch v LBTKL ta c:

=+++

=+

4 3,59 65,3 53 9.0 3,06 4.0 2,0

0 7,02

yx

yx

Gii h c: x = 0,03; y = 0,02.Chn p n D.

V d 29: Mt dung dch c cha 0,02 mol NH4+ , x mol Fe3+; 0,01 mol Cl -; 0,02 mol

SO4 2- , Khi c cn dung dch ny thu c lng mui khan l


Gii:p dng LBT in tch ta c 0,02 + 3x = 0,01 + 2. 0,02 x = 0,01 mol.

Vy tng khi lng mui khan thu c l: 0,02.18 + 0,01.56 + 0,01.35,5 + 0,02. 96 =

3,195.Chn p n B.

V d 30: Cho m gam bt Fe vo 800 ml dung dch gm Cu(NO3)2 0,2M v H2SO4 0,25

M. Sau khi cc phn ng xy ra hon ton, thu c 0,6m gam hn hp bt kim loi v V

lt kh NO (sn phm kh duy nht, ktc). Gi tr ca m v V ln lt l

A. 10,8 v 4,48. B. 10,8 v 2,24. C. 17,8 v 2,24 D. 17,8 v

4,48.

( Trch TSH B 2009)

Gii:n Cu(NO3)2 = 0,2.0,8 = 0,16 (mol); n H2SO4 = 0,25.0,8 = 0,2 (mol).


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Phng trnh in ly: Cu(NO3)2 Cu 2+ + 2 NO3- ; H2SO4 2H+ + SO4 2-.

Mol: 0,16 0,16 0,32 0,2 0,4 0,2

Qu trnh oxihoa- kh xy ra theo th t sau:

Fe + 4H+ + NO3- Fe3+ + NO+ 2H2O

Mol: 0,1 0,4 0,1 0,1 0,1

Fe + Cu2+ Fe2+ + Cu

Mol: 0,16 0,16 0,16

Fe + 2 Fe3+ 3 Fe2+

Mol: 0,05 0,1

p dng LBT in tch v LBTKL ta c: m 0,31.56 + 0,16. 64 = 0,6m m = 17,8

(g).

V = 0,1.22,4 = 2,24 (l).Chn p n C.

XI. Phng php s dng phn ng dng Ion thu gn:

V d 31: Trn 100 ml dung dch (gm Ba(OH)2 0,1M v NaOH 0,1M) vi 400 ml dung

dch (gm H2SO4 0,0375M v HCl 0,0125M) thu c dung dch X. Gi tr pH ca dung

dch X l:

A. 2. B.1. C.6. D. 7.

( Trch TSH B 2007)

Gii:Ta c: n OH- = 0,1.0,1.2 + 0,1.0,1 = 0,03 (mol);

n H+= 0,0375.0,4.2 +0,0125. 0,4 = 0,035 (mol).

Khi trn hai dung dch xy ra phn ng trung ha dng ion l: H+ + OH- H2O.

T phn ng ta c: n H+(phn ng) = n OH- = 0,03 (mol) n H+d = 0,035 0,03 = 0,005

(mol).

[ H+] =5,0

005,0= 0,01 = 10 -2 M. Vy: pH = 2.

Chn p n A.V d 32: Trn 100 ml dung dch hn hp gm H 2SO4 0,05M v HCl 0,1M vi 100 ml

dung dch hn hp gm NaOH 0,2M v Ba(OH)2 0,1M, thu c dung dch X. Dung

dch X c pH l:

A. 13,0. B. 1,2. . C. 1,0. D. 12,8.


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( Trch TSH B 2009)

Gii: Tng t VD 31.

Chn p n A.

XII. S dng s bin ha:

V d 33: Oxi ha hon ton m gam bt Fe thu c hn hp ba oxit st. Ha tan hon

ton hn hp ba oxit st bng dung dch HCl d, tip tc cho dung dch thu c phn

ng hon ton vi NaOH d, thu c kt ta X. Ly kt ta X, ra sch, ri em nung

ngoi khng kh n khi lng khng i cn nng 16 gam. Gi tr ca m l:

A. 5,6. B. 11,2. C. 8,4. D. 16,8.

Gii: S bin ha: Fe + O x i { FeO ; Fe2O3; Fe3O4 }

+H C l {FeCl2 ; FeCl3} +NaOH

X={ Fe(OH)2; Fe(OH)3} ca ot0 Fe2O3.

p dng LBT nguyn t ta c: n Fe = 2 nFe2O3 = 2.160

16= 0,2 (mol). Vy: m = 0,2.56 =

11,2 (g).

Chn p n B.

V d 34: Oxi ha mt dung dch cha 1000 gam ancol C 2H5OH bng oxi vi xc tc l

men gim. Thm tip mt lng dung dch NaOH vo dung dch thu c ri un nng

( c mt ca CaO) thu c kh X v dung dch mui cacbonnat. Ly kh X cho lm lnh

nhanh t nhit 15000C, thu c kh Y. Cho kh Y tc dng vi nc ( 80 0C, xc tc

HgSO4) thu c cht Z c kh nng tham gia phn ng trng gng.Hirat ha cht Z

(xc tc Ni) thu c m gam ancol. Gi thit rng cc phn ng u xy ra vi hiu sut

60%. Gi tr ca m l:

A. 60,00. B. 36,00. C.46,56 D. 77,76

Gii: S chuyn ha:C2H5OH mengim CH3COOH +NaOH CH4 nhlamlanhnhaC,15000

C2H2 + OH2 CH3CHO

+ )(2 NiH C2H5OH. Nu hiu sut ca tt c cc phn ng l 100%th ta vn thu c 1000 g. Nhng v mi qu trnh ch t h = 60% nn ch thu c:

1000. (0,6)5 =

77,76 (g).Chn p n D.

BI TP T GII:


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Tuyn sinh i hc 2009( t 1 n 16):

1/ Cho hn hp X gm hai axit cacboxylic no, mch khng phn nhnh. t chy hon

ton 0,3 mol hn hp X, thu c 11,2 lt kh CO2 ( ktc). Nu trung ha 0,3 mol X th

cn dng 500ml dung dch NaOH 1M. Hai axit l

A. HCOOH, HOOC CH2 COOH. B. HCOOH, CH3COOH.

C. HCOOH, C2H5COOH D. HCOOH, HOOC COOH

2/ X phng ha hon ton 66,6 gam hn hp hai este HCOOC 2H5 v CH3COOCH3 bng

dung dch NaOH, thu c hn hp X gm hai ancol. un nng hn hp X vi H2SO4

c 1400C, sau khi phn ng xy ra hon ton thu c m gam nc. Gi tr ca m l

A. 18,00. B. 8,10. C. 16,20. D. 4,05.

3/ Khi t chy hon ton m gam hn hp hai ancol no, n chc, mch h thu c V

lt kh CO2 ( ktc) v a gam H2O. Biu thc lin h gia m, a, V l:

A. m = a -6,5

V. B. m = 2a -

2,11

V. C. m = 2a -

4,22

V. D. m = a +

6,5

V.

4/ Hp cht X mch h c cng thc phn t l C4H9NO2. Cho 10,3 gam X phn ng va

vi dung dch NaOH sinh ra mt cht kh Y v dung dch Z. Kh Y nng hn khng

kh, lm giy qu tm ammr chuyn mu xanh. Dung dch Z c kh nng lm mt mu

nc brom. C cn dung dch z thu c m gam mui khan. Gi tr ca m l

A. 8,2 B. 10,8 C. 9,4 D. 9,6.

5/ t chy hon ton 0,2 mol mt ancol X no, mch h, cn va 17,92 lt kh O 2 (

ktc). Mt khc, nu cho 0,1 mol X tc dng va vi m gam Cu(OH)2 th to thnh

dung dch c mu xanh lam. Gi tr ca m v tn gi ca X tng ng l

A. 4,9 v propan-1,2- iol. B. 9,8 v propan-1,2-iol.

C. 4,9 v glixerol. D. 4,9 v propan-1,3- iol.

6/ Hn hp X gm hai ancol no, n chc, mch h, k tip nhau trong dy ng ng.

Oxi ha hon ton 0,2 mol hn hp X c khi lng m gam bng CuO nhit thch


19/64

hp, thu c hn hp sn phm hu c Y. Cho Y tc dng vi mt lng d dung dch

AgNO3 tronh NH3, thu c 54 gam Ag. Gi tr ca m l:

A. 15,3. B. 13,5. C. 8,1. D. 8,5.

7/ Este X ( c khi lng phn t bng 103 vC) c iu ch t mt ancol n chc

( c t l khi so vi oxi ln hn 1) v mt amino axit. Cho 25,75 gam X phn ng va

ht vi 300ml dung dch NaOH 1M, thu c dung dch Y. C cn Y thu c m gam

cht rn. Gi tr m l

A. 27,75. B. 24,25. C. 26,25. D. 29,75.

8/ Hiro ha hon ton m gam hn hp X gm hai anehit no, n chc, mch h, k

tip nhau trong dy ng ng thu c (m + 1) gam hn hp hai ancol. Mt khc, khi

t chy hon ton cng m gam X th cn va 17,92 lt kh O2 ( ktc). Gi tr ca m

lA. 17,8. B. 24,8. C. 10,5. D. 8,8.

9/ Cho 0,02 mol amino axit X tc dng va vi 200ml dung dch HCl 0,1M thu c

3,67 gam mui khan. Mt khc 0,02 mol X tc dng va vi 40 gam dung dch NaOH

4%. Cng thc ca X l:

A. H2NC2H3(COOH)2 B. H2NC3H5(COOH)2 C. (H2N)2C3H5COOH. D.

H2NC3H6COOH.

10/ Hn hp X gm axit Y n chc v axit Z hai chc ( Y,Z c cng s nguyn t C ).

Chia X thnh hai phn bng nhau. Cho phn mt tc dng ht vi Na, sinh ra 4,48 lt kh

H2 ( ktc). t chy hon ton phn hai, sinh ra 26,4 gam CO 2. Cng thc cu to thu

gn v phn trm v khi lng ca Z trong hn hp X ln lt l:

A. HOOC CH2 COOH v 70,87%. B. HOOC CH2 COOH v

54,88%.

C. HOOC COOH v 60,00%. D. HOOC COOH v 42,86%.

11/ Cho hn hp gm 1,12 gam Fe v 1,92 gam Cu vo 400 ml dung dch cha hn hp

gm H2SO4 0,5M v NaNO3 0,2M. Sau khi cc phn ng xy ra hon ton, thu c

dung dch X v kh NO ( sn phm kh duy nht ). Cho V ml dung dch NaOH 1M vo

dung dch X th lng kt ta thu c l ln nht. Gi tr ti thiu ca V l:

A. 240. B. 120. C. 360. D. 400.


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12/ Nung 6,58 gam Cu(NO3)2 trong bnh kn khng cha khng kh, sau mt thi gian thu

c 4,96 gam cht rn v hn hp kh X. Hp th hon ton X vo nc c 300 ml

dung dch Y. Dung dch Y c pH bng:

A. 2 B. 3. C. 4. D. 1.

13/ Cho 3,024 gam mt kim loi M tan ht trong dung dch HNO 3 long, thu c 940,8

ml kh NxOy ( sn phm kh duy nht, ktc) c t khi i vi H2 bng 22. Kh NxOy v

kim loi M l:

A. NO v Mg. B. N2O v Al. C. N2O v Fe. D. NO2 v Al.

14/ Cho 6,72 gam Fe vo 400 ml dung dch HNO3 1M, n khi phn ng xy ra honton, thu c kh NO ( sn phm kh duy nht) v dung dch X. Dung dch X c th ha

tan ti a m gam Cu. Gi tr ca m l:

A. 1,92. B. 0,64. C. 3,84. D. 3,20.

15/ Cho 2,24 gam bt st vo 200ml dung dch cha hn hp gm AgNO 3 0,1M v

Cu(NO3)2 0,5M. Sau khi cc phn ng xy ra hon ton, thu c dung dch X v m gam

cht rn Y. Gi tr ca m l

A. 2,80. B. 2,16. C. 4,08. D. 0,64.

16/ Nung nng m gam hn hp gm Al v Fe 3O4 trong iu kin khng c khng kh.

Sau khi phn ng xy ra hon ton, thu c hn hp rn X. Cho X tc dng vi dung

dch NaOH (d) thu c dung dch Y, cht rn Z v 3,36 lt kh H 2 ( ktc). Sc kh

CO2 (d) vo dung dch Y, thu c 39 gam kt ta. Gi tr ca m l:

A. 45,6. B. 48,3. C.36,7. D. 57,0.

Tuyn sinh i hc 2010(t 17 n 73):

17/ Hn hp Z gm hai axit cacboxylic n chc X v Y (MX > MY) c tng khi

lng l 8,2 gam. Cho Z tc dng va vi dung dch NaOH, thu c dung dch

cha 11,5 gam mui. Mt khc, nu cho Z tc dng vi mt lng d dung dch

AgNO3 trong NH3, thu c 21,6 gam Ag. Cng thc v phn trm khi lng ca X

trong Z l

A. C2H5COOH v 56,10%. B. C3H5COOH v 54,88%.


21/64

C. HCOOH v 45,12%. D. C2H3COOH v 43,90%.

18/ t chy hon ton mt lng hn hp X gm 2 ancol (u no, a chc, mch h, c

cng s nhm -OH) cn va V lt kh O2, thu c 11,2 lt kh CO2 v 12,6 gam

H2O (cc th tch kh o ktc). Gi tr ca V lA. 11,20. B. 14,56. C. 4,48. D. 15,68.

19/ Hn hp kh X gm mt ankan v mt anken. T khi ca X so vi H2 bng 11,25.

t chy hon ton 4,48 lt X, thu c 6,72 lt CO2 (cc th tch kh o ktc). Cng

thc ca ankan v anken ln lt l

A. CH4 v C3H6. B. CH4 v C4H8. C. C2H6 v C2H4. D. CH4 v

C2H4.

20/ Cho 150 ml dung dch KOH 1,2M tc dng vi 100 ml dung dch AlCl3 nng x

mol/l, thu c dung dch Y v 4,68 gam kt ta. Loi b kt ta, thm tip 175 ml

dung dch KOH 1,2M vo Y, thu c 2,34 gam kt ta. Gi tr ca x l

A. 0,9. B. 1,2. C. 1,0. D. 0,8.

21/ Ha tan hon ton 2,45 gam hn hp X gm hai kim loi kim th vo 200 ml dung

dch HCl 1,25M, thu c dung dch Y cha cc cht tan c nng mol bng nhau. Hai

kim loi trong X l

A. Be v Mg. B. Mg v Ca. C. Be v Ca. D. Mg v Sr.

22/ Nung 2,23 gam hn hp X gm cc kim loi Fe, Al, Zn, Mg trong oxi, sau mt thi

gian thu c 2,71 gam hn hp Y. Ho tan hon ton Y vo dung dch HNO3 (d), thu

c 0,672 lt kh NO (sn phm kh duy nht, ktc). S mol HNO3 phn ng l

A. 0,12. B. 0,14. C. 0,18. D. 0,16.

23/ Cho hn hp M gm anehit X (no, n chc, mch h) v hirocacbon Y, c tng smol l 0,2 (s mol ca X nh hn ca Y). t chy hon ton M, thu c 8,96 lt kh

CO2 (ktc) v 7,2 gam H2O. Hirocacbon Y l

A. C3H6. B. C2H2. C. C2H4.

D. CH4.


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24/ Cho 13,74 gam 2,4,6-trinitrophenol vo bnh kn ri nung nng nhit cao. Sau khi

phn ng xy ra hon ton, thu c x mol hn hp kh gm: CO2, CO, N2 v H2. Gi tr

ca x l

A. 0,36. B. 0,54. C. 0,45.

D. 0,60.

25/ ipeptit mch h X v tripeptit mch h Y u c to nn t mt aminoaxit (no,

mch h, trong phn t cha mt nhm -NH2 v mt nhm -COOH). t chy hon ton

0,1 mol Y, thu c tng khi lng CO2 v H2O bng 54,9 gam. t chy hon ton

0,2 mol X, sn phm thu c cho li t t qua nc vi trong d, to ra m gam kt ta.

Gi tr ca m l

A. 120. B. 45. C. 30.

D. 60.

26/ Ho tan hon ton 2,44 gam hn hp bt X gm FexOy v Cu bng dung dch

H2SO4 c nng (d). Sau phn ng thu c 0,504 lt kh SO2 (sn phm kh duy

nht, ktc) v dung dch cha 6,6 gam hn hp mui sunfat. Phn trm khi lng ca

Cu trong X l

A. 65,57%. B. 39,34%. C. 26,23%.

D. 13,11%.

27/ Hai hp cht hu c X v Y c cng cng thc phn t l C3H7NO2, u l cht rn

iu kin thng. Cht X phn ng vi dung dch NaOH, gii phng kh. Cht Y c

phn ng trng ngng. Cc cht X v Y ln lt l

A. axit 2-aminopropionic v axit 3-

aminopropionic.

B. axit 2-aminopropionic v amoni acrylat.

C. vinylamoni fomat v amoni acrylat.

D. amoni acrylat v axit 2-

aminopropionic.

28. Hn hp X gm alanin v axit glutamic. Cho m gam X tc dng hon ton vi dung

dch NaOH (d), thu c dung dch Y cha (m+30,8) gam mui. Mt khc, nu cho m

gam X tc dng hon ton vi dung dch HCl, thu c dung dch Z cha (m+36,5) gam


23/64

mui. Gi tr ca m l

A. 171,0. B. 165,6. C. 123,8.

D. 112,2.

29. Kh hon ton m gam oxit MxOy cn va 17,92 lt kh CO (ktc), thu c a gam

kim loi M. Ho tan ht a gam M bng dung dch H2SO4 c nng (d), thu c 20,16

lt kh SO2 (sn phm kh duy nht, ktc). Oxit MxOy l

A. FeO. B. CrO. C. Cr2O3.

D. Fe3O4.

30. Hn hp X gm 1 ancol v 2 sn phm hp nc ca propen. T khi hi ca X so vi

hiro bng 23. Cho m gam X i qua ng s ng CuO (d) nung nng. Sau khi cc phn

ng xy ra hon ton, thu c hn hp Y gm 3 cht hu c v hi nc, khi lng

ng s gim 3,2 gam. Cho Y tc dng hon ton vi lng d dung dch AgNO 3 trong

NH3, to ra 48,6 gam Ag. Phn trm khi lng ca propan-1-ol trong X l

A. 16,3%. B. 83,7%. C. 65,2%.

D. 48,9%.31. t chy hon ton 0,1 mol mt amin no, mch h X bng oxi va , thu c 0,5 mol

hn hp Y gm kh v hi. Cho 4,6 gam X tc dng vi dung dch HCl (d), s mol HCl

phn ng l

A. 0,3. B. 0,1. C. 0,4.

D. 0,2.

32. Hn hp X gm axit panmitic, axit stearic v axit linoleic. trung ho m gam X cn

40 ml dung dch NaOH 1M. Mt khc, nu t chy hon ton m gam X th thu c

15,232 lt kh CO2 (ktc) v 11,7 gam H2O. S mol ca axit linoleic trong m gam hn hp

X l

A. 0,005. B. 0,020. C. 0,010.

D. 0,015.

33. Trn 10,8 gam bt Al vi 34,8 gam bt Fe3O4 ri tin hnh phn ng nhit nhm


24/64

trong iu

kin khng c khng kh. Ho tan hon ton hn hp rn sau phn ng bng dung dch

H2SO4 long (d), thu c 10,752 lt kh H2 (ktc). Hiu sut ca phn ng nhit nhm

l

A. 80%. B. 60%. C. 70%.

D. 90%.

34. Cho dung dch X cha KMnO4 v H2SO4 (long) ln lt vo cc dung dch:FeCl2,

FeSO4, CuSO4, MgSO4, H2S, HCl (c). S trng hp c xy ra phn ng oxi ho - kh

l

A. 6. B. 4. C. 3.

D. 5.

35. Mt loi phn supephotphat kp c cha 69,62% mui canxi ihirophotphat, cn li

gm cc cht khng cha photpho. dinh dng ca loi phn ln ny l

A. 42,25%. B. 39,76%. C. 48,52%.

D. 45,75%.

36. in phn (vi in cc tr) 200 ml dung dch CuSO4 nng x mol/l, sau mt thi

gian thu c dung dch Y vn cn mu xanh, c khi lng gim 8 gam so vi dung dch

ban u. Cho 16,8 gam bt st vo Y, sau khi cc phn ng xy ra hon ton, thu c

12,4 gam kim loi. Gi tr ca x l

A. 1,50. B. 3,25. C. 2,25.

D. 1,25.

37. t chy hon ton m gam FeS2 bng mt lng O2 va , thu c kh X. Hp th

ht X vo 1 lt dung dch cha Ba(OH)2 0,15M v KOH 0,1M, thu c dung dch Y v

21,7 gam kt ta. Cho Y vo dung dch NaOH, thy xut hin thm kt ta. Gi tr ca m l

A. 23,2. B. 18,0. C. 12,6.

D. 24,0.38. Hn hp X gm CuO v Fe2O3. Ho tan hon ton 44 gam X bng dung dch HCl

(d), sau phn ng thu c dung dch cha 85,25 gam mui. Mt khc, nu kh hon

ton 22 gam X bng CO (d), cho hn hp kh thu c sau phn ng li t t qua dung

dch Ba(OH)2 (d) th thu c m gam kt ta. Gi tr ca m l


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26/64

C. Gly-Ala-Val-Val-Phe. D. Val-Phe-

Gly-Ala-Gly.

43: Axeton c iu ch bng cch oxi ho cumen nh oxi, sau thu phn trong

dung dch H2SO4 long. thu c 145 gam axeton th lng cumen cn dng (gi s

hiu sut qu trnh iu ch t 75%) l

A. 300 gam. B. 600 gam. C. 500 gam. D. 400

gam.

44: Cho 7,1 gam hn hp gm mt kim loi kim X v mt kim loi kim th Y tc dng

ht vi lng d dung dch HCl long, thu c 5,6 lt kh (ktc). Kim loi X, Y l

A. kali v bari. B. liti v beri. C. natri v magie. D. kali v

canxi.

45: Thu phn hon ton 0,2 mol mt este E cn dng va 100 gam dung dch NaOH

24%, thu c mt ancol v 43,6 gam hn hp mui ca hai axit cacboxylic n chc.

Hai axit l

A. HCOOH v C2H5COOH. B. HCOOH v CH3COOH.

C. C2H5COOH v C3H7COOH. D. CH3COOH v C2H5COOH.

46: un nng hn hp kh X gm 0,02 mol C2H2 v 0,03 mol H2 trong mt bnh kn

(xc tc Ni), thu c hn hp kh Y. Cho Y li t t vo bnh nc brom (d), sau khi

kt thc cc phn ng, khi lng bnh tng m gam v c 280 ml hn hp kh Z (ktc)thot ra. T khi ca Z so vi H2 l 10,08. Gi tr ca m l

A. 0,585. B. 0,620. C. 0,205. D. 0,328.

47: Cho 19,3 gam hn hp bt gm Zn v Cu c t l mol tng ng l 1 : 2 vo dung

dch cha 0,2 mol Fe2(SO4)3. Sau khi cc phn ng xy ra hon ton, thu c m gam

kim loi. Gi tr ca m l

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

48: Hn hp kh X gm imetylamin v hai hirocacbon ng ng lin tip. t chy

hon ton 100 ml hn hp X bng mt lng oxi va , thu c 550 ml hn hp Y


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gm kh v hi nc. Nu cho Y i qua dung dch axit sunfuric c (d) th cn li 250

ml kh (cc th tch kh v hi o cng iu kin). Cng thc phn t ca hai

hirocacbon l

A. CH4 v C2H6. B. C2H4 v C3H6. C. C2H6 v C3H8. D. C3H6

v C4H8.

49: t chy hon ton mt este n chc, mch h X (phn t c s lin kt nh hn

3), thu c th tch kh CO2 bng 6/7 th tch kh O2 phn ng (cc th tch kh o

cng iu kin). Cho m gam X tc dng hon ton vi 200 ml dung dch KOH 0,7M thu

c dung dch Y. C cn Y thu c 12,88 gam cht rn khan. Gi tr ca m l

A. 10,56. B. 7,20. C. 8,88. D. 6,66.

50: Hn hp M gm ancol no, n chc X v axit cacboxylic n chc Y, u mch h

v c cng s nguyn t C, tng s mol ca hai cht l 0,5 mol (s mol ca Y ln hn s

mol ca X). Nu t chy hon ton M th thu c 33,6 lt kh CO2 (ktc) v 25,2 gam

H2O. Mt khc, nu un nng M vi H2SO4 c thc hin phn ng este ho (hiu

sut l 80%) th s gam este thu c l

A. 22,80. B. 34,20. C. 27,36. D. 18,24.

51: Ho tan hon ton 8,94 gam hn hp gm Na, K v Ba vo nc, thu c dung dch

X v 2,688 lt kh H2 (ktc). Dung dch Y gm HCl v H2SO4, t l mol tng ng l 4 :

1. Trung ho dung dch X bi dung dch Y, tng khi lng cc mui c to ra lA. 13,70 gam. B. 12,78 gam. C. 18,46 gam. D. 14,62

gam.

52: Hn hp kh X gm N2 v H2 c t khi so vi He bng 1,8. un nng X mt thi

gian trong bnh kn (c bt Fe lm xc tc), thu c hn hp kh Y c t khi so vi He

bng 2. Hiu sut ca phn ng tng hp NH3 l

A. 25%. B. 50%. C. 36%. D. 40%.

53: Cho m gam NaOH vo 2 lt dung dch NaHCO3 nng a mol/l, thu c 2 lt dungdch X. Ly 1 lt dung dch X tc dng vi dung dch BaCl2 (d) thu c 11,82 gam kt

ta. Mt khc, cho 1 lt dung dch X vo dung dch CaCl2 (d) ri un nng, sau khi kt

thc cc phn ng thu c 7,0 gam kt ta. Gi tr ca a, m tng ng l

A. 0,08 v 4,8. B. 0,04 v 4,8. C. 0,14 v 2,4. D. 0,07 v

3,2.


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54: Ho tan hon ton m gam ZnSO4 vo nc c dung dch X. Nu cho 110 ml dung

dch KOH 2M vo X th thu c 3a gam kt ta. Mt khc, nu cho 140 ml dung dch

KOH 2M vo X th thu c 2a gam kt ta. Gi tr ca m l

A. 17,71. B. 16,10. C. 32,20. D. 24,15.

55: t chy hon ton m gam hn hp 3 ancol n chc, thuc cng dy ng ng, thu

c 3,808 lt kh CO2 (ktc) v 5,4 gam H2O. Gi tr ca m l

A. 5,42. B. 5,72. C. 4,72. D. 7,42.

56: Cho 0,15 mol H2NC3H5(COOH)2 (axit glutamic) vo 175 ml dung dch HCl 2M,thu c dung dch X. Cho NaOH d vo dung dch X. Sau khi cc phn ng xy ra hon

ton, s mol NaOH phn ng l

A. 0,70. B. 0,50. C. 0,65. D. 0,55.

57: Cho dung dch X gm: 0,007 mol Na+; 0,003 mol Ca2+; 0,006 mol Cl; 0,006 mol

HCO3 v 0,001 mol NO3. loi b ht Ca2+ trong X cn mt lng va dung

dch cha a gam Ca(OH)2. Gi tr ca a l

A. 0,180. B. 0,120. C. 0,444. D. 0,222.

58: Dung dch X c cha: 0,07 mol Na+; 0,02 mol v x mol . Dung dch Y c cha , v y

mol H+; tng s mol v l 0,04. Trn X v Y c 100 ml dung dch Z. Dung dch Z c

pH (b qua s in li ca H2O) l 24SOOH4ClO3NO4ClO3NO

A. 1. B. 12. C. 13. D. 2.

59: Oxi ho ht 2,2 gam hn hp hai ancol n chc thnh anehit cn va 4,8 gam

CuO. Cho ton b lng anehit trn tc dng vi lng d dung dch AgNO3 trong

NH3, thu c 23,76 gam Ag. Hai ancol l:

A. CH3OH, C2H5CH2OH. B. CH3OH, C2H5OH.

C. C2H5OH, C3H7CH2OH. D. C2H5OH, C2H5CH2OH.

60: Cho x mol Fe tan hon ton trong dung dch cha y mol H2SO4 (t l x : y = 2 : 5),

thu c mt sn phm kh duy nht v dung dch ch cha mui sunfat. S mol electron

do lng Fe trn nhng khi b ho tan l


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A. 2x. B. 3x. C. 2y. D. y.

61: Hn hp X gm 1 mol aminoaxit no, mch h v 1 mol amin no, mch h. X c kh

nng phn ng ti a vi 2 mol HCl hoc 2 mol NaOH. t chy hon ton X thu c 6

mol CO2, x mol H2O v y mol N2. Cc gi tr x, y tng ng l

A. 7 v 1,0. B. 8 v 1,5. C. 8 v 1,0. D. 7 v

1,5.

62: Cho m gam hn hp etanal v propanal phn ng hon ton vi lng d dung dch

AgNO3 trong NH3, thu c 43,2 gam kt ta v dung dch cha 17,5 gam mui amoni

ca hai axit hu c. Gi tr ca m l

A. 9,5. B. 10,9. C. 14,3. D. 10,2.

63: T 180 gam glucoz, bng phng php ln men ru, thu c a gam ancol etylic

(hiu sut 80%). Oxi ho 0,1a gam ancol etylic bng phng php ln men gim, thu

c hn hp X. trung ho hn hp X cn 720 ml dung dch NaOH 0,2M. Hiu sut

qu trnh ln men gim l

A. 90%. B. 10%. C. 80%. D. 20%.

64: Nh t t tng git n ht 30 ml dung dch HCl 1M vo 100 ml dung dch cha

Na2CO3 0,2M v NaHCO3 0,2M, sau phn ng thu c s mol CO2 l

A. 0,020. B. 0,030. C. 0,015. D. 0,010.

65: Hn hp gm 0,1 mol mt axit cacboxylic n chc v 0,1 mol mui ca axit vi

kim loi kim c tng khi lng l 15,8 gam. Tn ca axit trn l

A. axit propanoic. B. axit etanoic. C. axit metanoic. D. axit

butanoic.

66: t chy hon ton mt lng hirocacbon X. Hp th ton b sn phm chy vo

dung dch Ba(OH)2 (d) to ra 29,55 gam kt ta, dung dch sau phn ng c khi lng

gim 19,35 gam so vi dung dch Ba(OH)2 ban u. Cng thc phn t ca X l

A. C3H8. B. C2H6. C. C3H4. D. C3H6.

67: Xt cn bng: N2O4 (k) 2NO2 (k) 25oC. Khi chuyn dch sang mt trng thi

cn bng mi nu nng ca N2O4 tng ln 9 ln th nng ca NO2


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A. tng 9 ln. B. tng 3 ln. C. tng 4,5 ln. D. gim 3

ln.

68: Cho hn hp X gm ancol metylic v hai axit cacboxylic (no, n chc, k tip nhau

trong dy ng ng) tc dng ht vi Na, gii phng ra 6,72 lt kh H2 (ktc). Nu un

nng hn hp X (c H2SO4 c lm xc tc) th cc cht trong hn hp phn ng va

vi nhau to thnh 25 gam hn hp este (gi thit phn ng este ho t hiu sut 100%).

Hai axit trong hn hp X l

A. C3H7COOH v C4H9COOH. B. CH3COOH v C2H5COOH.

C. C2H5COOH v C3H7COOH. D. HCOOH v CH3COOH.

69: in phn (in cc tr) dung dch X cha 0,2 mol CuSO4 v 0,12 mol NaCl bng

dng in c cng 2A. Th tch kh (ktc) thot ra anot sau 9650 giy in phn l

A. 1,344 lt. B. 2,240 lt. C. 1,792 lt. D. 2,912

lt.

70: Cho m gam hn hp bt X gm ba kim loi Zn, Cr, Sn c s mol bng nhau tc dng

ht vi lng d dung dch HCl long, nng thu c dung dch Y v kh H2. C cn

dung dch Y thu c 8,98 gam mui khan. Nu cho m gam hn hp X tc dng hon

ton vi O2 (d) to hn hp 3 oxit th th tch kh O2 (ktc) phn ng l

A. 2,016 lt. B. 1,008 lt. C. 0,672 lt. D. 1,344

lt.71: t chy hon ton V lt hi mt amin X bng mt lng oxi va to ra 8V lt hn

hp gm kh cacbonic, kh nit v hi nc (cc th tch kh v hi u o cng iu

kin). Amin X tc dng vi axit nitr nhit thng, gii phng kh nit. Cht X l

A. CH3-CH2-CH2-NH2. B. CH2=CH-CH2-NH2.

C. CH3-CH2-NH-CH3. D. CH2=CH-NH-CH3.

72: Tch nc hn hp gm ancol etylic v ancol Y ch to ra 2 anken. t chy cng s

mol mi ancol th lng nc sinh ra t ancol ny bng 5/3 ln lng nc sinh ra tancol kia. Ancol Y l

A. CH3-CH2-CH(OH)-CH3. B. CH3-CH2-CH2-OH.

C. CH3-CH2-CH2-CH2-OH. D. CH3-CH(OH)-CH3.


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73: Cho 0,448 lt kh NH3 (ktc) i qua ng s ng 16 gam CuO nung nng, thu c

cht rn X (gi s phn ng xy ra hon ton). Phn trm khi lng ca Cu trong X l

A. 85,88%. B. 14,12%. C. 87,63%. D.

12,37%.

p n: 1D, 2B, 3A, 4C, 5A, 6D, 7C, 8A, 9B, 10D, 11C, 12D, 13B, 14A, 15C, 16B, 17B, 18A,

19C, 20A, 21D, 22D, 23D, 24C, 25A, 26C, 27B, 28A, 29C, 30B, 31D, 32A, 33A, 34C, 35B,

36C, 37C, 38B, 39D, 40C, 41C, 42C, 43D, 44C, 45B, 46D, 47C, 48B, 49C, 50D, 51C, 52A,

53A, 54B, 55C, 56C, 57D, 58A, 59A, 60D, 61A, 62B, 63A, 64D, 65B, 66A, 67B, 68B, 69C,

70B, 71A, 72C, 73D.

S IN PHN

A.KIN THC C BN:


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1. nh ngha s in phn: S in phn l qu trnh oxi ha - kh xy ra trn b

mt in cc, di tc dng ca dng in mt chiu chy qua cht in li trng

thi nng chy hay dung dch.

2. Phn ng oxi ha-kh xy ra in cc khi in phn:

a) Cation ( ion dng)v catot ( in cc m), ti cation nhn electron ( cht oxi

ha) to ra sn phm.

b) Anion (ion m) v anot ( in cc dng), ti anion nhng electron ( cht

kh) to ra sn phm.

Phn ng oxi ha - kh xy ra in cc l giai on quan trng nht, cn xc nh

r ion no u tin nhn hoc nhng electron v to ra sn phm g?

3. S oxi ho kh trn b mt in cc:

a) in phn cc cht nng chy ( mui, Al2O3) catot: ion dng kim loi nhn electron.

anot: ion m nhng electron.

b) in phn dung dch:

Khi in phn dung dch c nhiu cht oxi ha v cht kh th xy ra s oxi ha

kh ln lt cc in cc theo th t u tin.

vit phng trnh in phn, cn xt ring r cc qu trnh xy ra catot v

anot.c)Th t nhn electron:

cc m c cc ion H+ (H2O)cation kim loi. Cation kim loi nhn electron theo th t

u tin t sau ra trc:

Li+, K+,Ba2+, Ca2+, Na+,Mg2+, Al3+, H+ (H2O), Mn2+, Zn2+,Cr3+, Fe2+, Ni2+, Sn2+, Pb2+, H+

(axit), Cu2+, Fe3+,Hg+,Ag+, Hg2+ ,Pt2+,Au3+

Sn phm to thnh: M n+ + ne M; 2H+( axit) + 2e H2; 2H2O + 2e H2+

2OH-.

cc dng c cc anion v nhng ectron theo th t:

Cl-> Br-> S2-> CH3COO-> OH- > SO42-.

Sn phm to thnh: S2- - 2e S; 2O2- - 4e O2; 2Cl- - 2e Cl2; 2SO42- - 2e

S2O82-


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2CH3COO- - 2e CH3 CH3 + 2CO2; 2OH- (baz) 2e O2 + H2O; H2O - 2e

O2 + 2H+.

4. Hin tng dng cc tan:

Khi in phn dung dch mui trong nc, cc dng lm bng kim loi ca mui ha

tan th cc dng b n mn, gi l hin tng dng cc tan.

( Vt liu lm anot tr , khng b ha tan thng l: graphit, platin)

5. Tnh lng sn phm in phn thu c:

a) Tnh khi lng n cht:

p dng cng thc Faraday: m =n

AIt

96500hay s mol:

A

m=

n

It

96500.

b) Tnh khi lng hp cht:

Da vo cng thc Faraday tnh lng n cht trc ri suy ra lng hp cht bngphng trnh in phn.

B.CC DNG TON THNG GP:

DNG 1: Vit phng trnh in phn, gii thch qu trnh in phn.

V d 1:Vit phng trnh in phn Al2O3 nng chy vi in cc bng than ch.

Gii: Al2O3 ca ot0 2Al3+ + 3O2-.

Catot: 2Al3+ + 6e 2A; Anot: 3O2- - 6e 3/2 O2

Phng trnh in phn : Al2O3 p n c 2Al + 3/2 O2.

Nu in cc bng than, anot: C + O2 CO; CO2 nn anot b n mn dn.

V d 2:Vit phng trnh in phn NaOH nng chy.

Gii:NaOH ca ot0 Na+ + OH-.

Catot: Na+ + 1e Na; Anot: 2OH- - 2e H2O + O2.

Phng trnh in phn: 2NaOH p n c 2Na + H2O+ O2

V d 3:Gii thch qu trnh in phn dung dch CuSO4 vi in cc bng Cu.

Gii: CuSO4 Cu2+ + SO42-; H2O H+ + OH- .


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Catot: Cu2+, H+(H2O). Cu2+ + 2e Cu;

Anot: SO42-, OH-( H2O).

H2O 2e O2 + 2H+;

Cu + O2 CuO;

CuO + H2SO4 CuSO4 + H2O.

Xy ra hin tng dng cc tan.

V d 4:Gii thch qu trnh in phn dung dch NiSO4 vi anot tr.

Gii: NiSO4 Ni2+ + SO42-; H2O H+ + OH- .

Catot: Ni2+, H+(H2O). Ni2+ + 2e Ni.

Anot: SO42-, OH-( H2O). H2O 2e O2 + 2H+;

Phng trnh in phn: NiSO4 + H2O Ni + O2 + H2SO4.

V d 5: Cho dung dch ca hn hp NaCl v CuSO4.

a) Vit phng trnh in phn dung dch.

b) Gii thch ti sao dung dch sau in phn ha tan c Al2O3.

Gii:

a) NaCl Na+ + Cl- ; CuSO4 Cu2+ + SO42-; H2O H+ + OH-.

Catot: Cu2+, H+(H2O), Na+. Anot: Cl- , SO42- , OH-( H2O)

Cu2+ + 2e Cu 2Cl- - 2e Cl2

Phng trnh in phn: 2NaCl + CuSO4 p d d

Cu + Na2SO4 + Cl2

.b) Dung dch sau khi in phn ha tan Al2O3 nn c hai kh nng xy ra:

* Khi in phn c CuSO4 d:

CuSO4 + H2O p d d Cu + H2SO4 + O2

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

* Khi in phn c NaCl d:

2NaCl + 2H2O p d d 2NaOH + Cl2+ H2

Al2O3 + NaOH 2NaAlO2 + H2O.

DNG 2: Tnh khi lng kim loi v th tch cc cht kh thot ra in cc.


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V d 6: in phn 200ml dung dch CuSO4 vi cc in cc tr bng dng in mt

chiu I = 9,65 A. Khi th tch cc kh thot ra c hai in cc u bng 1,12 lt ( ktc)

th ngng in phn. Kim loi sinh ra bm vo catot c khi lng l:

A. 6,4 gam. B. 3,2 gam. C. 9,6 gam. D. 12,8

gam.

Gii: Phng trnh in phn CuSO4 + H2O p d d Cu + H2SO4 + O2 (1).

Sau khi CuSO4 b in phn ht, H2O b in phn:

catot: 2H+ + 2e H2.

anot: H2O 2e O2+ 2H+

H2O p H2+ O2 (2).

Theo bi ra: n(H2 thot ra catot) = 1,12/22,4 = 0,05 (mol) n(O2 thot ra anot

trong (2)) = 0,025

nCu( catot) = 2 n( O2 thot ra anot trong (1)) = 2.(0,05 0,025) = 0,05 mol.

Vy: mCu(catot) = 0,05.64 = 3,2 gam.

Chn p n B.

DNG 3: Tnh khi lng cc cht in phn.

V d 7: in phn c mng ngn 500 ml dung dch cha hn hp gm CuCl 2 0,1M v

NaCl 0,5M (in cc tr, hiu sut in phn 100%) vi cng dng in 5A trong

3860 giy. Dung dch thu c sau in phn c kh nng ho tan m gam Al. Gi tr lnnht ca m l

A. 4,05. B. 2,70. C. 1,35. D.

5,40.

( Trch TSH B 2009)

Gii:

n(CuCl2) = 0,05 (mol); n(NaCl) = 0,25 (mol).

Phng trnh in phn: 2NaCl + CuSO4

p d d

Cu + Na2SO4 + Cl2

(1).Mol: 0,1 0,05 0,05

Thi gian in phn (1) l t1 = 96500.2.0,05/ 5 = 1930 giy.

Sau phn ng d NaCl: n(NaCl d) = 0,15 mol.

2NaCl + 2H2O p d d 2NaOH + Cl2+ H2 (2)


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Al + H2O + NaOH NaAlO2 + 3/2 H2. (3)

Thi gian in phn (2) l t2 = 3860 1930 = 1930 giy nNaOH(2) =2nH2=96500

1930.5=

0,1 mol

Vy m = 0,1.27 = 2,7 gam.

Chn p n B.

V d 8:in phn nng chy Al2O3 vi anot than ch (hiu sut in phn 100%) thu

c m kg Al catot v 67,2 m3 ( ktc) hn hp kh X c t khi so vi hiro bng 16.

Ly 2,24 lt ( ktc) hn hp kh X sc vo dung dch nc vi trong (d) thu c 2

gam kt ta. Gi tr ca m l

A. 108,0. B. 75,6. C. 54,0. D.

67,5.( Trch TSH B 2009)

Gii:

Al2O3 c aot0 2Al3+ + 3O2-.

Catot: 2Al3+ + 6e 2A; Anot: 3O2- - 6e 3/2 O2

Phng trnh in phn : Al2O3 p n c 2Al + 3/2 O2 (1).

Nu in cc bng than, anot: C + O2 CO; CO2 (2) nn anot b n mn dn.

Hn hp kh X gm CO v CO2 ,O2(d) vi tng s mol l 67,2.1000/22,4 = 3000 (mol).

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

Trong 2,24 lt (0,1 mol) hn hp X:

nCO2 (3) = nCaCO3 =2/100 = 0,02 (mol) nCO = 0,06 (mol); nO2 = 0,02 (mol) .

Hn hp X gm nCO2 =nO2(d)= 600 mol; nCO= 1800mol nO2(1) = 600+600+ 900 =

2100 (mol)

nAl = 4/3.2100 = 2800 (mol).

Vy m = 2800.27 = 75600 gam = 75,6kg.

Chn p n B.

DNG 4: Tnh khi lng dung dch v nng dung dch cc cht sau in phn.


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V d 9: in phn 200ml dung dch NaCl 1M ( d = 1,15 g/ml) c mng ngn xp.Sau

khi thu c 1,12 lit kh ( ktc) thot ra catot th ngng in phn.Tnh nng %

cc cht trong dung dch sau in phn.

Gii: Phng trnh in phn: 2NaCl + 2H2O p d d 2NaOH + Cl2+ H2 (1)

H2O pddNaOH H2+ O2 (2)

Gi s ch c phn ng (1) xy ra v NaCl in phn ht: nCl2= nH2 = nNaCl = 0,1 mol.

Theo bi th nCl2 = nH2 = 0,05 (mol)< 0,1 (mol) phn ng (1) cha hon thnh v

khng xy ra phn ng (2).Dung dch sau in phn gm: 0,1 mol NaOH v 0,1 mol

NaCl (d).

Tng khi lng dung dch sau in phn: m = 1,15.200 0,05(71+2) = 226,35 (gam).

Vy: C%(NaOH) = 4/226,35 = 1,767%. C%(NaCl) = 5,85/226,35 = 2,584%.

V d 10: in phn 200ml dung dch NaCl 1M c mng ngn xp. sau khi thu c

4,48 lt kh (ktc) thot ra catot th ngng in phn. Tnh nng mol ca cc cht

trong dung dch sau in phn (Coi th tch dung dch khng i).

Gii: Xy ra hai qu trnh nh VD9. Dung dch sau in phn ch c 0,2 mol NaOH.

Vy CM(NaOH) = 0,2/ 0,2 = 1(M).

C. CU HI V BI TP TRC NGHIM T GII:

1. Dy cc kim loi u c th c iu ch bng phng php in phn dung dchmui ca chng l:

A. Fe, Cu, Ag. B. Mg, Zn, Cu. C. Al, Fe, Cr. D. Ba,

Ag, Au.

( Trch TSH A 2009)

2. in phn dung dch cha a mol CuSO4 v b mol NaCl ( vi in

cc tr, c mng ngn xp) dung dch sau in phn lm

phenolphtalein chuyn sang mu hng th iu kin ca a v b l:

A. b < 2a B. b = 2a C. b > 2a D. 2b = a

3. Trong cng nghip, natri hiroxit c sn xut bng phng php:

A. in phn dung dch NaCl, khng c mng ngn in cc.

B. in phn dung dch NaCl, c mng ngn in cc.

C. in phn dung dch NaNO3 , khng c mn ngn in cc.


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D. in phn NaCl nng chy.

4. Cho cc ion: Na+, Al3+, Ca2+, Cl-, SO42-, NO3-. Cc ion khng b in phn khi trng

thi dung dch l:

A. Na+, Al3+, SO42-, NO3-. B. Na+, Al3+, SO42-, Cl-.

C. Na+, Al3+, Cl-, NO3-. D. Al3+, Cu2+, Cl-, NO3-.

5. in phn ( in cc tr, c mng ngn xp) mt dung dch c cha cc cation: Fe2+,

Fe3+, Cu2+. Th t xy ra s kh catot l:

A. Fe2+, Cu2+, Fe3+. B. Fe3+, Cu2+, Fe2+. C. Fe3+, Fe2+, Cu2+. D. Fe2+,

Fe3+,Cu2+.

6. in phn ( in cc tr, c mng ngn xp) mt dung dch c cha cc anion: S 2-,

SO2-. Th t xy ra s oxi ha anot l:

A. S2-, OH-, SO42-. B. S2-, SO42-, OH-. C. OH-, S2-, SO42-. D. OH-,SO2-, S2-.

7. Cho cc dung dch ring bit sau: KCl, NaCl, CaCl2, Na2SO4, ZnSO4, H2SO4, KNO3,

AgNO3, NaOH. Dung dch khi in phn thc cht ch l in phn nc l:

A. NaOH, NaCl, ZnSO4, KNO3, AgNO3. B. NaOH, Na2SO4, H2SO4, KNO3,

CaCl2.

C. NaOH, Na2SO4, H2SO4, KNO3. D. Na2SO4, KNO3, KCl.

8. Khi in phn dung dch NaCl ( in cc tr, khng c mng ngn xp) th sn phm

thu c gm:

A. H2, Cl2, NaOH. B. H2, Cl2, nc Javen.

C. H2, nc Javen. D. H2,Cl2, NaOH, nc Javen.

9. Cho cc dung dch: KCl, NaCl, CaCl2, Na2SO4, ZnSO4, H2SO4, KNO3, AgNO3, NaOH.

Sau khi in phn, cc dung dch cho mi trng baz l:

A. KCl, KNO3, NaCl, Na2SO4. B. KCl, NaCl, CaCl2, NaOH.

C. NaCl, CaCl2, NaOH, H2SO4. D. NaCl, NaOH, ZnSO4, AgNO3.

10. Pht biu no sau y khng ng?

A. Khi in phn cc cht nng chy th catot cc cation kim loi nhn electron.

B. Khi in phn cc cht nng chy th anot cc anion nhng electron.

C.Khi in phn th trn cc b mt in cc xy ra qu trnh oxi ha kh.

D. Khi in phn cc dung dch mui trong nc th cc dng b n mn.


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11. in phn 200ml dung dch CuSO4 nng x M (in cc tr). Sau mt thi gian th

thy khi lng dung dch gim 8 gam v lm kt ta ht ion Cu2+ cn d trong dung

dch cn dng 100ml dung dch NaOH 0,5 M. Gi tr ca x l:

A. 0,5. B. 1,0. C. 1,5. D. Kt

qu khc.

12. in phn dung dch cha NaOH 0,01M v dung dch Na 2SO4 0,01M. Th tch dung

dch trong qu trnh in phn thay i khng ng k. pH ca dung dch sau in phn

l:

A. 2. B. 8. C.10. D.12.

13. in phn 400ml dung dch AgNO3 0,2M v Cu(NO3)2 0,1M ( h= 100%, in cc Pt)

vi cng dng in I = 10A. Sau thi gian t giy th ngng in phn, khi lng Cu

thot ra bm vo catot l 1,28 gam. Gi tr ca t l:A. 1158. B. 2316. C. 9650. D. 4825.

14. in phn dung dch NaOH vi cng dng in l I = 10A trong thi gian 268

giy. Sau khi in phn cn li 100 gam dung dch NaOH 24%. Nng % ca dung

dch NaOH trc khi in phn l:

A. 20. B. 25. C. 16. D. Kt

qu khc.

15. in phn 500ml dung dch AgNO3 0,1M v Cu(NO3)2 0,2M ( in cc platin) vi

cng dng in I = 10 A.Dung dch sau in phn c [H+] = 0,16M.Gi s hiu sut

in phn l 100% v th tch dung dch khng thay i. Nng mol ca mui nitrat

trong dung dch sau in phn l:

A. 0,2M. B. 0,3M. C. 0,1M. D.

0,17M.

BI TP CHUNG V KIM LOI


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I BI TP V XC NH TN KIM LOI

1) C th tnh c khi lng mol nguyn t kim loi M theo cc cch sau:

- T khi lng (m) v s mol (n) ca kim loi M = m/n- T Mhp cht Mkim loi

- T cng thc Faraday M =It

mnF(n l s electron trao i mi in cc)

- T a < m < b v < n < tm M tha mn trong khong xc

nh

- Lp hm s M = f(n) trong n l ha tr ca kim loi M (n = 1, 2, 3), nu trong bi

ton tm oxit kim loi MxOy th n = 2y/x kim loi M

- Vi hai kim loi k tip nhau trong mt chu k hoc phn nhm tm tn 2 kim

loi

2) Mt s ch khi gii bi tp:

- Bit s dng mt s nh lut bo ton nh bo ton khi lng, bo ton nguyn t,

bo ton mol electron, Bit vit cc phng trnh ion thu gn, phng php ion

electron

- Khi bi khng cho kim loi M c ha tr khng i th khi kim loi M tc dng vi

cc cht khc nhau c th th hin cc s oxi ha khc nhau t kim loi M c cc ha

tr khc nhau

- Khi hn hp u c chia lm hai phn khng bng nhau th phn ny gp k ln phn

kia tng ng vi s mol cc cht phn ny cng gp k ln s mol cc cht phn kia 3)


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Mt s v d minh ha:

V d 1: Cho 3,024 gam mt kim loi M tan ht trong dung dch HNO3 long, thu c

940,8 ml kh NxOy (sn phm kh duy nht, ktc) c t khi i vi H2 bng 22. Kh

NxOy v kim loi M l:

A. NO v Mg B. NO2 v Al C. N2O v Al D.

N2O v Fe

Hng dn:

M(NxOy) = 44 nN2O = 0,042 mol

M Mn+ + ne 2NO3- + 8e + 10H+ N2O + 5H2O

Theo lbt mol electron: ne cho = ne nhn n

M.8.42,0=3,024 M = 9n n = 3 v M

= 27 Al p n C

V d 2: Hn hp X gm Mg v kim loi M. Ha tan hon ton 8 gam hn hp X cn

va 200 gam dung dch HCl 7,3 %. Mt khc cho 8 gam hn hp X tc dng hon

ton vi kh Cl2 cn dng 5,6 lt Cl2 ( ktc) to ra hai mui clorua. Kim loi M v phn

trm v khi lng ca n trong hn hp X l:

A. Al v 75 % B. Fe v 25 % C. Al v 30 % D. Fe v 70 %

Hng dn:

nHCl = 0,4 mol ; nCl2 = 0,25 mol ; nMg = x mol ; nM = y mol 24x + My = 8 (1)

- X tc dng vi dung dch HCl (M th hin ha tr n) 2x + ny = 0,4 (2)

- X tc dng vi Cl2 (M th hin ha tr m) 2x + my = 0,5 (3)

- T (2) ; (3) y(m n) = 0,1 m > n No duy nht m = 3 v n = 2 x = y = 0,1

mol

- T (1) M = 56 Fe v % M = 70 % p n D


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V d 3: Hn hp X gm hai mui cacbonat ca 2 kim loi kim th hai chu k lin

tip. Cho 7,65 gam X vo dung dch HCl d. Kt thc phn ng, c cn dung dch th thu

c 8,75 gam mui khan. Hai kim loi l:

A. Mg v Ca B. Ca v Sr C. Be v Mg D. Sr v Ba

Hng dn:

- t cng thc chung ca hai mui l CO3. Phng trnh phn ng:

CO3 + 2HCl Cl2 + CO2 + H2O

- T phng trnh thy: 1 mol CO3 phn ng th khi lng mui tng: 71 60 = 11

gam

- Theo bi khi lng mui tng: 8,75 7,65 = 1,1 gam c 0,1 mol CO3 tham

gia phn ng

+ 60 = 76,5 = 16,5 2 kim loi l Be v Mg p n C

V d 4: Ha tan hon ton 6 gam hn hp X gm Fe v mt kim loi M (ha tr II) vo

dung dch HCl d, thu c 3,36 lt kh H2 ( ktc). Nu ch ha tan 1,0 gam M th dng

khng n 0,09 mol HCl trong dung dch. Kim loi M l:A. Mg B. Zn C. Ca D. Ni

Hng dn: nH2 = 0,15 mol

- nX = nH2 = 0,15 mol X = 40

- ha tan 1 gam M dng khng n 0,09 mol HCl 2/M< 0,09 22,2 < M < 40 nHCl = 0,1 loi

- Nu M l kim loi c hiroxit lng tnh (n = 2 hoc 3):

M + (4 n)OH + (n 2)H2O MO2n 4 + H2

y (4 n)y ny/2

- Do OH d nn kim loi M tan ht v nOH d = x (4 n)y mol x (4 n)y = 0,1

(2) v x + ny = 0,5 (3) y = 0,1 mol

- Thay ln lt n = 2 hoc 3 vo (1) ; (2) ; (3) ch c n = 3 ; x = 0,2 ; M = 27 l tha

mn %M = 36,9 % p n B


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III BI TON V KIM LOI TC DNG VI DUNG DCH AXIT

1) Kim loi tc dng vi dung dch axit:

a) i vi dung dch HCl, H2SO4 long:

M + nH+ Mn+ + n/2H2

(M ng trc hiro trong dy th in cc chun)

b) i vi H2SO4 c, HNO3 (axit c tnh oxi ha mnh):

- Kim loi th hin nhiu s oxi ha khc nhau khi phn ng vi H2SO4 c, HNO3 s t

s oxi ha cao nht

- Hu ht cc kim loi phn ng c vi H2SO4 c nng (tr Pt, Au) v H2SO4 c

ngui (tr Pt, Au, Fe, Al, Cr), khi S+6 trong H2SO4 b kh thnh S+4 (SO2) ; So hoc

S-2 (H2S)

- Hu ht cc kim loi phn ng c vi HNO3 c nng (tr Pt, Au) v HNO3 c

ngui (tr Pt, Au, Fe, Al, Cr), khi N+5 trong HNO3 b kh thnh N+4 (NO2)

- Hu ht cc kim loi phn ng c vi HNO3 long (tr Pt, Au), khi N+5 trong

HNO3 b kh thnh N+2 (NO) ; N+1 (N2O) ; No (N2) hoc N-3 (NH4+)

c) Kim loi tan trong nc (Na, K, Ba, Ca,) tc dng vi axit: c 2 trng hp

- Nu dung dch axit dng d: ch c phn ng ca kim loi vi axit

- Nu axit thiu th ngoi phn ng gia kim loi vi axit (xy ra trc) cn c phn ng

kim loi d tc dng vi nc ca dung dch



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- Kim loi tc dng vi hn hp axit HCl, H2SO4 long (H+ ng vai tr l cht oxi ha)

th to ra mui c s oxi ha thp v gii phng H2: M + nH+ Mn+ + n/2H2 (nH+ =

nHCl + 2nH2SO4)

- Kim loi tc dng vi hn hp axit HCl, H2SO4 long, HNO3 vit phng trnh phn

ng di dng ion thu gn (H+ ng vai tr mi trng, NO3 ng vai tr cht oxi ha)

v so snh cc t s gia s mol ban u v h s t lng trong phng trnh xem t s

no nh nht th cht s ht trc ( tnh theo)

- Cc kim loi tc dng vi ion NO3 trong mi trng axit H+ xem nh tc dng vi

HNO3

- Cc kim loi Zn, Al tc dng vi ion NO3 trong mi trng kim OHgii phng NH3

4Zn + NO3 + 7OH

4ZnO22

+ NH3 + 2H2O(4Zn + NO3 + 7OH+ 6H2O 4[Zn(OH)4]2+ NH3)

8Al + 3NO3 + 5OH + 2H2O 8AlO2 + 3NH3

(8Al + 3NO3 + 5OH + 18H2O 8[Al(OH)4] + 3NH3

- Khi hn hp nhiu kim loi tc dng vi hn hp axit th dng nh lut bo ton mol

electron v phng php ion electron gii cho nhanh. So snh tng s mol electron

cho v nhn bin lun xem cht no ht, cht no d

- Khi hn hp kim loi trong c Fe tc dng vi H2SO4 c nng hoc HNO3 cn ch xem kim loi c d khng. Nu kim loi (Mg Cu) d th c phn ng kim loi kh

Fe3+ v Fe2+. V d: Fe + 2Fe3+ 3Fe2+ ; Cu + 2Fe3+ Cu2+ + 2Fe2+

- Khi ha tan hon hon hn hp kim loi trong c Fe bng dung dch HNO3 m th

tch axit cn dng l nh nht mui Fe2+

- Kim loi c tnh kh mnh hn s u tin phn ng trc

- Nu bi yu cu tnh khi lng mui trong dung dch, ta p dng cng thc sau:

mmui = mcation + manionto mui = mkim loi + manion to mui

(manion to mui = manion ban u manion to kh)

- Cn nh mt s cc bn phn ng sau:

2H+ + 2e H2 NO3- + e + 2H+ NO2 + H2O


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SO42+ 2e + 4H+ SO2 + 2H2O NO3- + 3e + 4H+ NO + 2H2O

SO42+ 6e + 8H+ S + 4H2O 2NO3- + 8e + 10H+ N2O + 5H2O

SO42+ 8e + 10H+ H2S + 4H2O 2NO3- + 10e + 12H+ N2 + 6H2O

NO3- + 8e + 10H+ NH4+ + 3H2O

- Cn nh s mol anion to mui v s mol axit tham gia phn ng:

nSO42to mui = . nX (a l s electron m S+6 nhn to sn phm kh X)

nH2SO4phn ng = 2nSO2 + 4nS + 5nH2S

nNO3to mui = a.nX (a l s electron m N+5 nhn to ra sn phm kh X)

nHNO3 phn ng = 2nNO2 + 4nNO + 10nN2O + 12nN2

3) Mt s v d minh ha

V d 1: Cho 3,68 gam hn hp gm Al v Zn tc dng vi mt lng va dung dch

H2SO4 10 %, thu c 2,24 lt kh H2 ( ktc). Khi lng dung dch thu c sau phn

ng l:

A. 101,68 gam B. 88,20 gam C. 101,48

gam D. 97,80 gam

Hng dn: nH2 = nH2SO4 = 0,1 mol m (dung dch H2SO4) = 98 gam m (dung dchsau phn ng) = 3,68 + 98 - 0,2 = 101,48 gam p n C

V d 2: Ho tan hon ton 14,6 gam hn hp X gm Al v Sn bng dung dch HCl (d),

thu c 5,6 lt kh H2 ( ktc). Th tch kh O2 ( ktc) cn phn ng hon ton vi

14,6 gam hn hp X l:

A. 2,80 lt B. 1,68 lt C. 4,48 lt D. 3,92

lt

Hng dn: Gi nAl = x mol ; nSn = y mol 27x + 119y = 14,6 (1) ; nH2 = 0,25 mol

- Khi X tc dng vi dung dch HCl:


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V d 3: Cho 7,68 gam hn hp X gm Mg v Al vo 400 ml dung dch Y gm HCl 1M

v H2SO4 0,5M. Sau khi phn ng xy ra hon ton thu c 8,512 lt kh ( ktc). Bit

trong dung dch, cc axit phn li hon ton thnh cc ion. Phn trm v khi lng ca Al

trong X l:

A. 56,25 % B. 49,22 % C. 50,78 % D. 43,75

%

Hng dn: nH+ = 0,8 mol ; nH2 = 0,38 mol nH+phn ng = 0,76 mol < 0,8 mol axit

d, kim loi ht

- Gi nMg = x mol ; nAl = y mol % Al =

%

p n A

V d 4: Cho 0,10 mol Ba vo dung dch cha 0,10 mol CuSO4 v 0,12 mol HCl. Sau khi

cc phn ng xy ra hon ton, lc ly kt ta nung nhit cao n khi lng khng

i thu c m gam cht rn. Gi tr ca m l:

A. 23,3 gam B. 26,5 gam C. 24,9 gam D. 25,2

gam

Hng dn: Cc phn ng xy ra l:

Ba + 2HCl BaCl2 + H2 BaCl2 + CuSO4 BaSO4 +


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CuCl2

0,06 0,12 0,06 0,06 0,06 0,06

Ba + 2H2O Ba(OH)2 + H2 Ba(OH)2 + CuSO4 BaSO4 +

Cu(OH)2

0,04 0,04 0,04

0,04 0,04 0,04

Cu(OH)2 CuO + H2O

0,04 0,04

m (cht rn) = mBaSO4 + mCuO = (0,06 + 0,04).233 + 0,04.80 = 26,5 gam p n

B

V d 5: Th tch dung dch HNO3 1M (long) t nht cn dng ho tan hon ton 18

gam hn hp gm Fe v Cu trn theo t l mol 1 : 1 l: (bit phn ng to cht kh duy

nht l NO)

A. 1,0 lt B. 0,6 lt C. 0,8 lt D.

1,2 lt

Hng dn: nFe = nCu = 0,15 mol

- Do th tch dung dch HNO3 cn dng t nht mui Fe2+ ne cho = 2.(0,15 +

0,15) = 0,6 mol

- Theo lbt mol electron nH+ = nHNO3 = mol VHNO = 0,8 lt p n C

V d 6: Ha tan 9,6 gam Cu vo 180 ml dung dch hn hp HNO3 1M v H2SO4 0,5M,

kt thc phn ng thu c V lt ( ktc) kh khng mu duy nht thot ra, ha nu ngoi

khng kh. Gi tr ca V l:

A. 1,344 lt B. 4,032 lt C. 2,016 lt D.

1,008 lt


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Hng dn: nCu = 0,15 mol ; nNO3 = 0,18 mol ; nH+ = 0,36 mol

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Do

H+

ht ; Cu d0,36 0,09

VNO = 0,09.22,4 = 2,016 lt p n C

V d 7: Cho hn hp gm 1,12 gam Fe v 1,92 gam Cu vo 400 ml dung dch cha hn

hp gm H2SO4 0,5M v NaNO3 0,2M. Sau khi cc phn ng xy ra hon ton, thu c

dung dch X v kh NO (sn phm kh duy nht). Cho V ml dung dch NaOH 1M vo

dung dch X th lng kt ta thu c l ln nht. Gi tr ti thiu ca V l:

A. 360 ml B. 240 ml C. 400 ml D. 120 ml

Hng dn: nFe = 0,02 mol ; nCu = 0,03 mol n e cho = 0,02.3 + 0,03.2 = 0,12 mol ;

nH+ = 0,4 mol ; nNO3 = 0,08 mol (Ion NO3 trong mi trng H+ c tnh oxi ha mnh

nh HNO3)

- Bn phn ng: NO3 + 3e + 4H+ NO + 2H2O Do kim loi

kt v H+

d0,12 0,16

nH+ d = 0,4 0,16 = 0,24 mol nOH (to kt ta max) = 0,24 + 0,02.3 + 0,03.2

= 0,36 V = 0,36 lt hay 360 ml p n A

V d 8: Cho 24,3 gam bt Al vo 225 ml dung dch hn hp NaNO3 1M v NaOH 3M

khuy u cho n khi kh ngng thot ra th dng li v thu c V lt kh ( ktc).Gi

tr ca V l:

A. 11,76 lt B. 9,072 lt C. 13,44 lt D.

15,12 lt

Hng dn: nAl = 0,9 mol ; nNO3 = 0,225 mol ; nOH = 0,675 mol

8Al + 3NO3 + 5OH + 18H2O 8[Al(OH)4] + 3NH3 (1) Do


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NO3 ht

B: 0,9 0,225 0,675

P: 0,6 0,225 0,375 0,225

D: 0,3 0 0,3

Al + OH (d) + H2O AlO2 + H2 (2)

0,3 0,3 0,45

T (1) ; (2) V = (0,225 + 0,45).22,4 = 15,12 lt p n D

V d 9: Ha tan hon ton 100 gam hn hp X gm Fe, Cu , Ag trong dung dch HNO3

(d). Kt thc phn ng thu c 13,44 lt hn hp kh Y gm NO2, NO, N2O theo t l

s mol tng ng l 3 : 2 : 1 v dung dch Z (khng cha mui NH4NO3). C cn dung

dch Z thu c m gam mui khan. Gi tr ca m v s mol HNO3 phn ng ln lt

l:

A. 205,4 gam v 2,5 mol B. 199,2 gam v 2,4 mol

C. 205,4 gam v 2,4 mol D. 199,2 gam v 2,5 mol

Hng dn: nY = 0,6 mol nNO2 = 0,3 mol ; nNO = 0,2 mol ; nN2O = 0,1 mol

- nNO to mui = nNO + 3.nNO + 8.nN O = 0,3 + 3.0,2 + 8.0,1 = 1,7 mol mZ = mKl + mNO

to mui = 100 + 1,7.62 = 205,4 gam (1)

- nHNO phn ng = 2.nNO + 4.nNO + 10.nN O = 2.0,3 + 4.0,2 + 10.0,1 = 2,4 mol (2)

- T (1) ; (2) p n C

V d 10: Cho 6,72 gam Fe vo 400 ml dung dch HNO3 1M, n khi phn ng xy ra

hon ton, thu c kh NO (sn phm kh duy nht) v dung dch X. Dung dch X c

th ho tan ti a m gam Cu. Gi tr ca m l:

A. 1,92 gam B. 3,20 gam C. 0,64 gam D.

3,84 gam


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Hng dn: nFe = 0,12 mol ne cho = 0,36 mol; nHNO3 = 0,4 mol ne nhn = 0,3 mol

- Do ne cho > ne nhn Fe cn d dung dch X c Fe2+ v Fe3+

- Cc phn ng xy ra l:

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

0,1 0,4 0,1

Fe (d) + 2Fe3+ 3Fe2+

0,02 0,04

Cu + 2Fe3+ (d) Cu2+ + 2Fe2+

0,03 0,06

mCu = 0,03.64 = 1,92 gam p n A

V d 11: Ho tan hon ton 12,42 gam Al bng dung dch HNO3 long (d), thu cdung dch X v 1,344 lt ( ktc) hn hp kh Y gm hai kh l N2O v N2. T khi ca

hn hp kh Y so vi kh H2 l 18. C cn dung dch X, thu c m gam cht rn khan.

Gi tr ca m l:

A. 38,34 gam B. 34,08 gam C. 106,38

gam D. 97,98 gam

Hng dn: nAl = 0,46 mol ne cho = 1,38 mol ; nY = 0,06 mol ; Y = 36

- D dng tnh c nN2O = nN2 = 0,03 mol ne nhn = 0,03.(8 + 10) = 0,54 mol < ne

cho dung dch X cn cha mui NH4NO3 nNH4+ = NO3 = mol

- Vy mX = mAl(NO ) + mNH NO = 0,46.213 + 0,105.80 = 106,38 gam p n C

(Hoc c th tnh mX = mKl + mNO to mui + mNH = 12,42 + (0,03.8 + 0,03.10 + 0,105.8

+ 0,105).62 + 0,105.18 = 106,38 gam)

III BI TP V KIM LOI TC DNG VI DUNG DCH MUI

1) Kim loi tc dng vi dung dch mui:


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- iu kin kim loi M y c kim loi X ra khi dung dch mui ca n:

xM (r) + nXx+ (dd) xMn+ (dd) + nX (r)

+ M ng trc X trong dy th in cc chun+ C M v X u khng tc dng c vi nc iu kin thng

+ Mui tham gia phn ng v mui to thnh phi l mui tan

- Khi lng cht rn tng: m = mXto ra mMtan

- Khi lng cht rn gim: m = mMtan mX to ra

- Khi lng cht rn tng = khi lng dung dch gim

- Ngoi l:

+ Nu M l kim loi kim, kim th (Ca, Sr, Ba) th M s kh H+ ca H2O

thnh H2 v to thnh dung dch baz kim. Sau l phn ng trao i gia mui v

baz kim

+ trng thi nng chy vn c phn ng: 3Na + AlCl3 (khan) 3NaCl +

Al

+ Vi nhiu anion c tnh oxi ha mnh nh NO3-, MnO4-,th kim loi M

s kh cc anion trong mi trng axit (hoc baz)

- Hn hp cc kim loi phn ng vi hn hp dung dch mui theo th t u tin: kim

loi kh mnh nht tc dng vi cation oxi ha mnh nht to ra kim loi kh yu

nht v cation oxi ha yu nht

- Th t tng dn gi tr th kh chun (Eo) ca mt s cp oxi ha kh:

Mg2+/Mg < Al3+/Al < Zn2+/Zn < Cr3+/Cr < Fe2+/Fe < Ni2+/Ni < Sn2+/Sn < Pb2+/Pb < 2H+/H2

< Cu2+/Cu < Fe3+/Fe2+ < Ag+/Ag < Hg2+/Hg < Au3+/Au


- Phn ng ca kim loi vi dung dch mui l phn ng oxi ha kh nn thng s

dng phng php bo ton mol electron gii cc bi tp phc tp, kh bin lun nh

hn hp nhiu kim loi tc dng vi dung dch cha hn hp nhiu mui. Cc bi tp

n gin hn nh mt kim loi tc dng vi dung dch mt mui, hai kim loi tc dng

vi dung dch mt mui,c th tnh ton theo th t cc phng trnh phn ng xy ra


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- S dng phng php tng gim khi lng tnh khi lng thanh kim loi sau phn

ng,

- T s mol ban u ca cc cht tham gia phn ng bin lun cc trng hp xy ra

- Nu cha bit s mol cc cht phn ng th da vo thnh phn dung dch sau phn ng

v cht rn thu c bin lun cc trng hp xy ra

- Kim loi kh anion ca mui trong mi trng axit (baz) th nn vit phng trnh

dng ion thu gn

- Kim loi (Mg Cu) y c Fe3+ v Fe2+. V d: Fe + 2Fe3+ 3Fe2+ ; Cu + 2Fe3+

Cu2+ + 2Fe2+

- Fe + 2Ag+ Fe2+ + 2Ag. Nu Fe ht, Ag+ cn d th: Fe2+ + Ag+ Fe3+ + Ag

3) Mt s v d minh ha:

V d 1: Nhng mt thanh kim loi M ha tr II nng m gam vo dung dch Fe(NO3)2 th

khi lng thanh kim loi gim 6 % so vi ban u. Nu nhng thanh kim loi trn vo

dung dch AgNO3 th khi lng thanh kim loi tng 25 % so vi ban u. Bit gim

s mol ca Fe(NO3)2 gp i gim s mol ca AgNO3 v kim loi kt ta bm ht ln

thanh kim loi M. Kim loi M l:

A. Pb B. Ni C. Cd D. Zn

Hng dn: Gi nFe2+p = 2x mol nAg+p = x mol

M + Fe2+ M2+ + Fe

2x 2x 2x

m = 2x.(M 56) %mKl gim = (1)

M + 2Ag+ M2+ + 2Ag

0,5x x x

m = 0,5x.(216 M) %mKl tng = (2)

- T (1) ; (2) M = 65 Zn p n D


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V d 4: Cho 2,24 gam bt st vo 200 ml dung dch cha hn hp gm AgNO3 0,1M v

Cu(NO3)2 0,5M. Sau khi cc phn ng xy ra hon ton, thu c dung dch X v m gam

cht rn Y. Gi tr ca m l:

A. 2,80 gam B. 4,08 gam C. 2,16

gam D. 0,64 gam

Hng dn: nFe = 0,04 mol ; nAg+ = 0,02 mol ; nCu2+ = 0,1 mol

Th t cc phn ng xy ra l: (Fe2+/Fe < Cu2+/Cu < Fe3+ < Fe2+ < Ag+ < Ag)

Fe + 2Ag+ Fe2+ + 2Ag (1)

0,01 0,02 0,02

Fe + Cu2+ Fe2+ + Cu (2)

0,03 0,03T (1) ; (2) mY = 0,02.108 + 0,03.64 = 4,08 gam p n B

V d 5: Cho hn hp gm 1,2 mol Mg v x mol Zn vo dung dch cha 2 mol Cu2+ v 1

mol Ag+ n khi cc phn ng xy ra hon ton, thu c mt dung dch cha ba ion kim

loi. Trong cc gi tr sau y, gi tr no ca x tho mn trng hp trn:

A. 1,8 B. 1,5 C. 1,2 D. 2,0

Hng dn:

- Dung dch cha 3 ion kim loi Mg2+, Zn2+, Cu2+

- ne cho = (2,4 + 2x) mol v ne nhn = 1 + 2.2 = 5 mol

- Yu cu bi ton tha mn khi ne cho < ne nhn hay (2,4 + 2x) < 5 x < 1,3 x

=1,2 p n C

V d 6: Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO3)2 0,2M v

H2SO4 0,25M. Sau khi cc phn ng xy ra hon ton, thu c 0,6m gam hn hp bt

kim loi v V lt kh NO (sn phm kh duy nht, ktc). Gi tr ca 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


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Hng dn: nCu2+ = 0,16 mol ; nNO3 = 0,32 mol ; nH+ = 0,4 mol

- Cc phn ng xy ra l:

Fe + 4H+ + NO3 Fe3+ + NO + 2H2O (1)

0,1 0,4 0,1 0,1 0,1

VNO = 0,1.22,4 = 2,24 lt (*)

Fe + 2Fe3+ 3Fe2+ (2)

0,05 0,1

Fe + Cu2+ Fe2+ + Cu (3)

0,16 0,16

- T (1) ; (2) ; (3) nFep = 0,1 + 0,05 + 0,16 = 0,31 mol

- Hn hp bt kim loi gm Fe d v Cu (m 0,31.56) + 0,16.64 = 0,6m m = 17,8

gam (**)- T (*) ; (**) p n B

IV BI TP V KIM LOI TC DNG VI OXIT KIM LOI (PHN NG

NHIT NHM)


- Phn ng nhit nhm: Al + oxit kim loi oxit nhm + kim loi(Hn hp X) (Hn hp Y)

- Thng gp:

+ 2Al + Fe2O3 Al2O3 + 2Fe

+ 2yAl + 3FexOy y Al2O3 + 3xFe

+ (6x 4y)Al + 3xFe2O3 6FexOy + (3x 2y)Al2O3

- Nu phn ng xy ra hon ton, ty theo tnh cht ca hn hp Y to thnh binlun. V d:

+ Hn hp Y cha 2 kim loi Al d ; oxit kim loi ht

+ Hn hp Y tc dng vi dung dch baz kim (NaOH,) gii phng H2 c Al

d


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- nSO2 = 1,2 mol nFe = mol

- mAl2O3 = 92,35 0,8.56 0,25.27 = 40,8 gam (1) nAl2O3 = 0,4 mol

- Theo lbt nguyn t i vi O nO(Fe O ) = 0,4.3 = 1,2 mol

- Ta c: cng thc oxit st l Fe2O3 (2)

- T (1) ; (2) p n C

V d 4: Trn 5,4 gam bt Al vi 17,4 gam bt Fe3O4 ri tin hnh phn ng nhit nhm

(trong iu kin khng c khng kh). Gi s ch xy ra phn ng kh Fe3O4 thnh Fe.

Ha tan hon ton cht rn sau phn ng bng dung dch H2SO4 long (d) thu c5,376 lt kh H2 ( ktc). Hiu sut phn ng nhit nhm v s mol H2SO4 phn ng

l:

A. 75 % v 0,54 mol B. 80 % v 0,52 mol

C. 75 % v 0,52 mol D. 80 % v 0,54 mol

Hng dn: nAl = 0,2 mol ; nFe3O4 = 0,075 mol ; nH2 = 0,24 mol

- Phn ng xy ra khng hon ton: 8Al + 3Fe3O4 4Al2O3 + 9Fe

x 0,5x (mol)

- Hn hp cht rn gm:

- Ta c phng trnh: .2 + (0,2 x).3 = 0,24.2 x = 0,16 mol Hphn ng =


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% (1)

- nH+phn ng = 2.nFe + 3.nAl + 6.nAl2O3 + 8.nFe3O4 = 0,36 + 0,12 + 0,48 + 0,12 = 1,08

mol

nH2SO4phn ng = mol (2)

- T (1) ; (2) p n D

V MT BI TON KINH IN

1) Ni dung tng qut:

M hn hp rn (M, MxOy) M+n + sn phm kh

m gam m1 gam (n l s oxi ha cao nht

ca M)

(M l kim loi Fe hoc Cu v dung dch HNO3 (H2SO4 c nng) ly va hoc d)

- Gi: nM = x mol ; ne (2) nhn = y mol ne nhng = x.n mol

- Theo lbt khi lng t (1) nO = mol

- ne nhn = ne(oxi) + ne (2) = .2 + y = + y mol

- Theo lbt mol electron: ne nhng = ne nhn x.n = + y

- Nhn c hai v vi M ta c: (M.x).n = + M.y m.n =

m. = m = (*)


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- Thay M = 56 (Fe) ; n = 3 vo (*) ta c: m = 0,7.m1 + 5,6.y (1)

- Thay M = 64 (Cu) ; n = 2 vo (*) ta c: m = 0,8.m1 + 6,4.y (2)

(Khi bit 2 trong 3 i lng m, m1, y ta s tnh c i lng cn li)

2) V d minh ha:

V d 1: Cho 11,36 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung

dch HNO3 long (d), thu c 1,344 lt kh NO (sn phm kh duy nht, ktc) v

dung dch X. C cn dung dch X thu c m gam mui khan. Gi tr m l:


Hng dn: nNO = 0,06 mol y = 0,06.3 = 0,18 mol

Theo cng thc (1) ta c: nFe = mol nFe(NO3)3 = 0,16

mol

mmui khan = 0,16.242 = 38,72 gam p n A

V d 2: kh hon ton 3,04 gam hn hp X gm FeO, Fe3O4, Fe2O3 cn 0,05 mol H2.

Mt khc, ha tan hon ton 3,04 gam hn hp X trong dung dch H2SO4 c thu c V

ml kh SO2 (sn phm kh duy nht ktc). Gi tr ca V l:A. 112 ml B. 224 ml C. 336 ml D. 448

ml

Hng dn: Thc cht phn ng kh cc oxit l: H2 + O(oxit) H2O. V vy nO(oxit) = nH2

= 0,05 mol mFe = 3,04 0,05.16 = 2,24 gam

Theo cng thc (1) ta c: ne nhn (S+6 S+4) = y = mol nSO2

= 0,01 mol V = 0,01.22,4 = 0,224 lt hay 224 ml p n B

V d 3: Nung m gam bt Cu trong oxi thu c 37,6 gam hn hp rn X gm Cu, CuO

v Cu2O. Ha tan hon ton X trong dung dch H2SO4 c, nng (d) thy thot ra 3,36 lt

kh ( ktc). Gi tr ca m l:


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A. 25,6 gam B. 32 gam C. 19,2 gam D. 22,4

gam

Hng dn: nSO2 = 0,15 mol y = 0,15.2 = 0,3 mol

Theo cng thc (2) ta c: m = 0,8.37,6 + 6,4.0,3 = 32 gam p n B

ng hnh cng cc s thttp://ngocbinh.dayhoahoc.com

MỘT SỐ PHƯƠNG PHÁP GIẢI BÀI TOÁN HÓA HỌ1

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