C+íc dß¦íng to+ín h+¦a hß+ìc v+¦ c¦í - NGß+îC QUANG

8/8/2019 C+c dng to+n h+a h+c v+ c - NG+C QUANG

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CC DNG TON HA V C

PHNG PHP BO TON NGUYN TVD1 :Cho chui phn ng nh sau :

Fe Fe3O4 Fe2O3 Fe2(SO4)3 Fe(NO3)3 Fe(OH)3 Fe2O3 Fe3O4

Vit cc phng trnh phn ng xy ra . V tnh s mol ca cc nguyn t trong hp cht . a ra nhnxtVD2 :Cho hn hp gm 0,1 mol Fe ; 0,2 mol FeO , O,2 mol Fe2O3 v 0,1 mol Fe3O4 tc dng hon tonvi dung dch H2SO4 c nng . Vit cc phng trnh phn ng xy ra v tnh khi lng mui thu c .

Bi ton ny nu cc em gii bnh thng s phi vit rt nhiu phng trnhNu dng phng php bo ton nguyn t :S mol Fe c trong FeO = 0,2 ; Fe2O3 = 0,4 ; Fe3O4 = 0,3 mol , Fe ban u 0,1 Tng s mol Fe l : 0,2 + 0,4 + 0,3 + 0,1 = 1 mol(Fe , FeO , Fe2O3 , Fe3O4 ) + H2SO4 Fe2(SO4)3 + SO2 + H2O1 mol 0,5 mol (Bo ton nguyn t Fe ) Khi lng ca mui l : 0,5.400 = 200 gam

Cu 1:Cho 1,6 gam bt Fe2O3 tc dng vi axit d HCl .Khi lng mui trong dung dch sau phn ngl ?

0,1 mol Fe2O3 c 0,2 mol Fe FeCl30,2 0,2

Khi lng muiCu 2 :Ho tan hon ton hn hp gm Fe , FeO , Fe2O3 , Fe3O4 trong dung dch HNO3 long nng d thuc 4,48 lt kh NO duy nht ktc v 96,8 gam mui Fe(NO3)3 phn ng . Tnh s mol HNO3s : 1,4 mol( Fe , FeO , Fe2O3 , Fe3O4) + HNO3 Fe(NO3)3 + NO + H2O

0,4 0,2

S mol mui : Fe(NO3)3 = 0,4 mol S mol N l : 3.0,4 = 1,2 molS mol N : 4,48/22,4 = 0,2 mol Tng s mol N v phi : 1,2 + 0,2 = 1,4 S mol N v tri = 1,4 S mol HNO3 = 1,4 mol

Cu hi ph : Tnh m H2O to thnh , m xit ban u .Cu 3 :Hn hp X gm mt xit ca st c khi lng 2,6 gam . Cho kh CO d i qua X nung nng , khi ra hp th vo dung dch nc vi trong d th thu c 10 gam kt ta . Tnh tng khi lng ca Fec trong X l ?S : 1 gam .

(Fe , FexOy ) + CO Fe + CO22,6 gam

CO2 + Ca(OH)2 CaCO3 + H2O0,1 0,1 (tnh c t 10 gam kt ta ) S mol CO = CO2 = 0,2 (Bo ton C ) S mol O c trong FexOy + O trong CO = O trong CO2 n O (FexOy ) = 0,1.2 0,1.1 = 0,1 Khi lng ca O = 1.16 = 1,6 m Fe = 2,6 1,6 = 1 gam

Cu 4 :Ho tan hon ton m gam Fe trong dung dch HCl thu c x gam mui clorua . Nu ho tan honton m gam Fe trong dung dch HNO3 long d th thu c y gam mui nitrat . Khi lng 2 mui chnh

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CC DNG TON HA V C

lch nhau 23 gam . Ga tr ca m l ?

Fe + HCl FeCl2a aFe + HNO3 Fe(NO3)3a a

Khi lng FeCl2 = x = 127a Khi lng Fe(NO3)3 = y = 242ay x = 23 242a 127a = 115a = 23 a = 0,2m Fe = 0,2.56 = 11,2 gam

Cu 5 :Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo axit HNO3 va thu cdung dch X ch cha hai mui sunfat v kh duy nht NO . Ga tr ca a l ?S : a = 0,06 mol

Dng phng php bo ton nguyn t :

FeS2 Fe2(SO4)3 (1)

0,12 0,06 (Bo ton nguyn t Fe )

Cu2S CuSO4 (2)

a 2a (Bo ton nguyn t Cu )

Bo ton nguyn t S :

V tri :0,24 + a v V phi :0,18 + 2a

0,24 + a = 0,18 + 2a a = 0,06 mol

Chn p n D

Cu 6 :Cho kh CO i qua ng s cha 16 gam Fe 2O3 un nng sau phn ng thu c hn hp rn Xgm Fe , FeO , Fe3O4 , Fe2O3 . Ho tan hon ton X bng H2SO4 c nng thu c dung dch Y . C cndung dch Y thu c lng mui khan l bao nhiuS : 40CO + Fe2O3 (Fe , FeO , Fe3O4 , Fe2O3 ) + CO2(Fe , FeO , Fe3O4 , Fe2O3 ) + H2SO4 Fe2(SO4)3 + SO2 + H2O

Nh vy : Fe2O3 Fe2(SO4)30,1 0,1 ( Bo ton nguyn t Fe )

Khi lng Fe2(SO4)3 = 400.0,1 = 40Cu hi thm : Nu cho bit kh SO2 thu uc l 0,3 mol , Tnh n H2SO4 ,

Cu 7 :Ho tan hon ton hn hp X gm 6,4 gam Cu v 5,6 gam Fe bng dung dch HNO3 1M sau phnng thu c dung dch A v kh NO duy nht . Cho tip dung dch NaOH d vo dung dch A thu ckt ta B v dung dch C . Lc kt ta B ri em nung ngoi khng kh n khi lng khng i th khilng cht rn thu c l ?S : 16

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CC DNG TON HA V C

Cu Cu(NO3)2 Cu(OH)2 CuO0,1 0,1 ( Bo ton nguyn t Cu )Fe Fe(NO3)3 Fe(OH)3 Fe2O30,1 0,05 ( Bo ton nguyn t Fe )

Cht rn l : CuO 0,1 mol ; Fe2O3 0,05 mol

Khi lng : 80.0,1 + 0,05.160 = 16 gam

Cu 8 :Ho tan hon ton hn hp X gm 0,4 mol FeO v 0,1 mol Fe 2O3 vo dung dch HNO3 long d ,thu c dung dch A v kh NO duy nht . Dung dch A cho tc dng vi dung dch NaOH d thu ckt ta . Ly ton b kt ta nung trong khng kh n khi lng khng i thu c cht rn c khilng l ?S : 48 gam

FeO Fe(NO3)3 Fe(OH)3 Fe2O30,4 0,2 (bo ton Fe )Fe2O3 Fe(NO3)3 Fe(OH)3 Fe2O30,1 0,1

Cht rn sau khi nung l : Fe2O3 : 0,3 mol Khi lng 0,3.16 = 48 gam

Cu 9 :Ho tan hon ton m gam hn hp X gm Fe v Fe2O3 trong dung dch HCl thu c 2,24 lt khH2 ktc v dung dch B .Cho dung dch B tc dng vi dung dch NaOH d lc ly kt ta , nung trongkhng kh n khi lng khng i thu c cht rn c khi lng bng 24 gam . Tnh m ?s : 21.6 gam .

Fe + HCl H0,1 0,1Fe FeCl2 Fe(OH)2 Fe2O30,1 0,05 (bo ton Fe )Fe2O3 FeCl3 Fe(OH)3 Fe2O3

a a Theo phng trnh s mol Fe2O3 thu c l : 0,05 + a Cht rn sau khi nung l : Fe2O3 : 24/160 = 0,15 mol 0,05 + a = 0,15 a = 0,1

m = 0,1.56 + 0,1.160 = 21,6

Bi tp lm thm :

BI TON CO2

Dn kh CO2 vo dung dch kim NaOH , KOH th t xy ra phn ng :

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4/64

CC DNG TON HA V C

CO2 + KOH KHCO3 KHCO3 + KOH K2CO3 + H2ONgoi ra ta c th vit lun hai phn ng :

1---------- CO2 + KOH KHCO3 (1)

2---------- CO2 + 2KOH K2CO3 (2)

Xt t s mol : k = n KOH / n CO2Nu k 1 Ch c phn ng (1) , mui thu c l KHCO3 (D CO2 )Nu 1 < k < 2 C c hai phn ng , mui thu c l hai mui KHCO3 , K2CO3 (Sauphn ng c CO2 v KOH u ht ).Nu k 2 Ch c phn ng (2) , mui thu c l mui trung ha K2CO3 . (d KOH)

Cu 1:Cho 1,568 lt CO2 ktc li chm qua dung dch c ha tan 3,2 gam NaOH . Hy xc

nh khi lng mui sinh ra ?

n CO2 = 0,07 mol , n NaOH= 0,08 mol k = 0,08/0,07 = 1,14 > 1 v < 2 C c hai phnng (1) , (2) , hai mui to thnh v CO2 , NaOH u ht .

---------- CO2 + NaOH NaHCO3 (1)x x x

---------- CO2 + 2NaOH Na2CO3 (2)y 2y y

Gi s mol ca kh CO2 tham gia mi phn ng l : x , y

S mol CO2 phn ng : x + y = 0,07 , s mol NaOH : x + 2y = 0,08 x = 0,06 , y = 0,01 mol Mui thu c : NaHCO3 : 0,06 mol , Na2CO3 : 0,01 mol Khi lng mui : 0,06.84 + 0,01.56 = 6,1 gamDn kh CO2 vo dung dch kim Ca(OH)2 , Ba(OH)2 th t xy ra phn ng :CO2 + Ca(OH)2 CaCO3 + H2O (1) CO2 + Ca(OH)2 Ca(HCO3)2 (2)

Ngoi ra ta c th vit lun hai phn ng :

1---------- CO2 + Ca(OH)2 CaCO3 + H2O (1)

2---------- 2CO2 + Ca(OH)2 Ca(HCO3)2 (2)

Xt t s mol : k = n CO2 / n Ca(OH)2Nu k 1 Ch c phn ng (1) , mui thu c l CaCO3 (D Ca(OH)2)Nu 1 < k < 2 C c hai phn ng , mui thu c l hai mui CaCO3 , Ca(HCO3)2(Sau phn ng c CO2 v Ca(OH)2 u ht ).

Nu k 2 Ch c phn ng (2) , mui thu c l mui trung ha Ca(HCO3)2 . (d CO2 )

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CC DNG TON HA V C

Cu 2 :Cho 8 lt hn hp kh CO v CO2 trong CO2 chim 39,2 % i qua dung dchccha 7,4 gam Ca(OH)2 . Hy xc nh s gam kt ta thu c sau phn ng ?

Hn hp kh CO , CO2 ch c CO2 phn ng vi Ca(OH)2 Th tch kh CO2 = 8.39,2/100

= 313,6 n = 0,14 mol . nCa(OH)2 = 7,4/74 = 0,1 mol k = 1,4 C hai phn ng .

1---------- CO2 + Ca(OH)2 CaCO3 + H2O (1)x x x

2---------- 2CO2 + Ca(OH)2 Ca(HCO3)2 (2)y y/2

Gi s mol ca CO2 v Ca(OH)2 phn ng l x , y S mol CO2 : x + y = 0,14 , s mol Ca(OH)2 : x + y/2 = 0,1 mol x = 0,06 ; y = 0,08 mol Khi lng kt ta l : 0,06.100 = 6 gam

Cu 3 :Cho m gam tinh bt ln men thnh ancol etylc vi hiu sut 81% .Ton b lngCO2 vo dung dch Ca(OH)2 thu c 550 gam kt ta v dung dch Xun k dung dch Xthu thm c 100 gam kt ta . Ga tr ca m l ?

(C6H10O5)n + n H2O n C6H12O63,75/n 3,75

Phn ng ln men : C6H12O6 2CO2 + 2H2O3,75---------7,5

Dn CO2 vo dung dch nc vi trong m thu c dung dch X khi un nng X thu c kt ta chng trng c Ca(HCO3)2 to thnh .Ca(HCO3)2 CaCO3 + CO2 + H2O

1- ----------------------1 molCa(OH)2 + CO2 CaCO3 + Ca(HCO3)2 + H2O

5,5 1Bo ton nguyn t C S mol CO2 = 7,5 Khi lng tinh bt : 162.3,75.100/81 = 750

Cu 4 : Hp th hon ton 2,688 lt kh CO2 ktc vo 2,5 lt dung dch Ba(OH)2 nng amol/l , thu c 15,76 gam kt ta .Ga tr ca a l ?

n CO2 = 0,12 mol , n BaCO3 = 0,08 moln CO2 > n BaCO3 Xy ra 2 phn ng :CO2 + Ba(OH)2 BaCO3 + Ba(HCO3)2 + H2O0,12 0,08Bo ton nguyn t C S mol ca C Ba(HCO3)2 l : 0,12 0,08 = 0,04 mol S mol ca Ba(HCO3)2 = 0,02 molBo ton nguyn t Ba S mol Ba(OH)2 = 0,08 + 0,02 = 0,1 mol CM Ba(OH)2 = 0,1/2,5 = 0,04 M

Cu 5 :Cho 3,36 lt kh CO2 ktc vo 200 ml dung dch cha NaOH 1M v Ba(OH)2 0,5M .Khi lng kt ta thu c sau phn ng l ?

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CC DNG TON HA V C

n CO2 = 0,15 mol , n OH- = 0,2 + 0,2 = 0,4 mol , n Ba2+ = 0,1 molCO2 + OH- HCO3-

0,15 0,4 ------0,15OH- d = 0,4 0,15 = 0,25 mol ,HCO3- + OH- CO32- + H2O0,15 0,25 -0,15 molOH- d : 0,25 0,15 = 0,1 molBa2+ + CO32- BaCO3 0,1 0,15 ----0,1 mol Khi lng kt ta l : 0,1.197 = 19,7 gam

Cu 6 :Dn 5,6 lt CO2 hp th hon ton vo 500 ml dung dch Ca(OH)2 nng a Mth thu c 15 gam kt ta . Ga tr ca a l ?

Gii tng t bi 4 .

Cu 7 :Dn 112 ml CO2 ktc hp th han ton vo 200 ml dung dch Ca(OH)2 thu c0,1 gam kt ta . Nng mol ca nc vi trong l ?

Gii tng t bi 4 .

Cu 8 :Dn 5,6 lt CO2 hp th hon ton vo 200 ml dung dch NaOH nng a M thuc dung dch X c kh nng tc dng ti a 100 ml dung dch KOH 1M .Tnh a ?

n CO2 = 0,25 mol , n KOH= 0,1 molDung dch X c kh nng ha tan c KOH nn X phi c mui NaHCO3NaHCO3 + KOH Na2CO3 + K2CO3 + H2O

0,1-----0,1Ta c th vit tm tt phn ng :CO2 + NaOH Na2CO3 + NaHCO3 + H2O0,25 0,1Bo ton nguyn t C S mol Na2CO3 = 0,15 molBo ton nguyn t Na S mol NaOH = 0,1 + 0,15.2 = 0,4 CM NaOH= 0,4/0,2 = 2 M

Cu 9 . Cho ton 0,448 lt kh CO2 (ktc) hp th hon ton bi 200 ml dung

dch Ba(OH)2 thu c 1,97 gam kt ta. Hy la chn nng mol/l ca dungdch Ba(OH)2.A. 0,05M B. 0,1M C. 0,15M D. p n khc.

Gii tng t Cu 4

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CC DNG TON HA V C

Cu 10 . Cho 3,36lt kh CO2 (ktc) hp th bng 100 ml dung dch NaOH 1Mv Ba(OH)2 1M th thu c bao bao nhiu gam kt ta.A. 19,7 gam B. 24,625 gam C. 14,775 gam D.c A, B u ng.

Gii tng t cu 5

Cu 11. Hn hp X gm 2 mui cacbonat ca 2 kim loi thuc 2 chu k lin tipca phn nhm chnh nhm II. Ha tan ht 16,18 gam hn hp A trong dungdch HCl thu c kh B. Cho ton b kh B hp th vo 500 ml dung dchBa(OH)2 1M. Lc b kt ta , ly dung dch nc lc tc dng vi lng d dung dchNa2SO4 th thu c 11,65 gam kt ta. Xc nh cng thc ca 2 mui.A. BeCO3 v MgCO3 B. MgCO3 v CaCO3 C. CaCO3 v SrCO3 D.c A, B u ng

n Ba(OH)2 = 0,5mol , n BaSO4 = 0,05 mol

Xt bi ton t: phn dn kh CO2 qua dung dch Ba(OH)2 1M .Trng hp (1) ch xy ra mt phn ngCO2 + Ba(OH)2 BaCO3 + H2O (1)0,15---0,45Sau phn ng Ba(OH)2 dBa(OH)2 + Na2SO4 BaSO4 + 2NaOH

0,05------------------0,05 mol S mol Ba(OH)2 tham gia phn ng (1) = 0,5 0,05 = 0,45Gi cng thc trung bnh ca hai mui : MCO3MCO3 + 2HCl MCl2 + CO2 + H2O0,2---------------------------0,2

MCO3 = 16,18/0,2 = 80,9 M = 20,19 Mg , Be tha mn

Trng hp 2 : C 2 phn ng : Vit tm tt liBa(OH)2 + CO2 BaCO3 + Ba(HCO3)2 + H2O (1)0,5 0,05Ba(HCO3)2 + Na2SO4 BaSO4 + 2NaHCO30,05-----------------------0,05Bo ton s mol Ba (1) S mol BaCO3 = 0,045Bo ton C (1) S mol CO2 = 0,45 + 0,1 = 0,55MCO3 + 2HCl MCl2 + CO2 + H2O0,55-------------------------0,55 MCO3 = 16,18/0,55 = 29,4 Mg , Ca tha mn

Chn D .

Cu 1 2 : Mt bnh cha 15 lt dd Ba(OH)2 0,01M. Sc vo dd V ltkh CO2 (ktc) ta thu c 19,7g kt ta trng th gi tr ca V l:A. 2,24 lt B. 4,48 ltC. 2,24 lt v 1,12 lt D. 4,48 lt v 2,24 lt

n Ba(OH)2 = 0,15 mol , n BaCO3 = 0,1 molV CO2 l nhn t gy ra hai phn ng nn s c 2 p n ca CO2 tha mn u bi .

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CC DNG TON HA V C

Trng hp (1) ch c 1 phn ng :CO2 + Ba(OH)2 BaCO3 + H2O

0,15 0,1 S mol CO2 tnh theo BaCO3 : n CO2 = 0,1 mol VCO2 = 2,24 mol

Trng hp 2 : C c 2 phn ng : Vit tm tt li

CO2 + Ba(OH)2 BaCO3 + Ba(HCO3)2 + H2O0,15-----v----0,1 ----------0,05

Bo ton nguyn t Ba : S mol Ba(HCO3)2 l 0,05 molBo ton nguyn t C : S mol CO2 = 0,1 + 0,05.2 = 0,2 mol VCO2 = 4,48 ltChn p n C.

Cu 13 : t chy hon ton cht hu c X cn 6,72 lt kh O2 ktc , cho ton b sn phmchy vo bnh ng Ba(OH)2 thu c 19,7 gam kt ta v khi lng dung dch gim 5,5gam . Lc b kt ta un nng dung dch li thu c 9,85 gam kt ta na . Tm cng thcca X

Theo gi thit : S mol O2 = 0,3 mol ; n BaCO3 lc u = 0,1 mol ; n BaCO3sau khi un nng dd = 0,05 molCch tm khi lng nguyn t C nhanh nht cho bi ton ny :un nng dd :Ba(HCO3)2 BaCO3+ CO2+ H2O

0,05 0,05

CO2 + Ba(OH)2 BaCO3 + Ba(HCO3)2 + H2O0,1 0,05

Bo ton nguyn t CS mol CO2 = 0,1 + 0,05.2 = 0,2 mol

Khi lng dd gim = m BaCO3 ( m CO2 + m H2O ) = 5,5 m H2O = 5,4 n H2O = 0,3 mol ( >n CO2Hp cht no )T s mol H2O v CO2n H= 0,6 ; n C= 0,2 ,n Otrong CO2 + n O trong H2O = 0,4 + 0,3 = 0,7n O trong X = 0,7 0,6 = 0,1 ( 0,6 l s mol O trong O2 )Gi ctpt ca X l CxHyOzx : y : z = 2 : 6 : 1 C2H6O

Cu 14 : Hp th hon ton 3,584 lt kh CO2 ktc vo 2 lt dung dch Ca(OH)2 0,05M thuc kt ta X v dung dch Y . Khi khi lng ca dung dch Y so vi khi lng dungdch Ca(OH)2 s thay i nh th no ?

n CO2 = 0,16 mol , n Ca(OH)2 = 0,1 moln CO2 / n Ca(OH)2 = 1,6 C 2 phn ng .

---------- CO2 + Ca(OH)2 CaCO3 + H2O (1)x x x

---------- 2CO2 + Ca(OH)2 Ca(HCO3)2 (2) y y/2Gi s mol ca CO2 tham gia hai phn ng l x , y molx + y = 0,16 ; x + y/2 = 0,1 x = 0,04 , y = 0,12m CO2 a vo = 0,16.44 = 7,04Khi lng kt ta tch ra khi dung dch l : 0,04.100 = 4 gam Khi lng dung dch tng = m CO2 m = 3,04 gam .

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9/64

CC DNG TON HA V C

Cu 15 : Cho 4,48l kh CO2 (dktc) vao` 500ml hn hp NaOH 0,1M v Ba(OH)2 0,2M. thu c m gam ktta . Tnh mA:19,7 B:17,72 C:9,85 D:11,82

Bi gii :Gii bng phng php ion : n CO2 = 0,2 mol , n NaOH= 0,05 mol , n Ba(OH)2 = 0,1 mol Tng s mol OH- = 0,25 mol , s

mol Ba2+

= 0,2 molXt phn ng ca CO2 vi OH-

CO2 + OH- HCO3-

Ban u 0,2 0,25 Tnh theo CO2 : HCO3- = 0,2 mol , OH- d = 0,25 0,2 = 0,05Tip tc c phn ng :

HCO3- + OH- CO32- + H2O Ban u 0,2 0,05 Tnh theo HCO3- : S mol CO32- = 0,05 mol

Tip tc c phn ng: CO32- + Ba2+ BaCO3 Ban u 0,05 0,2 Tnh theo CO32- : BaCO3 = 0,05 mol m = 0,05.197 = 9,85 gam

Chn p n C.

Cu 16: Hp th hon ton 4,48 lt kh CO2 ( ktc) vo 500 ml dung dch hn hp gm NaOH 0,1M v Ba(OH)20,2M, sinh ra m gam kt ta. Gi tr ca m lA. 19,70. B. 17,73. C. 9,85. D. 11,82.

n NaOH= 0,5.0,1 = 0,05 mol,n Ba(OH)2 = 0,5.0,2 = 0,1 mol , n CO2 = 4,48/22,4 = 0,2 molNaOH Na+ + OH-0,05 0,05 mol

Ba(OH)2 Ba2+ + 2OH-0,1 0,1 0,2 moln OH- = 0,05 + 0,2 = 0,25 mol

CO2 + OH- HCO3-

Ban u 0,2 0,25CO2 ht , n OH-d : 0,25 0,2 = 0,025 mol , n HCO3- = 0,2 mol

OH- + HCO3- CO32- + H2OBan u 0,05 0,2 0,05HCO3- d : 0,2 0,05 = 0,15 mol , n CO32- : 0,05 mol

Ba2+ + CO32- BaCO3Ban u 0,1 0,05Ba2+ d : 0,1 0,05 = 0,05 mol , n BaCO3 = 0,05 molKhi lng kt ta : 197.0.05 = 9,85 gamChn C

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CC DNG TON HA V C

PHNG PHP QUY I

Cc em thn mn , gii nhng bi ton hn hp gm nhiu cht nu gii bng cch thng

thng , nh t n , bo ton e , bo ton khi lng th bi ton s rt phc tp . gip cc emgii nhanh nhng bi ton ny ti s gii thiu cho cc em phng php QUY Ihn hp v ccnguyn t .

V d nh hn hp FeS , FeS2 , S , Fe nu t s mol cc cht th s c 4 n , nhng nu cc em bitcch quy i hn hp trn v F , S th ch cn hai nguyn t Ch cn hai n V cn nhiu v d khc , hy vng cc v d di y s gip cc em nm bt c phng phpgii ton QUY I.

Cu 1: ho tan hon ton 2,32 gam hn hp gm FeO, Fe3O4 v Fe2O3 (trong s mol FeO bng s mol Fe2O3),cn dng va V lt dung dch HCl 1M. Gi tr ca V lA. 0,23. B. 0,18. C. 0,08. D. 0,16.

V s mol FeO , Fe2O3 bng nhau nn ta c thquy i chng thnh Fe3O4 . Vy hn hp trn ch gm Fe3O4 .n Fe3O4 = 2,32 : 232 = 0,01 molFe3O4+ 8HCl 2FeCl3 + FeCl2 + 4H2O0,01 0,08 moln HCl= 0,08 molVHCl= 0,08/1 = 0,08 ltChn C .

Cu 2: Cho 11,36 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung dch HNO3 long (d), thu c1,344 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 ca m l

A. 38,72. B. 35,50. C. 49,09. D. 34,36.n NO = 1,344/22,4 = 0,06 mol , gi x , y l s mol ca Fe(NO3)3 , H2O(Fe , FeO , Fe2O3 , Fe3O4) + HNO3 Fe(NO3)3 + NO + H2O

x 0,06 y11,36 2y.63 242x 0,06.30 18.y

Trong : x mol Fe(NO3)3 c : 3x mol N , Trong 0,06 mol NO c 0,06 mol NS mol N v tri : 3x + 0,06 molTheo nh lut bo ton nguyn t:n Nv tri = 3x + 0,06n HNO3 = 3x + 0,06(v 1 mol HNO3 tng ng vi 1 mol N ) .Mt khc trong y mol H2O c 2y mol HS H HNO3 cng l 2y ( Bo ton H ) n HNO3 = 2y mol3x + 0,06 = 2y 3x 2y= -0,06 (1)

p dng nh lut bo ton khi lng: 11,36 + 126y = 242x + 1,8 + 18y242x 108y = 9,56 (2)Giai (1) , (2) x = 0,16 mol , y = 0,27m Fe(NO3)3 = 242.0,16 = 38,72 gam

Chn ACch 2 :Hn hp gm Fe , FeO , Fe2O3 , Fe3O4 c to thnh t hai nguyn t O , Fe .Ga s coi hn hp trn l hn hp ca hai nguyn t O , Fe . Ta quan st s bin i s oxi ha :Fe 3eFe+3x 3xO + 2e O-2y 2yN+5 + 3e N+2

0,18 0,06Theo nh lut bo ton e : 3x 2y = 0,18

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CC DNG TON HA V C

Tng khi lng hn hp l : 56x + 16y = 11,36Gii hx = 0,16 mol , y = 0,15 molVit phn ng ca Fe vi HNO3Fe + HNO3Fe(NO3)3 + NO + H2O0,16 0,16 molkhi lng mui l : 0,16.242 = 38,72 gam

Cu 3 : Cho 9,12 gam hn hp gm FeO, Fe 2O3 , Fe3O4 tc dng vi dung dch HCl (d). Saukhi cc phn ng xy ra hon ton, c dung dch Y; c cn Y thu c 7,62 gam FeCl 2 vm gam FeCl 3 . Gi tr ca m l :A. 9,75 B . 8,75 C . 7 ,80 D . 6,50

Ta c thquy i Fe 3 O 4 = FeO + Fe 2 O 3 Lc ny hn hp cht rn ch cn FeO , Fe 2 O 3

FeO + 2HCl FeCl2 + H2Ox xFe 2O 3 + 6HCl 2FeCl3 + 3H2OY 2y

Gi x , y l s mol ca cht FeO , Fe 2 O 3 . m cht rn = 72x + 160y = 9,12 gamKhi lng mui FeCl 2 l : 127x = 7,62Giai h : x = 0,06 mol , y = 0,03 mol Khi lng mui FeCl 3 = 2.0,03.162,5 = 9,75 gamChn p n A .

Cu 4: Ho tan hon ton 2,9 gam hn hp gm kim loi M v oxit ca n vo nc, thu c 500 mldung dch cha mt cht tan c nng 0,04M v 0,224 lt kh H2 ( ktc). Kim loi M l

A. Ca B. Ba C. K D. NaDng phng php quy inguyn t : M , M2On quy i thnh M , OTa c : Mx + 16y = 2,9x = 0,5.0,04 = 0,02

M + nH2OM(OH)n + n/2H2x xMMn+ + ne; O + 2e O2-; 2H+ + 2e H2x----------nx y--2y 0,020,01Bo ton s mol e nx = 2y + 0,02Gii h ta c : n = 2 ; M = 137 Bap n B

Cu 5: Cho 61,2 gam hn hp X gm Cu v Fe3O4 tc dng vi dung dch HNO3 long, un nng vkhuy 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. Gitr ca m l

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

n NO = 0,15 mol ; kim loi d l Cu Ch to thnh mui st IIQuy ihn hp thnh 3 nguyn t Cu , Fe , OS cho nhn e :Cu 2e Cu2+ Fe 2e Fe2+ O + 2e O-2 N+5 + 3e N+2

x--2x y--2y z--2z 0,45-0,15Bo ton mol e 2x + 2y = 2z + 0,45Khi lng : 61,2 = 64x + 56y + 16z + 2,4 (Cu d )V Fe3O4 n Fe : n O = y : z = 3 : 4Giai h ta c : x = 0,375 ; y = 0,45 ; z = 0,6

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CC DNG TON HA V C

Mui khan : Cu(NO3)2 = 0,375.188 = 70,5 ; Fe(NO3)2 = 0,45.180 = 81 Tng khi lng : 151,5 gam .p n A

Cu 6 : Nung m gam bt Cu trong oxi thu c 24,8 gam hn hp cht rn X gm Cu , CuO , Cu2Ohon tan hon ton X trong H2SO4 c nng thot ra 4,48 lt kh SO2 duy nht ktc . Tnh m

Quy i Cu , CuO , Cu2O v hai nguyn t : Cu , OCu 2e Cu+2

O + 2e O-2

S+6+ 2e S+4

Gi x , y l s mol ca Cu , O 2x = 2y + 0,4 x y = 0,264x + 16y = 24,8x = 0,35 , y = 0,15m = 64.0,35 = 22,4

Cu 7 :Ha tan hon ton m gam hn hp X gm Fe , FeCl2 , FeCl3 trong H2SO4 c nng thot ra4,48 lt kh SO2 duy nht ktc v dung dch Y . Thm NH3 d vo Y thu c 32,1 gam kt ta .Tnh m

Quy i hn hp Fe , FeCl2 , FeCl3 thnh Fe , ClS cho nhn e :Fe 3e Fe+3

Cl- + e Cl-1

S+6+ 2e S+4

Gi x , y l s mol ca Fe , Cln SO2 = 0,2 mol3x = y + 0,4Fe Fe+3Fe(OH)3Kt ta l Fe(OH)3 c n = 0,3 x = 0,3 y = 0,5Do m = 0,3.56 + 0,5.35,5 = 34,55

Cu 8 : 2002 ACho 18,5 gam hn hp Z gm Fe , Fe3O4 tc dng vi 200 ml dung dch HNO3 long un nng v khuy u .Sau phn ng xy ra han ton thu c 2,24 lt kh NO duy nht ktc , dung dch Z1 v cn li 1,46 gam kimloi .Tnh nng mol/lit ca dung dch HNO3

Dng phng php quy i nguyn t :Hn hp z ch c hai nguyn t Fe , O .V Z + HNO3 cn d kim loi Fe d , vy Z1 ch c mui st II

Fe - 2e Fe+2

x 2xO + 2e O-2 y 2yN+5 + 3e N+2

0,3 0,1Theo nh lut bo ton e :2x 2y = 0,3Tng khi lng Z : 56x + 16y = 18,5 - 1,46Gii h : x = 0,27 , y = 0,12C phng trnh :Fe + HNO3 Fe(NO3)2 + NO + H2O (1)0,27 0,27 0,1

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CC DNG TON HA V C

Bo ton nguyn t N (1) s mol HNO3 = 2.0,27 + 0,1 = 0,64 Nng mol ca HNO3 : 0,64 / 0,2 = 3,2

Cu 9 : m gam bt Fe trong kh oxi thu c 7,36 gam cht rn X gm Fe , Fe2O3 , FeO , Fe3O4 . ha tan hon ton hn hp X cn va ht 120 ml dung dch H2SO4 1M to thnh 0,224 lt kh H2 iukin tiu chun . Tnh m

Li gii :

Quy i hn hp X thnh 2 nguyn t Fe , O ; gi s mol st to thnh mui Fe II , st to thnh mui Fe III ,O l a ,b ,cTa c : 56a + 56b + 16c = 7,36S cho nhn e :Fe 2e Fe2+Fe 3e Fe3+

O + 2e O-2

2H+1 + 2e H22a + 3b 2c = 0,02Tm tt s phn ng :

Fe ( a mol ) Fe ( b mol ) + H2SO4 (0,12 mol ) FeSO4 (a mol ) + Fe2(SO4)3 (b/2 mol )Bo ton nguyn t S: a + 3/2 b = 0,12Gii h 3 phng trnh : a = 0,06 , b = 0,04 , c = 0,11Tng s mol Fe : 0,06 + 0,04 = 0,1 Khi lng Fe : 0,1 . 56 = 5,6 gam

Cu 10 . Ho tan 20,8 gam hn hp bt gm FeS, FeS2, S bng dung dch HNO3 c nng d thu c 53,76lt NO2 (sn phm kh duy nht, kC. v dung dch A. Cho dung dch A tc dng vi dung dch NaOH d,lc ly ton b kt ta nung trong khng kh n khi lng khng i th khi lng cht rn thu c l:A. 16 gam B. 9 gam C. 8,2 gam D. 10,7 gam

Hn hp FeS , FeS2 , S c to ra t cc nguyn t : Fe , SS mol NO2 = 53,76/22,4 = 2,4 mol

S bin i e :Fe 3eFe+3x 3xS - 6e S+6

y 6yN+5 + 1eN+4

2,4 2,4 molTng s mol e cho = tng s mol e nhn3x + 6y = 2,4Tng khi lng hn hp ban u : 56x + 32y = 20,8Gii h ta c : x = 0,2 , y = 0,3 molTheo di s bin i cc nguyn t :

Fe

Fe

3+

Fe(OH)3

Fe2O30,2 0,1 ( Bo ton nguyn t Fe )S SO42-Na2SO4 tanKhi lng cht rn l : khi lng Fe2O3 : 0,1.160 = 16 gamChn A

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CC DNG TON HA V C

PHNG PHP GII TON IONPhng trnh ion :iu kin c Phn ng gia cc ion : Mt trong 3 iu kin sau .

+ L phn ng ca Axt v Baz

+ Sn phm sau phn ng c kt ta .+ Sn phm sau phn ng c kh

V d :

H+ + OH- H2O ( Phn ng AXT BAZ TRUNG HO )

CO32- + 2H+ CO2 + H2O ( Phn ng Axt Bazo TRUNG HO )

HCO3- + H+ CO2 + H2O ( Phn ng AXT BAZ TRUNG HO )

HCO3- + OH- CO32- + H2O ( Phn ng AXT BAZ TRUNG HO )

CO32- + Ba2+ BaCO3 ( To kt ta )

NH4+ + OH- NH3 + H2O ( To kh )Cc dng ton nn gii theo phng php ion :

+ Nhiu axit + Kim loi

+ Nhiu baz + Nhm , Al3+ , H++ Nhiu mui CO32- , HCO3- + OH-

+ Cu + HNO3 (KNO3 , NaNO3 ) + H2SO4 .Bi tp :

Phn I : Vit cc phng trnh phn ng di dng ion trong cc trng hp sau :1.Trn dung dch gm NaOH , Ba(OH)2 , KOH vi dung dch gm HCl , HNO3 .

NaOH Na+ + OH-Ba(OH)2 Ba2+ + 2OH-

KOH K+ + OH-

HCl H+ + Cl-

HNO3 H+ + NO3-

Phng trnh ion : H+ + OH- H2O

2.Trn dung dch gm NaOH , Ba(OH)2 , KOH vi dung dch gm HCl , H2SO4NaOH Na+ + OH-Ba(OH)2 Ba2+ + 2OH-

KOH K+ + OH-

HCl H+ + Cl-

H2SO4 2H+ + SO42-

Phng trnh ion :H+ + OH- H2OBa2+ + SO42- BaSO4 ( v BaSO4 l cht kt ta nn c phn ng )

3.Ho tan hn hp kim loi gm Na , Ba vo dung dch gm NaCl , Na2SO4Na + H2O NaOH + H2 Ba + H2O Ba(OH)2 + H2

NaOH Na+ + OH-

Ba(OH)2 Ba2+ + 2OH-NaCl Na+ + Cl-

Na2SO4 2Na+ + SO42-

Ba2+ + SO42- BaSO4

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CC DNG TON HA V C

4.Ho tan hn hp kim loi gm Na, Ba vo dung dch gm HCl , H2SO4HCl H+ + Cl-

H2SO4 2H+ + SO42-

Ba + H+ Ba2+ + H2Na + H+ Na+ + H2

Ba2+

+ SO42-

BaSO4

5.Ho tan hn hp kim loi gm Na , Ba vo dung dch cha (NH4)NO3Na + H2O NaOH + H2Ba + H2O Ba(OH)2 + H2

NaOH Na+ + OH-

Ba(OH)2 Ba2+ + 2OH-NH4NO3 NH4+ + NO3-

OH- + NH4+ NH3 + H2O ( c kh bay ln NH3 )

6.Ho tan hn hp kim loi gm K , Ca vo dung dch cha (NH4)2CO3

Na + H2O NaOH + H2Ca + H2O Ca(OH)2 + H2

NaOH Na+ + OH-

Ca(OH)2 Ca2+ + 2OH-(NH4)2CO3 2NH4+ + CO32-

OH- + NH4+ NH3 + H2OCa2+ + CO32- CaCO3 (Kt ta )

7.Ho tan hn hp K , Ca vo dung dch hn hp cha NH4HCO3Na + H2O NaOH + H2Ca + H2O Ca(OH)2 + H2

NaOH Na+ + OH-

Ca(OH)2 Ca2+ + 2OH-NH4HCO3 NH4+ + HCO32-

OH- + NH4+ NH3 + H2OOH- + HCO3- CO32- + H2OCa2+ + CO32- CaCO3 (Kt ta )

8.Ho tan K , Na, Al vo ncKOH K+ + OH-

NaOH Na+ + OH-

Al + OH-

+ H2O AlO2-

+ 3/2 H2

9.Ho tan hn hp gm Al , Fe vo dung dch hn hp gm HCl , H2SO4 .HCl H+ + Cl-

H2SO4 2H+ + SO42-

Al + 3H+ Al3+ + 3/2 H2Fe + 3H+ Fe2+ + H2

10.Trn NaOH , KOH vi NaHCO3 v Ca(HCO3)2KOH K+ + OH-

NaOH Na+ + OH-

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CC DNG TON HA V C

NaHCO3 Na + HCO3-Ca(HCO3)2 Ca2+ + 2HCO3-

OH- + HCO3- CO32- + H2OCO32- + Ca2+ CaCO3

11.Trn dung dch gm Na2CO3 , K2CO3 vi dung dch cha CaCl2 , MgCl2 , Ba(NO3)2Na2CO3 2Na+ + CO32-

K2CO3 2K+ + CO32-

CaCl2 Ca2+ + 2Cl-

Ba(NO3)2 Ba2+ + 2NO3-

Ca2+ + CO32- CaCO3 Ba2+ + CO32- BaCO3

Nhn xt : Bi ton s rt phc tp nu cc em khng bit a v dng ion gii chng , thng thcc em phi vit rt nhiu phng trnh S gy kh khn khi gi n .

Nu dng phng php ion ch cn 1,2 phng trnh Rt thun tin

Phn II : Bi tp

Cu 1 :Trn 200 ml dung dch hn hp NaOH 1M , Ba(OH)2 0,5M vo 300 ml dung dch hn hpHCl 0,5M , H2SO4 1M Tnh nng ca cc ion cn li sau phn ng v Khi lng kt ta to thnh

NaOH Na+ + OH-

0,2 0,2Ba(OH)2 Ba2+ + 2OH-0,1 0,1 0,2

HCl H+ + Cl-

0,15 0,15H2SO4 2H+ + SO42-0,3 0,6 0,3

Tng s mol H+ : 0,75 mol , tng s mol OH - 0,4 mol , SO42- : 0,3 mol , Ba2+ : 0,1 mol ( cc ion tham giaphn ng )Cc ion khng tham gia phn ng : Na+ : 0,2 mol , Cl- 0,15 molPhng trnh ion :

H+ + OH- H2OBan u 0,75 0,4Phn ng 0,4 0,4Kt thc 0,35 0

Ba2+ + SO42- BaSO4 Ban u 0,1 0,3Phn ng 0,1 0,1 0,1Kt thc 0 0,2 Kt thc phn ng cn li cc ion : H+ d 0,35 mol , SO42- d : 0,3 mol , Na+ : 0,2 mol , Cl- 0,15 molTng th tch dung dch sau phn ng : 200 + 300 = 500 ml = 0,5 (l )p dng cng thc tnh nng [Na+ ] = 0,2/0,5 = 0,4 M , [Cl- ] = 0,15/0,5 = 0,3 , [H+ ] = 0,35/0,5 = 0,7 ,[SO42- ] = 0,3/0,5 = 0,6

Khi lng kt ta : 233.0,1 = 23,3 gam

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17/64

CC DNG TON HA V C

Cu 2 :Trn 200 ml dung dch NaHSO4 0,075M v Ba(HSO4)2 0,15M vi V lit dung dch hn hpNaOH 1M v Ba(OH)2 1M thu c dung dch c PH = 7 . Tnh V v khi lng kt ta to thnh .

NaOH Na+ + OH-

x xBa(OH)2 Ba2+ + 2OH-

y y 2yNaHSO4 Na+ + HSO4-

0,015 0,015Ba(HSO4)2 Ba2+ + 2HSO4-

0,03 0,03 0,06

Tng s mol ca cc ion tham gia phn ng : OH- : x + 2y , HSO4- : 0,075 , Ba2+ : y + 0,03Phng trnh ion :OH- + HSO4- SO42- + H2O (1)

0,075 0,075Ba2+ + SO42- BaSO4

dung dch thu c c mi trng PH = 7 (1) Phn ng va , x + 2y = 0,075x = 0,1V ; y = 0,1V 0,3V = 0,075 V = 0,25 ltThay gi tr V Ba2+ : 0,025 + 0,03 = 0,055S mol SO42- : 0,075 mol

Ba2+ + SO42- BaSO4 Ban u 0,055 0,075Phn ng 0,055 0,055 0,055Kt thc 0 0,2 Khi lng kt ta l : 233.0,055 = 12,815 gam

thi tuyn sinh C 2007 :Cu 3 :Thm m gam K vo 300 ml dung dch cha Ba(OH) 2 0,1M v NaOH 0,1M thu c dungdch X . Cho t t 200 ml dung dch Al2(SO4)3 0,1M thu c kt ta Y . thu uc lng kt ta Yln nht th gi tr ca m lA.1,17 B.1,71 C.1,95 D.1,59

Gi x l s mol K tham gia phn ng :NaOH Na+ + OH-0,03 0,03Ba(OH)2 Ba2+ + 2OH-

0,03 0,03 0,062K + H2O 2KOH + H2 x x

KOH K+ + OH-2x xAl2(SO4)3 2Al3+ + 3SO42-0,02 0,04 0,06 Tng s mol OH- : 0,09 + x ; Al3+ : 0,04 mol

Al3+ + 3OH- Al(OH)3 (1)0,04 0,09 + x

Ba2+ + SO42- BaSO4 (2)

Phn ng (2) cho kt ta khng i , kt ta cc i th phn ng (1) va 3.0,04 = 0,09 + x x = 0,03 m = 39.0,03 = 1,17 gam

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CC DNG TON HA V C

Cu 4 :Trn dung dch Ba2+ ; OH- : 0,06 v Na+ 0,02 mol vi dung dch cha HCO3- 0,04 mol ;(CO3)2- 0,04 mol v Na+.Khi lng (g)kt ta thu c sau phn ng l ?

Dng nh lut bo ton in tch :Dung dch (1)

Ba2+

: a mol OH-

: 0,06 mol Na+

: 0,02 mol2a + 0,02 = 0,06 a = 0,02 molDung dch (2)HCO3- : 0,04 mol CO3- : 0,03 mol Na+ : b mol0,04 + 2.0,03= bb = 0,1Phn ng ha hc ca cc ion

OH- + HCO3-CO32- + H2OBan u 0,06 0,04Phn ng 0,04 0,04 0,04Kt thc 0,02 0 0,04

Tng s mol CO32- : 0,04 + 0,03 = 0,07 molBa2+ + CO32-BaCO3

Ban u 0,02 0,04Phn ng 0,02 0,02 0,02Kt thc 0 0,02 0,02Khi lng kt ta thu c l : 0,02.197 = 3,94 gam

Cu 5 :Cho m gam hn hp Mg , Al vo 250 ml dung dch X cha hn hp axit HCl 1M v axitH2SO4 0,5M thu c 5,32 lt kh H2 ktc v dung dch Y . Tnh PH ca dung dch Y ( Coi dung dchc th tch nh ban u ) .

HClH+ + Cl-0,25 0,25H2SO42H+ + SO42-0,125 0,25Tng s mol H+ : 0,5 molTheo gi thit n H2 = 0,2375 molPhng trnh ionMg + 2H+Mg2+ + H22Al + 6H+ 2Al3+ + 3H2H+ phn ng = 2.n H2 = 0,475 molH+ d : 0,5 0,475 = 0,025 mol [H+] = 0,025/0,25 = 0,1 PH = 1s : PH = 1

Cu 6 :Cho hn hp X cha Na2O , NH4Cl , NaHCO3 v BaCl2 c s mol mi cht u bng nhau .Cho hn hp X vo H2O d un nng dung dch thu c cha .

A.NaCl B.NaCl , NaOH C.NaCl , NaOH , BaCl2 D.NaCl , NaHCO3 , NH4Cl , BaCl2

Na2O + H2O 2NaOHa 2aNaOHNa+ OH-2a 2aNH4ClNH4+ + Cl-a aNaHCO3Na+ + HCO3-a a

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CC DNG TON HA V C

BaCl2Ba2+ + 2Cl-a aCc phng trnh ion :OH- + NH4+NH3 + H2Oa a OH- d a mol , NH4+ htOH- + HCO3- CO32- + H2Oa ac OH- , HCO3- u htCO32- + Ba2+BaCO3a aC CO32- , Ba2+ u htVy dung dch ch cn ion Cl- , Na+ Mui NaCl

Cu 7 :Trn 100 ml dung dch gm Ba(OH)2 0,1M v NaOH 0,1M vi 400 ml dung dch gm H2SO40,0375M v HCl 0,0125M thu c dung dch X . Tnh PH ca dung dch X .

Ba(OH)2Ba2+ + 2OH-0,01 0,01 0,02NaOHNa+ + OH-0,01 0,01H2SO42H+ + SO42-0,015 0,03 0,03HClH+ + Cl-0,005 0,005Tng s mol OH- : 0,03 mol , H+ 0,035 molMc d c phn ng : Ba2+ + SO42-BaSO4 nhng n khng nh hng n PHKhng xt n .

OH- + H+H2OBan u 0,03 0,035Phn ng 0,03 0,03Kt thc0 0,005Tng th tch ca dung dch sau phn ng : 100 + 400 = 500 ml = 0,5 lt[H+] = 0,01PH =2

Cu 8 :Cho dung dch cha a mol Ca(HCO3)2 vo dung dch cha a mol Ca(HSO4)2. Hin tng quanst c l ?A. Si bt kh B. vn cC. Si bt kh v vn c D. Vn c, sau trong sut tr li.

Ca(HCO3)2Ca2+ + 2HCO3-a a 2aCa(HSO4)2Ca2+ + 2HSO4-a a 2a

HCO3- + HSO4-CO2+ SO42- + H2Oa a a aCa2+ + SO42-CaSO4C kt ta to thnh v si bt kh

Cu 10 :Trn V1 ml dung dch gm NaOH 0,1M , v Ba(OH)2 0,2 M vi V2 ml gm H2SO4 0,1 M vHCl 0,2 . M thu oc dung dch X c gi tr PH = 13 . Tnh t s V1 : V2A.4/5 B.5/4 C.3/4 D.4/3

NaOH Na+ + OH-0,1V1 0,1V1Ba(OH)2 Ba2+ + 2OH-

0,2V1 0,4V1

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CC DNG TON HA V C

Tng s mol OH- : 0,5V1

H2SO4 2H+ + SO42-0,1V2 0,2V2HCl H+ + Cl-0,2V2 0,2V2

Tng s mol H+

: 0,4V2Phng trnh ion : H+ + OH-H2OV dung dch thu c c PH = 13 OH- d : 0,5V1 0,4V2PH = 13 [OH-] = 0,1 , Tng th tch sau phn ng : V1 + V2n OH- = 0,1V1 + 0,1V20,5V1 0,4V2 = 0,1V1 + 0,1V20,4V1 = 0,5V2V1 : V2 = 5 : 4Cu 12 .Mt dung dch cha a mol NaHCO3 v b mol Na2CO3. Khi thm (a+b) mol CaCl2 hoc (a+b) molCa(OH)2 vo dung dch th lng kt ta thu c trong hai trng hp c bng nhau khng ?A. Lng kt ta trong hai trng hp c bng nhau.B. Lng kt ta trong trng hp 2 gp i vi trng hp 1.C. Trng hp 1 c b mol kt ta, trng hp 2 c (a+b) mol kt ta.D. Trng ,hp 1 c a mol kt ta, trng hp 2 c (a+b) mol kt ta.

Trng hp 1 :Na2CO3 + CaCl2CaCO3 + 2NaCl

Ban u a a + bPhn ng a a aKt thc 0 a aKt ta thu c l : a mol

Trng hp 2Ca(OH)2Ca2+ + 2OH-a + b a + b 2(a+b)NaHCO3Na+ + HCO3-A a a

Na2CO32Na+ + CO32-B 2b b

Phng trnh ion :OH- + HCO3-CO32- + H2O

Ban u a + b aPhn ng a a aKt thc b 0 aTng s mol CO32- sau : a + bTng s mol Ca2+ : a + bPhng trnh ion khc : Ca2+ + CO32-CaCO3 Kt ta thu c l : a + b mol

Cu 1 3 :Thc hin hai th nghim sau :1.Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO3 1M thot ra V1 lt kh NO .2.Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO4 0,5M thot ra V2 lt khNOSo snh V1 v V2

Trng hp 1 :n Cu = 3,84/64 = 0,06 molHNO3H+ + NO3-0,08 0,08 0,08

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CC DNG TON HA V C

3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + H2OBan u 0,06 0,08 0,08Phn ng 0,08 0,02

Nhn thy : 0,06/3 > 0,08>8 > 0,08/2

H+

ht

Tnh theo H+

Th tch kh NO : 0,02.22,4

Trng hp 2 :n Cu = 0,06 molH2SO42H+ + SO42-0,04 0,08HNO3H+ + NO3-0,08 0,08 0,08Tng s mol ca H+ : 0,16 mol

3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + H2OBan u 0,06 0,16 0,08Phn ng 0,16 0,04

Nhn thy : 0,06/3 = 0,16>8 > 0,08/2 Cu v H+

htTnh theo H+

Th tch ca NO : 0,04.22,4Th tch V2 = 2V1

Cu 15 :Cho 0,3 mol Na vo 100 ml dung dch cha CuSO4 1M v H2SO4 2M . Hin tng quan stc l .A.C kh bay ln B.C kh bay ln v c kt ta xanhC.C kt ta D.C kh bay ln v c kt ta mu xanh sau kt ta li tan .

CuSO4Cu2+ + SO42-0,1 0,1H2SO42H+ + SO42-0,2 0,2

Khi cho Na vo dung dch Na phn ng vi H+Na + H+Na+ + H2

Ban u 0,3 0,2Phn ng 0,2 0,2 0,2Kt thc0,1 0D Na : 0,1 molNa + H2ONaOH + H20,1 0,1NaOHNa+ + OH-0,1 0,1

Cu2+ + 2OH-Cu(OH)2Ban u 0,1 0,1Phn ng 0,05 0,05 0,05 mol

Kt ta Cu(OH)2 : 0,05 mol

Cu 16 :Dung dch X c cc ion Mg2+ , Ba2+ , Ca2+ V 0,1 mol Cl- , 0,2 mol NO3- . Thm dn V ltdung dch K2CO3 1M vo dung dch X n khi c lng kt ta ln nht . Ga tr ca V l ?

Dng nh lut bo ton in tchMg2+ : a mol Ba2+ : b mol Ca2+ : c mol

Cl- : 0,1 mol NO3- : 0,2 mol

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CC DNG TON HA V C

2a + 2b + 2c = 0,1 + 0,2 = 0,3 mol

K2CO32K+ + CO32-V.1 1.V

Mg2+ + CO32-MgCO3A aBa2+ + CO32-BaCO3B bCa2+ + CO32-CaCO3C cTng s mol CO32- phn ng : a + b + c = 0,15 molV = 150 ml

Cu 17 :Cho 4,48l kh C02 (dktc) vao` 500ml hn hp NaOH 0,1M v Ba(OH)2 0,2M. thu c m gam kt ta . Tnh mA:19,7 B:17,72 C:9,85 D:11,82Bi gii :Gii bng phng php ion : n CO2 = 0,2 mol , n NaOH= 0,05 mol , n Ba(OH)2 = 0,1 mol Tng s mol OH- = 0,25

mol , s mol Ba2+

= 0,2 molXt phn ng ca CO2 vi OH-

CO2 + OH- HCO3-

Ban u 0,2 0,25 Tnh theo CO2 : HCO3- = 0,2 mol , OH- d = 0,25 0,2 = 0,05Tip tc c phn ng :

HCO3- + OH- CO32- + H2O Ban u 0,2 0,05 Tnh theo HCO3- : S mol CO32- = 0,05 mol

Tip tc c phn ng: CO32- + Ba2+ BaCO3 Ban u 0,05 0,2 Tnh theo CO32- : BaCO3 = 0,05 mol m = 0,05.197 = 9,85 gam

Chn p n C.

Cu 18 : Cn trn dung dch A cha HCl 0,1M v H2SO4 0,2M vi dung dch B cha NaOH 0,3M v KOH0,2M theo t l mol no thu c dung dch c PH l 7Gi th tch tch mi dung dch tng ng l A , BTrong A c (0,1A + 0,4A) = 0,5A (mol) H+

Trong B c (0,3B + 0,2B) = 0,5B (mol) OH-

H+ + OH- H2O (1)V PH = 7 nn l mi trng trung tnh (1) va 0,5A = 0,5B A/B = 5/5 = 1

Cu 19 : Cho 10,1 gam hn hp X gm Na , K vo 100 ml dung dch HCl 1,5M v H2SO4 0,5M , thu c dung

dch Y v 3,36 lt kh H2 ktc . Tnh khi lng cht rn thu c khi c cn dung dch .

S mol H+ trong dung dch l : 0,15 + 0,1 = 0,25 molS mol kh thu c : 3,36/22,4 = 0,15 mol2H+ + 2e H20,25 0,125

Do H+ htNa v K ( Gi chung l R ) phn ng vi H2O thu c 0,15 0,125 = 0,025 mol :2R + H2O 2ROH + H2

0,025 S mol OH- = 0,05 molKhi lng ca cht rn thu c gm : Na+ , K+ , OH- , Cl- , SO42-

m = m Na+K+ m OH- + m SO42- + m OH- = 10,1 + 0,05.17 + 0,15.35,5 + 0,05.96 = 21,075 gam

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CC DNG TON HA V C

Cu 20: Cho t t 200 ml dung dch HCl vo 100 ml dung dch cha Na2CO3 , K2CO3 , NaHCO3 1M thu c1,12 lt kh CO2 ktc v dung dch X . Cho nc vi trong d vo dung dch X thu c 20 gam kt ta . Tnhnng mol HCl

Gi s mol CO32- trong dung dch l x molHCO3- : 0,1 molH+ + CO32- HCO3- + H2O (1)

x x nHCO3- thu c sau (1) v ban u c l : 0,1 + x , s mol H+ phn ng l xV c kh nn H+ d nn :H+ + HCO3- CO2 + H2O (2) 0,05phn ng (2) :HCO3- l 0,05 mol ,H+ l 0,05 molV dung dch c phn ng vi nc vi trong d to kt ta nn HCO3- d

HCO3- + OH- CO32- + H2O (3) (s mol CO32- = S mol kt ta = 0,2 )0,2 HCO3- = 0,2 phng trnh : 0,05 + 0,2 = x + 0,1 x = 0,15 mol

Vy H+ tham gi (1) , (2) l : 015 + 0,05 = 0,2CM= 1

Cu 21 : Cho hn hp X gm 2 kim loi kim thuc hai chu k lin tip nhau vo 200 ml dung dch cha BaCl2

0,3M v Ba(HCO3)2 0,8M thu c 2,8 lt kh H2 v m gam kt ta . Xc nh m .

S mol H2 = 0,125 molGi cng thc ca kim loi : 2R + H2O 2ROH + H2S mol OH- = 0,25 molBa2+ : 0,3.0,2 + 0,8.0,2 = 0,22 molHCO3- : 0,32 molPhn ng : OH- + HCO3- CO32- + H2O

0,250,25 0,25Kt ta BaCO3 tnh theo Ba2+ : 0,22.197 = 43,34

Cu 22 : Cho 200 ml gm HNO3 0,5M v H2SO4 0,25M tc dng vi Cu d c V lit NO (ktc) c cn

dung dch sau phn ng c m gam mui khan . V v m c gi tr ln lt l :A.2,24; 12,7 B.1,12 ; 10,8C.1,12 ; 12,4 D.1,12 ; 12,7

HNO3 H+ + NO3-0,1 0,1 0,1H2SO4 2H+ + SO42-

0,05 0,1 Tng s mol ca H+ : 0,2 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O

0,2 0,1 Tnh theo H+0,075 0,2 0,05 0,05

NO3- d : 0,05 mol Khi lng mui : = Cu2+ + NO3- d + SO42- = 64.0,075 + 0,05.62 + 0,05.96 = 12,7

Th tch kh NO l : 0,05.22,4 = 11,2 lt

Cu 23 : Cho 20 gam hn hp mt kim loi M ha tr II v Al vo dung dch cha hai axit HCl v H2SO4 , bits mol H2SO4 bng 1/3 ln s mol HCl , thu c 11,2 lt kh H2 v 3,4 gam kim loi . Lc ly phn dung dchri em c cn . s gam mui khan thu c l bao nhiu .S : 57,1 gam

Xt trng hp kim loi M ha tr II c phn ng vi axit

HCl H+ + Cl-3x 3x 3x ( x l s mol ca H 2SO4 )

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CC DNG TON HA V C

H2SO4 2H+ + SO42- x 2x xTng s mol H+ l : 5xV kim loi d nn H+ ht2H+ + 2e H25x 2,5x

2,5x = 0,5 molx = 0,2 molV kim loi d : 3,4 gam nn khi lng kim loi phn ng : 20 3,4 = 16,6 gamKhi lng mui thu c = khi lng kim loi phn ng + khi lng (SO42- , Cl- )= 16,6 + 96.0,2 + 35,5.0,6 =57,1 gam

Cu 24: Dung dch X cha cc ion: Fe3+, SO42-, NH4+, Cl-. Chia dung dch X thnh hai phn bngnhau:- Phn mt tc dng vi lng d dung dch NaOH, un nng thu c 0,672 lt kh ( ktc) v 1,07gam kt ta;- Phn hai tc dng vi lng d dung dch BaCl2, thu c 4,66 gam kt ta.Tng khi lng cc mui khan thu c khi c cn dung dch X l (qu trnh c cn ch c nc bayhi)A. 3,73 gam. B. 7,04 gam. C. 7,46 gam. D. 3,52 gam.

NaOHNa+ + OH-

BaCl2Ba2+ + 2Cl-

n kh = 0,672/22,4 = 0,03 molOH- + NH4+ NH3 + H2O (1)

0,03 0,03Fe3+ + 3OH- Fe(OH)3 (2)0,01 0,01ssKt ta l : Fe(OH)3n kt ta : 1,07/107 = 0,01 molBa2+ + SO42- BaSO4 (3)

0,02 0,02 moln BaSO4 = 0,02 mol ,

T (1) , (2) , (3) n NH4+ = 0,03 mol , n Fe3+ = 0,01 mol , n SO42- = 0,02 molGi s mol ca Cl- l xp dng nh lut bo ton in tch : 0,03.1 + 0,01.3 = 0,02.2 + x x = 0,02 molKhi lng cht rn khan thu c = Tng khi lng ca cc ionm mui = 0,03.18 + 0,01.56 + 0,02.96 + 0,02.35,5 = 3,73 gam Chn A

Cu 25: Ho tan ht 7,74 gam hn hp bt Mg, Al bng 500 ml dung dch hn hp HCl 1M vH2SO4 0,28M thu c dung dch X v 8,736 lt kh H2 ( ktc). C cn dung dch X thu c lngmui khan lA. 38,93 gam. B. 77,86 gam. C. 103,85 gam. D. 25,95 gam.

Dng phng php bo ton electron + ionn HCl= 0,5.1 = 0,5 moln H2SO4 = 0,28.0,5 = 0,14 moln H2 = 8,736/22,4 = 0,39 mol

HCl H+ + Cl-0,5 0,5

H2SO4 2H+ + SO42-0,14 0,28Tng s mol ca H+ l : 0,5 + 0,28 = 0,78 molS nhn e :2H+ + 2e H2

0,39

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CC DNG TON HA V C

S mol H+ nhn l : 0,39.2 = 0,78H+ phn ng va dung dch khng cn H+

Tm ttDo 7,74 gam bt Mg , Al phn ng ht nn khi lng mui khan thu c bng :m kim loi + m Cl- + m SO42- = 7,74 + 17,75 + 13,44 = 38,93 gam

Chn ACu 26 : Trn 100 ml dung dch hn hp gm H 2SO4 0,05M v HCl 0,1M vi 100 ml dung dchhn hp gm NaOH 0,2M v Ba(OH)2 0,1M thu c dung dch X. Dung dch X c pH l

A. 1,2 B. 1,0 C. 12,8 D. 13,0Tng s mol H+ : nH+ = 0,1(2CM(H2SO4) + CM(HCl) )= 0,02; nNaOH= 0,1[CM(NaOH) + 2CM(Ba(OH)2)] = 0,04.

H+ + OH-H2O Ban u 0,02 0,04 d 0,02 mol OH-. [OH-] = 0,02/(0,1+0,1) = 0,1 = 10-1. [H+] = 10-13. pH = 13p n D

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CC DNG TON HA V C

BI TON LIN QUAN N PH

Cc cng thc cn nh :p H = -lg[H+][H+].[OH-] = 10-14

p H < 7 axitp H > 7 Bazo ( phi tnh theo s mol OH-

Cu 1 : Cho hng s axit ca CH3COOH l = 1,8.10-5 . PH ca dung dch CH3COOH 0,4Ml .A.0,4 B.2,59 C.5,14 D.3,64

CH3COOH CH3COO- + H+B 0,4 Ply 0,4. 0,4. 0,4.

[H+] = 0,4.1,8.10-5 = 0,72. 10-5 p H = 5,14

Cu 2 : Pha thm 40 cm3 nc vo 10 cm3 dung dch HCl c pH = 2 c mt dung dchmi c PH bng bao nhiu .A.2,5 B.2,7 C.5,2 D.3,5

Dung dch HCl c p H = 2 [H+

] = 0,01 mol n H+ = 0,01.0,01 = 0,0001Thm 40 cm3 vo c th tch : 50 ml [H+] = 0,0001/0,05 = 2.10-3 pH = 2,7

Cu 3 : Cho 150 ml dung dch HCl 0,2M tc dng vi 50 ml dung dch NaOH 0,56M . Dungdch sau phn ng c PH bng bao nhiu .A.2,0 B.4,1 C.4,9 D.1,4

HCl H+ + Cl- NaOH Na+ + OH-0,03 0,03 0,03 0,028 0,028

H

+

+ OH

-

H2O0,03 0,028 H+ d = 0,002 mol [H+] = 0,002/0,2 = 0,01 p H = 2

Cu 4 : Trn 500 ml dung dch HCl 0,02M tc dng vi 500 ml dung dch NaOH 0,018Mdung dch sau phn ng c PH l bao nhiu .

Gii tng t cu 3 .

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CC DNG TON HA V C

Cu 5 : Cn thm bao nhiu th tch nc V2 so vi th tch ban u V1 pha long dungdch HCl c PH = 3 thnh dung dch c PH = 4 .A.V2 = 9V1 B.V1 = 1/3 V2 C.V1 = V2 D.V1 = 3V2

Ban u th tch V1 : PH = 3 [H+] = 0,001 mol n H+ = 0,001V1

Thm vo th tch V2 Tng th tch V1 + V2 , PH = 4 [H+] = 0,0001 n H+ =0,0001(V1+ V2)V s mol H+ khng i 0,001V1 = 0,0001(V1+V2) 10V1 = V1 + V2 V2 = 9V1Cu 6 : Trn 100 ml dung dch Ba(OH)2 0,5M vi 100 ml dung dch KOH 0,5M c dungdch A . Nng mol/l ca ion trong dung dch .A.0,65M B.0,55M C.0,75M D.0,85M

n Ba(OH)2 = 0,05 mol ; n KOH = 0,05 molBa(OH)2 Ba2+ + 2OH- KOH K+ + OH-

0,05 ------------------0,1 0,05----------0,05 Tng s mol OH- = 0,15 mol [OH-] = 0,15/0,2 = 0,75 M

Cu 7 : C dung dch H2SO4 vi PH = 1 . Khi rt t t 50 ml dung dch KOH 0,1M vo 50ml dung dch trn . Nng mol./l ca dung dch thu c lA.0,005M B.0,003M C.0,25M D.0,025M

Dung dch H2SO4 c pH = 1 [H+] = 0,1 M 50 ml dung dch trn c n H+= 0,005 mol ,n SO42- = 0,0025 mol

KOH K+ + OH-

0,005 --0,0005------0,005 molPhn ng : H+ + OH- H2O Ban u 0,005 0,005 C H+ , v OH- u ht . Dung dch thu c ch c mui K2SO4 : 0,0025 mol , tng thtch V = 50 + 50 = 100 ml CM = 0,0025/0,1 = 0,025 mol

Cu 8 :Trn V1 ml dung dch gm NaOH 0,1M , v Ba(OH)2 0,2 M vi V2 ml gm H2SO40,1 M v HCl 0,2 M thu oc dung dch X c gi tr PH = 13 . Tnh t s V1 : V2A.4/5 B.5/4 C.3/4 D.4/3

NaOH Na+ + OH- Ba(OH)2 Ba2+ + 2OH-

0,1V1 ------------0,1V1 0,2V1 ---------------0,4V1Tng s mol OH- = 0,5V1

H2SO4 2H+ + SO42- HCl H+ + Cl-

0,1V2------0,2V2 0,2V2------0,2V2 Tng s mol H+ = 0,4V2

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CC DNG TON HA V C

Phng trnh : H+ + OH- H2O Ban u 0,4V2 0,5V1

pH = 13 OH- d = 0,5V1 0,4V2pH = 13 [H+] = 10-13 [OH-] = 10-1 n OH- = 0,1.(V1 + V2)

0,5V1 0,4V2 = 0,1V1 + 0,1V2 V1/V2 = 5/4

Bi 9 :Cho m gam hn hp Mg , Al vo 250 ml dung dch X cha hn hp axit HCl 1M vaxit H2SO4 0,5M thu c 5,32 lt kh H2 ktc v dung dch Y . Tnh PH ca dung dch Y( Coi dung dch c th tch nh ban u )A. 1. B. 6. C. 7. D. 2.

n HCl = 0,25.1 = 0,25 mol , n H2SO4 = 0,5.0,25 = 0,125 mol , n H2 = 5,32/22,4 = 0,2375 molHCl H+ + Cl-0,25 0,25H2SO4 2H+ + SO42-

0,125 0,25Tng s mol H+ = 0,25 + 0,25 = 0,5 M

Phng trnh phn ng :Al + 6H+ Al3+ + 3H2

Mg + 2H+ Mg2+ + H2S mol H+ phn ng l : 2.n H2 = 2.0,2375 = 0,475 molH+ d = 0,5 0,475 = 0,025 mol[H+] = 0,025/0,25 = 0,1 PH = 1( Do th tch dung dch khng i ) Chn A

Bi 10 :Trn 100 ml dung dch gm Ba(OH)2 0,1M v NaOH 0,1M vi 400 ml dung dchgm H2SO4 0,0375 M v HCl 0,0125 M thu c dung dch X . Tnh PH ca dung dch X .

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

n Ba(OH)2 = 0,1.0,1 = 0,01 mol , n NaOH= 0,1.0,1 = 0,01 mol , n H2SO4 = 0,4.0,0375 = 0,015 mol , n HCl =0,0125.0,4 = 0,025 molBa(OH)2 Ba2+ + 2OH-0,01 0,02NaOH Na+ + OH-0,01 0,01Tng s mol OH- : 0,02 + 0,01 = 0,03 mol

H2SO4 2H+ + SO42-0,015 0,03HCl H+ + Cl-0,005 0,005Tng s mol ca H+ : 0,035 molPhn ng :

H+ + OH- H2O Ban u 0,035 0,03 Phn ng 0,03 0,03

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CC DNG TON HA V C

Kt thc 0,005 0Sau phn ng d 0,005 mol H+ , Tng th tch l 0,5 lit[H+] = 0,005/0,05 = 0,01 PH = -Lg[H+] = -lg0,01 = 2 Chn B

Cu 11 : X l dung dch H2SO4 0,02 M . Y l dung dch NaOH 0,035 M . Khi trn ln dungdch X vi dung dch Y ta thu c dung dch Z c th tch bng tng th tch hai dung dchmang trn v c PH = 2 . Coi H2SO4 in li hon ton hai nc . Hy tnh t l th tch giadung dch X v Y .

PH = 2 Dung dch sau phn ng d [H+] = 0,01 n H+ = 0,01(V1 + V2)Gi th tch cc dung dch l : V1 ; V2

H2SO4 2H+ + SO42- NaOH Na+ + OH-

0,02V1--0,04V1 0,035V2---------0,035V2

Phn ng : 2H+ + OH- H2O Ban u : 0,04V1 0,035V2 D H+ = 0,04V1 0,035V2 = 0,01V1 + 0,01V2 0,03V1 = 0,045V2 V1 = 1,5V2

Cu 12 : Thm t t 400 g dung dch H2SO4 49% vo nc v iu chnh lng nc c ng 2 lit dung dch A . Coi H2SO4 in hon ton c hai nc .1.Tnh nng mol ca ion H+ trong dung dch A .2.Tnh th tch dung dch NaOH 1,8M thm vo 0,5 lit dung dch A thu c .A.dung dch c PH = 1 .B.Dung dch thu c c PH = 13 .

n H2SO4 = 400.0,49/98 = 2 mol , V = 2 ltH2SO4 2H+ + SO42-2 ----------4 mol [H+] = 4/2 = 2 M2. dung dch thu c c pH = 1 H+ dn NaOH = V1.1,8 mol n OH- = 1,8.V1n H+ = 0,5.2 = 1 mol Phn ng : H+ + OH H2O Ban u 1 1,8V1PH = 1 [H+] = 0,1 n H+ d = 0,1(0,5 + V1)Theo phng trnh : n H+ d = 1 1,8V1 0,05 + 0,1V1 = 1 1,8V1 1,9V1 = 0,95 V1 = 0,5 lt

dung dch thu c c pH = 13 OH- d , [H+] = 10-13 [OH-] = 0,1 n OH- d = 0,1.(V1 + 0,5)n H+ = 0,5.2 = 1 moln NaOH = V1.1,8 mol n OH- = 1,8.V1 Phn ng : H+ + OH H2O Ban u 1 1,8V1 OH- d = 1,8V1 1 = 0,1(V1 + 0,5) 1,7V1 = 1,05 V1 =

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CC DNG TON HA V C

Cu 13 : Trn 250 ml dung dch hn hp HCl 0,08 mol/l v H 2SO4 0,01 mol /l vi 250 mldung dch Ba(OH)2 c nng x mol /l thu c m gam kt ta v 500 ml dung dch c p H= 12 . Hy tnh m v x . Coi Ba(OH)2 in li hon ton c hai nc .

Cu 14 :Trn 300 ml dung dch NaOH 0,1 mol/lit v Ba(OH)2 0,025 mol/l vi 200 ml dung

dch H2SO4 nng x mol /lit Thu c m gam kt ta v 500 ml dung dch c PH = 2 . Hytnh m v x . Coi H2SO4 in li hon ton c hai nc .

Cu 15 : Trn ln V ml dung dch NaOH 0,01M vi V ml dung dch HCl 0,03 M c 2V ml dung dch Y. Dungdch Y c pH lA. 4. B. 3. C. 2. D. 1.

n NaOH= V.0,01 = 0,01.V mol , n HCl = V.0,03 = 0,03.V molNaOH Na+ + OH-0 ,01V 0,01V 0,01V

HCl H+ + Cl-0,03V 0,03V 0,03V

H+ + OH- H2OBan u 0,01V 0,03VOH- d : 0,03V 0,01V = 0,02 V , Tng th tch l : 2V[H+] = 0,02V/2V = 0,01PH = - lg0,01 = 2Chn C .

Cu 16 : Trn 100 ml dung dch c pH = 1 gm HCl v HNO 3 vi 100 ml dung dch NaOHnng a (mol / l) thu c 200 ml dung d ch c pH = 12 . Gi t r ca a l (b i t t rong midung dch [H+][OH -] = 10-14)A . 0,15 B . 0,30 C . 0 ,03 D. 0,12

n HCl = x mol , n HNO3 = y mol

HCl H+ + Cl-x xHNO3 H+ + NO 3 -

y y Tng s mol H+ : x + yV PH = 1 [H+ ] = 0,1 n H + = 0,1. 0 ,1 = 0,01 mol x + y = 0,01 mol N NaOH = 0,1.aNaOH Na+ + OH-0,1.a 0,1.aC phn ng :

H+ + OH- H2 O (1) Ban u 0,01 mol 0,1.aV PH =12 OH- d,PH =12 [H+ ] = 10 - 1 2 [OH- ] = 10 - 2 n O H - = V.[OH- ] = 10 - 2 .0 ,2 = 0 ,002

molTheo (1) OH- d : 0,1.a 0,01 = 0,002 a = 0,12 MChn p n D

Cu 17 : Trn 100 ml dung dch hn hp gm H2SO4 0,05M v HCl 0,1M vi 100 ml dung dch hn hpgm NaOH 0,2M v Ba(OH)2 0,1M thu c dung dch X. Dung dch X c pH l

A. 1,2 B. 1,0 C. 12,8 D. 13,0Tng s mol H+ : nH+ = 0,1(2CM(H2SO4) + CM(HCl) )= 0,02; nNaOH= 0,1[CM(NaOH) + 2CM(Ba(OH)2)] = 0,04.

H+ + OH-H2O Ban u 0,02 0,04 d 0,02 mol OH-. [OH-] = 0,02/(0,1+0,1) = 0,1 = 10-1. [H+] = 10-13 pH = 13

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CC DNG TON HA V C

p n D

BI TP V HNO3 V MUI NITRATCu 1:Nung nng hn hp 27,3 gam hn hp NaNO3 , Cu(NO3)2 . Hn hp kh thot ra c dnvo 89,2 ml nc th cn d 1,12 lt kh ktc khng b hp th . Tnh khi lng ca mi mui tronghn hp u

NaNO3 NaNO2 + O2 (1)a a/2Cu(NO3)2 CuO + 2NO2 + O2 (2)

b 2b b

2NO2 + O2 + H2O 2HNO3 (3)

Kh NO2 , O2 phn ng vi nhau theo t l ca phng trnh (2) Kh thot ra l O2 = s mol O2 phn ng (1) a = 0,05 mol a = 0,1 mol Khi lng mui = 85.0,1 + 188b = 27,3 b = 0,1 mol

Khi lng NaNO3 : 8,5 gam , Cu(NO3)2 : 18,8 gam

Cu 2 : Cho bt Cu d vo V1 lt dung dch HNO3 4M v vo V2 lt dung dch HNO3 3M v H2SO41M . NO l kh duy nht thot ra . Xc nh mi quan h gia V1 v V2 bit rng kh thot ra hai thnghim l nh nhau .

Th nghim (1) :HNO3 H+ + NO3-4V1 4V1 4V13Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O

4V1 4V1 Tnh theo H+ V1

Th nghim (2) :HNO3 H+ + NO3-3V2 3V2 3V2

H2SO4 2H+ + SO42-V2 2V2

Tng s mol ca H+ : 5V2 , S mol ca NO3- : 3V23Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O5V2 3V2 Tnh theo H+ 1,25V2

V th tch kh NO c hai trng hp l nh nhau V1 = 1,25V2

Cu 3 : Thc hin hai th nghim:1) Cho 3,84 gam Cu phn ng vi 80 ml dung dch HNO3 1M thot ra V1 lt NO.2) Cho 3,84 gam Cu phn ng vi 80 ml dung dch cha HNO3 1M v H2SO4 0,5 M thot ra V2 ltNO.Bit NO l sn phm kh duy nht, cc th tch kh o cng iu kin. Quan h gia V1 v V2 l(cho Cu = 64)A. V2 = V1. B. V2 = 2V1. C. V2 = 2,5V1. D. V2 = 1,5V1.

Th nghim (1) :HNO3 H+ + NO3-0,08 0,08 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O0,06 0,08 0,08 Tnh theo H+ 0,02 molTh nghim (2) :HNO3 H+ + NO3-0,08 0,08 0,08

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CC DNG TON HA V C

H2SO4 2H+ + SO42-0,04 0,08 Tng s mol ca H+ : 0,16 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O0,06 0,16 0,08 0,04 Tnh theo H+ v Cu S mol k trong trng hp ny l : 0,04

V2 = 2V1

Cu 4 : Cho 200 ml gm HNO3 0,5M v H2SO4 0,25M tc dng vi Cu d c V lit NO (ktc)

c cn dung dch sau phn ng c m gam mui khan . V v m c gi tr ln lt l :A.2,24; 12,7 B.1,12 ; 10,8 C.1,12 ; 12,4 D.1,12 ; 12,7

HNO3 H+ + NO3-0,1 0,1 0,1H2SO4 2H+ + SO42-0,05 0,1 Tng s mol ca H+ : 0,2 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O

0,2 0,1 Tnh theo H+

NO3- d : 0,05 mol Khi lng mui : = Cu2+ + NO3- d + SO42- = 64.0,075 + 0,05.62 + 0,05.96 = 12,7Th tch kh NO l : 0,05.22,4 = 11,2 lt

Cu 5 : Cho 0,96 gam Cu vo 100ml dung dch cha ng thi KNO3 0,08M v H2SO4 0,2M sinh

ra V (lit ) mt cht kh c t khi so vi H2 l 15 v dung dch A . V c gi tr l :

A. 0,1792 lit B. 0,3584 lit C. 0,448 lit D. 0,336 lit

Kh c t khi so vi H2 l 15 NOKNO3 K+ + NO3-0,008 0,008

H2SO4 2H+ + SO42-0,02 0,02

Tng s mol ca H+ : 0,2 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O0,015 0,02 0,008

Tnh theo NO3- --------------------------- 0,008 mol

Th tch kh NO : 0,008.22,4 = 0,1792 lt

Cu 6 : em nung mt khi lng Cu(NO3)2 sau mt thi gian dng li , lm ngui ri em cn thykhi lng gim 0,54 gam . Vy 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

Cu(NO3)2 CuO + 2NO2 + O2a a 2a a

Cht rn c c Cu(NO3)2 d

Theo nh lut bo ton khi lng : khi lng cht rn gim chnh l ca NO2 v O2 thot ra 0,54 = 92a + 16a a = 0,005 mol Khi lng cht Cu(NO3)2 b nhit phn : 188.0,05 = 0,94 gam

Cu 7 : Hon tan hon ton 19,2 gam Cu trong dung dch HNO3 long nng d , kh sinh ra em trnvi O2 d thu c X , Hp th X vo nc chuyn ht NO2 thnh HNO3 . Tnh s mol O2 thamgia phn ng .S : 0,15 mol

Nhn thy : Cu - 2e Cu2+0,3 0,6 mol

NO3- NO NO2 NO3- Nh vy N khng thay i s oxi ha trong c qu trnh

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CC DNG TON HA V C

O2 4e 2O-2

0,6 mol s mol O2 phn ng : 0,6/4 = 0,15 VO2 = 3,36 lt

Cu 8 : Cho hn hp gm 1,12 gam Fe v 1,92 gam Cu vo 400 ml dung dch cha hn hp gmH2SO4 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 thuc l ln nht. Gi tr ti thiu ca V l

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

Cu 9 : Cho 2,16 gam Mg tc dng vi dung dch HNO3 (d). Sau khi phn ng xy rahon ton thu c 0,896 l t kh NO ( ktc) v dung dch X. Khi lng mui khanthu c khi lm bay hi dung dch X l :A . 8,88 gam B. 13,92 gamC . 6,52 gam D . 13,32 gam

3Mg + 8HNO 3 3Mg(NO 3) 2 + 2NO + 4H2 O (1)0,06 0,06 mol 0,04 mol 4Mg + 10HNO 3 4Mg(NO 3) 2 + NH4NO 3 + 3H2O (2)0,03 0 ,03 0,0075 mol n N O = 0 ,896/22 ,4 = 0 ,04 mol , n M g = 0,09 mol , t phn ng (1) n M g(1) = 3/2nN O = 3/2 .0 ,04= 0,06 mol . n M g(2) = 0,09 0,06 = 0,03 mol ,T p (1) n M g ( N O 3 ) 2 = 0,06 mol T p (2) n M g ( N O 3 ) 2 = 0,03 mol , n N H 4 N O 3 = 0,0075 mol Khi lng mui khan khi lm bay hi dung dch l : 148.0,09 + 0,0075.80 = 13,92 g Chn p n B

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CC DNG TON HA V C

Cu 10 : Th t ch dung dch HNO3 1M (long) t nht cn dng ho tan hon tonmt hn hp gm 0,15 mol Fe v 0,15 mol Cu l (bit phn ng to cht kh duy nhtl NO)A . 1,0 l t B . 0,6 l t C. 0,8 l t D . 1,2 l t

Nhn xt : Lng HNO3 ti thiu cn dung khi Fe Fe 2 + , Cu Cu 2 + S cho nhn e :Fe 2e Fe 2 + Cu 2e Cu 2 + N+ 5 + 3e N+ 2

0,15 0,3 0,15 0,3 3x x Theo nh lut bo ton mol e : 0,3 + 0,3 = 3x x = 0,2 molFe , Cu + HNO 3 Fe(NO3) 2 + Cu(NO3) 2 + NO + H2O0,15 0,15 0,15 0,15 0,2 Bo ton nguyn t N : S mol HNO 3 = 0,3 + 0,3 + 0,2 = 0,8 mol Chn C .Cu 11 : Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO 3)2 0,2M v H2SO40,25M. Sau khi cc phn ng xy ra hon ton, thu c 0,6m gam hn hp bt kim loi v Vlt 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.

n Cu(NO3)2 = 0,16 n Cu2+ = 0,16 , n NO3- = 0,32 moln H2SO4 = 0,2 n H+ = 0,4V thu c hn hp kim loi nn Ch c mui Fe2+ to thnh3Fe + 2NO3- + 8H+ 3Fe2+ + 2NO + 4H2O (1)0,15---------- 0,4---------------0,1Fe + Cu2+ Fe2+ + Cu (2)0,16-----0,16-----------0,16 Khi lng ng trong 0,6m gam hn hp sau phn ng l : 64.0,16 molBo ton st : m = 0,15.56p(1) + 0,16.56p(2) + (0,6m 0,16.64 ) d m = 17,8Mt khc VNO = 0,1.22,4 = 2,24

Chn p n B

Cu 17: Hn hp A gm 16,8 gam Fe ; 6,4 gam Cu v 2,7 gam Al . Cho A tc dng vi dung dch HNO3 chthot ra kh N2 duy nht , trong dung dch thu c khng c mui NH4NO3 . Th tch dung dch HNO3 2M tithiu cn dng ha tan hon ton hn hp A l .

A.660 ml B.720 ml C.780 ml D.840 ml

10Al + 36HNO3 10Al(NO3)3 + 3N2 + 18H2O0,1 0,3610Fe + 36HNO3 10Fe(NO3)3 + 3N2 + 18H2O x 3,6x x ( x l s mol Fe phn ng vi HNO3 )Fe + 2Fe(NO3)3 3Fe(NO3)2 y 2y ( y l s mol Fe phn ng vi mui FeIII )

Cu + 2Fe(NO3)3 2Fe(NO3)2 + Cu(NO3)20,1 0,2 lng HNO3 ti thiu th : x + y = 0,3 ; 2y + 0,2 = x x = 4/15 ; y = 1/30Tng s mol : HNO3 phn ng l : 0,36 + 3,6.4/15 = 1,32 molV = 1,32/2 = 0,66 lit = 660 ml

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CC DNG TON HA V C

BI TP V NHM

L thuyt :

AlCl3 + NaOH Al(OH)3 + 3NaClAl2(SO4)3 + 6KOH 2Al(OH)3 + 3K2SO4Al(NO3)3 + 3KOH Al(OH)3 + 3KNO32AlCl3 + 3Ca(OH)2 2Al(H)3 + 3CaCl2Al2(SO4)3 + 3Ba(OH)2 2Al(OH)3 + 3BaSO4

Phn ng nhit nhm : Al + FexOy Al2O3 + Fe

Sau phn ng nhit nhm :Ga thit cho phn ng xy ra hon ton Th cht rn chc chn c Al2O3 , Fe v c th Al hoc

FexOy d .Ga thit khng ni n hon ton , hoc bt tnh hiu xut th cc em nn nh n trng hp chtrn sau phn ng c c 4 cht Al , FexOy , Al2O3 , Fe .

Phng trnh ion :Al3+ + OH- Al(OH)3 (1)Al(OH)3 + OH- AlO2- + H2O (2)

Khi cho kim vo dung dch mui Al3+ , cc em nh phi xt n c hai phn ng (1) , (2) , ty iukin bi ton cho .Nu bi ton cho kt ta thu c m gam cc em ng nhm ln l ch c phn ng (1)M n c hai trng hp : TH1 c (1) ; TH2 c c (1) v (2) [trng hp ny s mol kt ta thu c

= (1) (2) ]Al + OH- + H2O AlO2-+ 3/2 H2AlO2- + H+ + H2O Al(OH)3

V d 1 : Cho t t 100 ml dung dch NaOH 7M vo 200 ml dung dch Al(NO3)3 1M Tnh khi lngca cc ion thu c sau phn ng .

n Al= 0,5 ; n Al(NO3)3 = 0,2

3NaOH + Al(NO3)3 NaNO3 + Al(OH)3 Ban u 0,7 0,2 Phn ng 0,6 0,2 0,2

Kt thc 0,1 0 0,2 C tip phn ng :

NaOH + Al(OH)3 NaAlO2 + H2O Ban u 0,1 0,2 Phn ng 0,1 0,1Kt thc 0,1

Vy kt ta Al(OH)3 thu c l 0,1 mol Khi lng 7,8 gam

Luyn tp :

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CC DNG TON HA V C

Cu 1 : Thm m gam K vo 300 ml dung dch cha Ba(OH)2 0,1M v NaOH 0,1M thu c dungdch X . Cho t t 200 ml dung dch Al2(SO4)3 0,1M thu c kt ta Y . thu uc lng kt ta Yln nht th gi tr ca m l .A.1,17 B.1,71 C.1,95 D.1,59

Dng phng php ion :Gi x l s mol K cn a vo :n Ba(OH)2 = 0,3.0,1 = 0,03 mol ; n NaOH = 0,3.0,1 = 0,03 mol , n Al2(SO4)3 = 0,02 mol

K + H2O KOH + H2x xCc phng trnh in ly :

KOH K+ + OH-

x xBa(OH)2 Ba2+ + 2OH-

0,03 0,06 molNaOH Na+ + OH-

0,03 0,03 molTng s mol OH- d : x + 0,06 + 0,03 = x + 0,09 mol

Al2(SO4)3 2Al3+ + 3SO42-0,02 0,04 0,06Cc phng trnh ion :

Ba2+ + SO42- BaSO4 (1)Al3+ + 3OH- Al(OH)3 (2)0,04 x + 0,09 c kt ta ln nht th phn ng (2) phi xy ra va 0,04/1 = ( x + 0,09 )/3 x = 0,03 mol Khi lng ca K l : 0,03.39 = 1,17 gamChn p s B

Cu 2 :Cho 200 ml dung dch AlCl3 1,5M tc dng vi V lt dung dch NaOH 0,1M . Lung kt tathu c l 15,6 gam . Tnh gi tr ln nht ca V ?

AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)Al(OH)3 + NaOH NaAlO2 + H2O (2)

n AlC3 = 0,3Kt ta thu c l 15,6 gam ( n = 0,2 mol ) , c hai trng hp xy ra :TH1 : AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1) n NaOH = 0,2.3 = 0,6 mol V = 0,6/0,1 = 6 lt

TH2: C c hai phn ng :

AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)0,3 0,9 0,3

Al(OH)3 + NaOH NaAlO2 + H2O (2)

x x x S mol kt ta thu c : 0,3 x = 0,2 x = 0,1 mol Tng s mol NaOH phn ng : 0,9 + 0,1 = 1 mol V NaOH dng = 1/0,1 = 10 ltVy gi tr ln nht ca V l 10 ltNhn xt : gi tr V ln nht khi xy ra hai phn ng .

Cu 3 :Th tch dung dch NaOH 2M l bao nhiu khi cho tc dng vi 200 ml dung dch X ( HCl1M AlCl3 0,5M ) th thu uc kt ta ln nht ?s : 250 ml

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37/64

CC DNG TON HA V C

n HCl= 0,2 mol ; n AlCl3 = 0,1 molHCl + NaOH NaCl + H2O (1)0,2 0,2

AlCl3 + 3NaOH Al(OH)3 + 3NaCl (2)0,1 0,3Phn ng (1) : Xy ra trc lng kt ta ln nht th (2) va Tng s mol NaOH phn ng : 0,5 mol V = 0,25 lt

Cu 4 :Cho V lt dung dch hn hp 2 mui MgCl2 1M v AlCl3 1M tc dng vi 1 lt NaOH 0,5Mth thu c kt ta ln nht . Tnh V.S : V = 100 m l

MgCl2 +2NaOH Mg(OH)2 + 2NaCl (1)V.1 2V.1

AlCl3 + 3NaOH Al(OH)3 + 3NaCl (2)V.1 3V.1 kt ta ln nht th (1) , (2) va Tng s mol NaOH phn ng : 5V = 0,5 V = 0,1 lt

Cu 5 : Cho V lt dung dch hn hp 2 mui MgCl2 1M v AlCl3 1M tc dng vi 1.2 lt NaOH 0,5Mthu c 9.7 gam kt ta . Tnh V ln nht .S : 100 ml .

MgCl2 +2NaOH Mg(OH)2 + 2NaCl (1)V.1 2V.1

AlCl3 + 3NaOH Al(OH)3 + 3NaCl (2)V.1 3V.1 V.1

Al(OH)3 + NaOH NaAlO2 + H2O (3)V.1 V.1 thu c mt lng kt ta 9,7 c hai kh nng xy ra : TH1 C (1) , (2) hoc TH2 c (1) , (2) , (3) NaOH ln nht TH2 : Tng s mol NaOH phn ng : 6V = 0,6 V =0,1 lt = 100 ml

Cu 6: Cho V lt dung dch NaOH 0,2M vo dung dch cha 0,15 mol AlCl3 thu c 9,36 gam kt ta . TnhV .


Kt ta thu c l 8,36 gam ( n = 0,12 mol ) , c hai trng hp xy ra :TH1 : AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1) n NaOH = 0,12.3 = 0,36 mol V = 0,36/0,2 = 1,8 lt

TH2 : C c hai phn ng :

AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)0,15 0,45 0,15

Al(OH)3 + NaOH NaAlO2 + H2O (2) x x x

S mol kt ta thu c : 0,15 x = 0,12 x = 0,03 mol Tng s mol NaOH phn ng : 0,45 + 0,03 = 0,48 mol VNaOHdng = 0,48/0,2 = 2,4 lt

S : 1,8 lt v 2,4 ltVi bi ton nh th ny : Khi hi n NaOH th c 2 p an , nu hi n AlCl3 th c 1 p n

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CC DNG TON HA V C

Cu 7 : Cho 1 lt dung dch HCl vo dung dch cha 0,2 mol NaAlO2 lc ,nung kt ta n khi lng khngi c 7,65 gam cht rn . Tnh nng ca dung dch HCl

n cht rn (Al2O3) = 0,075 mol

C th c hai phn ng :

HCl + NaAlO2 + H2O Al(OH)3 + NaCl .0,2 0,2 0,23HCl + Al(OH)3 AlCl3 + H2O3x x

V kt ta thu c c 0,075 mol Nn chc chn phi c hai phn ng .Al(OH)3 Al2O3 + H2O0,2 x 0,2 - x 0,2 x = 0,075 x = 0,125 molTng s mol HCl phn ng : 0,2 + 0,575 = 0,775 mol CM= 0,775 mol

Cu 8 : Hn hp X gm Na v Al. Cho m gam X vo lng d nc th thot ra 1 lt kh .Nu cng cho m gamX vo dung dch NaOH d th c 1,75 lt kh .Tnh thnh phn phn trm khi lng cacc cht trong hn hp X (bit cc kh o iu kin tiu chun ).

TN1 nhiu em ngh Al khng tham gia phn ng nhng thc t n c phn ng vi NaOH va to cNa + H2O NaOH + 1/2H2 x x 1/2xAl + NaOH + H2O NaAlO2 + 3H2

X 3/2xVH2 = 1 lt , khng bit Al c phn ng ht hay khng

TN2 : Chc chn Al ht v NaOH dNa + H2O NaOH + 1/2H2Al + NaOH + H2O NaAlO2 + 3/2H2

Th tch kh thu c : 1,75V TN1 Al dGi x , y l s mol Na , Al phn ng : TN1 : V = 22,4(x/2 + 3x/2 ) ; TN2 : 1,75V = 22,4(x/2 + 3y/2) Biu thc quan h gia x , y : y = 2x Tnh % Na = 23.x/(23x + 27y) = 29,87 %

Cu 9 : Chia m gam hn hp A gm Ba , Al thnh 2 phn bng nhau:-Phn 1: Tan trong nc d thu c 1,344 lt kh H2 (ktc) v dung dch B.-Phn 2: Tan trong dung dch Ba(OH)2 d c 10,416 lt kh H2(ktc)a/ Tnh khi lng kim loi Al trong hn hp ban u .

b/ Cho 50ml dung dch HCl vo B .Sau phn ng thu c 7,8 gam kt ta .Tnh nng mol ca dung dchHCl .

Bi ny ging bi trn : Al phn (1) dPhn I :Ba + H2O Ba(OH)2 + H2 x x x2Al + Ba(OH)2 + 2H2O Ba(AlO2)2 + 3/2 H2

x 3/2xTng s mol H2 = x + 3/2x = 2,5x = 0,06

Phn II :Ba + H2O Ba(OH)2 + H2 x x x

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39/64

CC DNG TON HA V C

2Al + Ba(OH)2 + 2H2O Ba(AlO2)2 + 3/2 H2 y 3/2y x + 3/2y = 0,465 x = 0,024 ; y = 0,294 Khi lng Al : 0,294.27 = 7,938 gam

Cu 10:Thm 240 ml dung dch NaOH vo cc ng 100 ml dung dch AlCl3 nng CM mol ,khuy u ti phn ng hon ton thy trong cc c 0,08 mol kt ta . Thm vo cc 100 ml dungdch NaOH 1M khuy u thy phn ng xy ra hon ton thu oc 0,06 mol kt ta . Tnh nng CMA.2M B.1,5M C.1M D.1,5M

Trng hp u : V kt ta lc u l 0,08 mol , sau thm 0,1 mol NaOH th cn kt ta l 0,06 mol Chng t rng . Ch c phn ng :

AlCl3 + 3NaOH Al(OH)3 + 3NaCl0,08 0,24 mol 0,08 molV AlCl3 d a mol

AlCl3 + 3NaOH Al(OH)3 + 3NaCl

a 3a aLc ny Al(OH)3 c to thnh l : 0,08 + aAl(OH)3 + NaOH NaAlO2 + H2O

Ban u 0,08 + a Phn ng x x Kt thc 0,08 + a x 0,08 + a x = 0,06 x a = 0,02

x + 3a = 0,1 x = 0,04 ; a = 0,02 mol Ton b s mol AlCl3 = a + 0,08 = 0,02 + 0,08 = 0,1 CM= 0,1/0,1 = 1M

Cu 11:Trong 1 cc ng 200 ml dd AlCl3 2M. Rt vo cc V ml dd NaOH nng a mol/l, ta thu c mt kt ta, em sy kh v nung n khi lng khng

i th c 5,1g cht rna) Nu V = 200 ml th a c gi tr no sau y:A. 2M B. 1,5M hay 3MC. 1M hay 1,5M D. 1,5M hay 7,5M

b) Nu a = 2 mol/l th gi tr ca V l:A. 150 ml B. 650 mlC. 150 ml hay 650 ml D. 150 ml hay 750 ml

Bi ny cc em t gii :AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)Al(OH)3 + NaOH NaAlO2 + H2O (2)

Phi chia lm hai trng hp :TH1: C 1 phn ngAlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)

TH2 : C c hai phn ng:


Bi ton nhit nhm :

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CC DNG TON HA V C

Cu 12: Khi cho 41,4 gam hn hp X gm Fe2O3, Cr2O3 v Al2O3 tc dng vi dung dch NaOH c(d), sau phn ng thu c cht rn c khi lng 16 gam. kh hon ton 41,4 gam X bng phnng nhit nhm, phi dng 10,8 gam Al. Thnh phn phn trm theo khi lng ca Cr2O3 trong hnhp X l (Cho: hiu sut ca cc phn ng l 100%; O = 16; Al = 27; Cr = 52; Fe = 56)A. 50,67%. B. 20,33%. C. 66,67%. D. 36,71%.

Cho X phn ng vi NaOHCr2O3 + 2NaOH 2NaCrO2+ H2OAl2O3 + 2NaOH 2NaAlO2 + H2O

V Fe2O3 khng phn ung nn m cht rn cn li = m Fe2O3 = 16 gam n Fe2O3 = 16/160 = 0,1 mol .Tin hnh phn ng nhit nhm X .Cr2O3 + 2Al Al2O3 + 2Cr (1)Fe2O3 + 2Al Al2O3 + 2Fe (2)0,1 molTheo gi thit s mol Al cn phn ng l 10,8/27 = 0,4 mol . Theo (2) n Al = 2 n Fe3O4 = 2.0,1 = 0,2 mol n Al (1) = 0,4 0,2 = 0,2 mol n Cr2O3 = n Al(1) / 2 = 0,1 mol m Cr2O3 = 152.0,1 = 15,2 gam %Cr2O3 =15,2.100/41,4 = 36,71%

Chn p n D .

Cu 13:Nung nng m gam hn hp Al v Fe2O3 (trong mi trng khng c khng kh) n khi phn ngxy ra hon ton, thu c hn hp rn Y. Chia Y thnh hai phn bng nhau:- Phn 1 tc dng vi dung dch H2SO4 long (d), sinh ra 3,08 lt kh H2 ( ktc);- Phn 2 tc dng vi dung dch NaOH (d), sinh ra 0,84 lt kh H2 ( ktc).Gi tr ca m lA. 22,75 B. 21,40. C. 29,40. D. 29,43.

Phn ng nhit nhm :2Al + Fe2O3 Al2O3 + 2Fe (1)Phn II tc dng vi NaOH :

Al + NaOH + H2O

NaAlO2 + 3/2H2 x 3/2 xS mol kh H2 thu c phn II : 3/2x = 0,84/ 22,4 = 0,0375 molx = 0,025 molPhn (1) tc dng vi H2SO4 :2Al + 3H2SO4 Al2(SO4)3 + 3H20.025 0,0375Fe + H2SO4 long FeSO4 + H2

y y y = 0,12Al + Fe2O3 Al2O3 + 2Fe (1)0,1 0,05 0,1

Khi lng ca Al : (0,1 +0,025).27 = 3,375 , khi lng ca Fe2O3 : 0,05.160 = 8 tng Khi lng = 11,375 m = 11,375.2 = 22,75 chn A

Cu 14: Cho V lt dung dch NaOH 2M vo dung dch cha 0,1 mol Al2(SO4)3 v 0,1 mol H2SO4 n khiphn ng hon ton, thu c 7,8 gam kt ta. Gi tr ln nht ca V thu c lng kt tatrn lA. 0,45. B. 0,35. C. 0,25. D. 0,05.

NaOH Na+ + OH-x xAl2(SO4)3 2Al3+ + 3SO42-0,1 0,2

H2SO4 2H+ + SO42-0,1 0,2

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CC DNG TON HA V C

n NaOH= x , n Al2(SO4)3 = 0,1 , n H2SO4 = 0,1 , n kt ta = 7,8/78 = 0,1 molKhi phn ng : H+ tc dng vi OH- trc ,H+ + OH- H2O0,2 0,2Khi phn ng vi Al3+ c hai kh nng :Ch c phn ng:Al3+ + 3OH- Al(OH)3

0,3 0,1Tng OH- = 0,2 + 0,3 = 0,5 molV = 0,5/2 = 0,25 ltC c hai phn ng :

Al3+ + 3OH- Al(OH)3 (1)0,2 0,6 0,2

Al(OH)3 + OH- AlO2- + H2O (2)a aLng kt ta thu c sau phn ng (1) l 0,2 mol nhng n s b phn ng mt phn a mol phn ng (2) , dosau khi kt thc (2) c 0,1 mol kt ta a = 0,2 0,1 = 0,1 molTng s mol OH- l : 0,2 + 0,6 + 0,1 = 0,9 molVNaOH= 0,9/2 = 0,45 lt Chn A

Cu 15: Ho tan hon ton 0,3 mol hn hp gm Al v Al4C3 vo dung dch KOH (d), thu c a mol

hn hp kh v dung dch X. Sc kh CO2 (d) vo dung dch X, lng kt ta thu c l 46,8 gam. Gitr ca a lA. 0,55. B. 0,60. C. 0,40. D. 0,45.

Al4C3 + 12H2O 4Al(OH)3 + 3CH4 x 4x 3xAl(OH)3 + KOH KAlO2 + H2O4x 4x

Al + KOH + H2O KAlO2 + 3/2 H2 y y 3/2yTng th tch kh l : 3x + 3/2yDung dch gm : KAlO2 : 4x + y mol v KOH dCO2 + KAlO2 + H2O Al(OH)3 + KHCO3

4x + y 4x + yn kt ta = 0,6 mol4x + y = 0,6 molx + y = 0,3x = 0,1 ; y = 0,2 molTng s mol ca kh : 3.0,1 + 3/2.0,2 = 0,6 mol Chn p n B

Cu 16: Cho hn hp gm Na v Al c t l s mol tng ng l 1 : 2 vo nc (d). Sau khi cc phnng xy ra hon ton, thu c 8,96 lt kh H2 ( ktc) v m gam cht rn khng tan. Gi tr cam lA. 10,8. B. 5,4. C. 7,8. D. 43,2.

Na + H2O NaOH + H2 x x x/2

Al + NaOH + H2O NaAlO2 + 3/2 H2 (2) Ban u 2x xNaOH ht , cht rn khng tan l Al d 2x x = x mol , n H2(2) = 3/2 x (mol)Gi s mol ca Na , Al l x , 2x (v n Na : n Al = 1 : 2 )n H2 = 8,96/22,4 = 0,4 mol Theo (1) , (2) Tng s mol H2 : x + 3/2 x = 2x2x = 0,4 x = 0,2 molVy Al d : 0,2.27 = 5,4 gam .Chn B

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CC DNG TON HA V C

Cu 17: t nng mt hn hp gm Al v 16 gam Fe2O3 (trong iu kin khng c khng kh) nkhi phn ng xy ra hon ton, thu c hn hp rn X. Cho X tc dng va vi V ml dung dchNaOH 1M sinh ra 3,36 lt H2 ( ktc). Gi tr ca V lA. 200. B. 100. C. 300. D. 150.

n Fe3O4 = 16/160 = 0,1 mol , n H2 = 3,36/22,4 = 0,15 mol

2Al + Fe2O3Al2O3 + Fe (1)0,2 0,1 0,1Phn ng xy ra hon ton nn cht rn X thu c sau (1) l : Al2O3 , Fe v AlV X c phn ng vi NaOH to ra H2 nn khng th Fe2O3 d (Phn ng xy ra hon ton khng c ngha l chai cht tham gia phn ng u ht m c th c mt cht d , khng khi no c c hai cht d )S mol Al = 2 n Fe2O3 = 0,2 mol , n Al2O3 = n Fe2O3 = 0,1 molAl2O3 + 2NaOH 2NaAlO2 + H2O (1)0,1 0,2Al + NaOH + H2ONaAlO2 + 3/2 H2 (2)0,1 0,1 mol 0,15 molT (2)n NaOH (2) = 2/3n H2 = 0,1 molT (1)n NaOH = 2n NaOH = 0,3 molTng s mol NaOH = 0,2 + 0,1 = 0,3 molV NaOH = 0,3/1 = 300 mlChon C

Cu 18 : in phn nng chy Al2O3 vi anot than ch (hiu sut in phn 100%) thu c mkg 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 cam l

A. 54,0 B. 75,6 C. 67,5 D. 108,0

Ti catot(-) :Al3+ + 3e Al Ti anot (+) : O-2 2e O2 Kh oxi sinh ra anot t chy dn C : C + O2 CO2

CO2 + C 2COCo 4e C+4 v Co 2e C+2

Phng trnh in phn :2Al2O3 + 3C 4Al + 3CO2 (1)

0,8---0,6Al2O3 + C 2Al + 3CO (2)

1,2


43/64

CC DNG TON HA V C

CC BI TP ST

C cc phn ng m cc em cn phi nh :

Fe + Fe3+ Fe2+Cu + Fe3+ Cu2+ + Fe2+

Fe2+ + Ag+ Fe3+ + Ag

Fe(NO3)2 nung Fe2O3 + NO2 + O2Fe(OH)2 nung trong khng kh Fe2O3 + H2O ( Khng c khng kh th ra FeO )FeCO3 nung trong khng kh + O2 Fe2O3 + CO2

Nu hn hp cho FeO , Fe2O3 , Fe3O4ta c th quy i v hn hp FeO , Fe2O3 hoc ch cnFe3O4 nu s mol ca chng bng nhau .

Cu 1 :Cho 16,8 gam bt st vo 800 ml dung dch HNO3 0,5 M thu c kh NO duy nht .Tnh : Th tch kh thu c Tnh khi lng kim loi cn d . Khi lng mui thu c

S : V = 2.24 lt , m kim loi d = 8.4 g , m Fe(NO3)2 = 27 gBi gi :n Fe = 0,3 mol , n HNO3 = 0,4 mol

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O B 0,3 0,4 P 0,1 0,4 0,1 0,1 Kt 0,2 0 0,1 0,1 2 Fe(NO3)3 + Fe d 3Fe(NO3)2 B 0,1 0,2 P 0,1 0,05 0,15 Kt 0 0,15 0,15Khi lng kim loi d : 0,15.56 = 8,4 gam , khi lng mui : 0,15.180 = 27 gam , th tch kh :

0,1.22,4 = 0,224 ltCu 2 :Cho 16,8 gam bt st vo V lt dung dch HNO3 0,5 M thu c 8,4 gam kim loi d . Tnhth tch kh thu c .S : V = 2.24 l

Bi gii :n Fe phn ng= (16,8 8,4)/56 = 0,15 molV kim loi d nn ch to thnh mui st II

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

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CC DNG TON HA V C

Fe + 2Fe(NO3)3 3Fe(NO3)2 x x

Tng s mol Fe phn ng : 3/2x = 0,15 x = 0,1 mol Th tch kh thu c : 2,24 lt

Cu 3 :Cho 16,8 gam bt Fe vo 400 ml dung dch HNO3 1M , thu c kh NO duy nht , lng

mui thu c cho vo dung dch NaOH d thu c kt ta . Nung nng kt ta m khng c khngkh thu c m gam cht rn . Tnh m ?S : m FeO = 10.8 gam

n Fe = 0,3 mol , n HNO3 = 0,4 molFe + 4HNO3 Fe(NO3)3 + NO + 2H2O

B 0,3 0,4 P 0,1 0,4 0,1 0,1 Kt 0,2 0 0,1 0,1

2 Fe(NO3)3 + Fe d 3Fe(NO3)2 B 0,1 0,2 P 0,1 0,05 0,15 Kt 0 0,15 0,15

Fe(NO3)2 Fe(OH)2 FeO0,15 0,15 Khi lng FeO thu c : 0,15.72 = 10,8 gam

Cu 4 :Cho m gam Fe vo 400 ml dung dch HNO3 1M thu c V lt kh NO duy nht v 14 gamkim loi . Tnh m ? V ?

S : m = 22.4 gam , V = 2,24 lt

Hc sinh t gii .

Cu 6 :Cho 5,6 gam bt Fe tc dng vi 500 ml dung dch AgNO3 0,5 M . Tnh : khi lng mui thu c khi lng kim loi thu c

S : m mui = 21.1 gam , m Ag= 27 gamBi gii :Cc em ch n phn ng : Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag

Fe + 2AgNO3 Fe(NO3)2 + 2Ag0,1 0,25 AgNO3 d : 0,05 mol , Fe(NO3)2 to thnh : 0,1 molFe(NO3)2 + AgNO3 Fe(NO3)3 + Ag0,1 0,05 Fe(NO3)2 d : 0,05 mol , Fe(NO3)3 to thnh 0,05 mol Tng s mol Ag hai phn ng : 0,25 mol m Ag = 0,25.108 = 27 gamKhi lng mui : 0,05.180 + 0,05.242 = 21,1 gam

Cu 7 :Cho m gam bt Fe tc dng vi 250 ml dung dch AgNO3 1M thu c dung dch A , choton vo dung dch A mt lng KOH d thu c kt ta , nung kt ta trong chn khng thu c7,6 gam cht rn .Tnh m ?Gi : Bi ton ny cc em xt hai kh nng :

(1) : St d(2) : st ht C phn ng :

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CC DNG TON HA V C

Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag

Cu 8 : 2002 ACho 18,5 gam hn hp Z gm Fe , Fe3O4 tc dng vi 200 ml dung dch HNO3 long un nng vkhuy u . Sau phn ng xy ra han ton thu c 2,24 lt kh NO duy nht ktc , dung dch Z1 vcn li 1,46 gam kim loi .

Tnh nng mol/lit ca dung dch HNO3

Dng phng php quy i nguyn t :Hn hp z ch c hai nguyn t Fe , O .V Z + HNO3 cn d kim loi Fe d , vy Z1 ch c mui st II

Fe - 2e Fe+2

x 2xO + 2e O-2

y 2yN+5 + 3e N+2

0,3 0,1Theo nh lut bo ton e :2x 2y = 0,3Tng khi lng Z : 56x + 16y = 18,5 - 1,46Gii h : x = 0,27 , y = 0,12C phng trnh :Fe + HNO3 Fe(NO3)2 + NO + H2O (1)a a 0,1

T a + a/2 = 0,27 a = 0,18 molBo ton nguyn t N (1) s mol HNO3 = 3a + 0,1 = 0,64 Nng mol ca HNO3 : 0,64 / 0,2 = 3,2

Cu 9 : Kh 4,8 gam mt oxit ca kim loi trong dy in ha nhit cao cn 2,016 lt kh H2ktc . Kim loi thu c em ha tan trong dung dch HCl thu c 1,344 lt kh H2 ktc . Hy xc

nh cng thc ha hc ca oxit dng .Ch : Bi ny cc em hay b nhm v khng ha tr thay i hai phng trnhOxt cha bit ca kim loi no Gi MxOy

MxOy + yH2 xM + yH2Oa ay ax ay = 0,09 mol2M + 2nHCl 2MCln + nH2ax nax nax / 2= 0,06 molM : 56ax + 16ay = 4,8 ax = 0,06 x : y = ax : ay = 0,06 : 0,09 = 3 : 2

n = 0,12 : 0,06 = 2 Ch c Fe tha mn v n c hai ha tr

Cu 10 : Mt dung dch c ha tan 1,58 gam KMnO4 tc dng vi dung dch hn hp c ha tan 9,12gam FeSO4 v 9,8 gam H2SO4 . Hy tnh s gam cc cht c trong dung dch sau phn ng .

n KMnO4 = 0,01 ; n FeSO4 = 0,06 ; n H2SO4 = 0,1 molPhn ng : 10FeSO4+ 2KMnO4 + 8H2SO4 5Fe2(SO4)3 + 2MnSO4 + K2SO4+ 8H2O

B 0,1 0,06 0,1 P 0,1 0,02 0,08 0,05 0,02 0,01 Kt 0 0,04 0,02 0,05 0,02 0,01

Dung dch sau phn ng gm cc cht : dng kt thc phn ng

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46/64

CC DNG TON HA V C

Cu 11:Hn hp X gm Cu v Fe c t l khi lng tng ng l 7:3 . Ly m gam X cho phn ngxy ra hon ton vi dung dch cha 44,1 gam HNO 3 sau phn ng cn li 0,75m gam cht rn v c0,56 lt kh Y gm NO v NO2 ktc . Ga tr ca m l ?

Ban u : Cu : 0,7m Fe : 0,3m .

Sau phn ng : Fe : 0,05m Cu : 0,7mV st d nn ch c mui Fe II .Fe + HNO3 Fe(NO3)2 + NO + NO2 + H2OV NO v NO2 cng c 1 nguyn t N nn tng s mol N trong hai kh l 0,56/22,4 = 0,25S mol HNO3 = 0,7 mol N trong HNO3 l 0,7Gi s mol Fe phn ng l xTheo nh lut bo ton nguyn t N 0,7 = 2x + 0,25 x = 0,225 mol .Khi lng Fe phn ng : 0,225.56 = 12,6V st phn ng : 0,3m 0,05 m = 0,25m = 12,6 m = 50,4 gam

Cu 12 : Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO 3)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 Vlt kh NO (sn phm kh duy nht, ktc). Gi tr ca m v V ln lt lA. 17,8 v 4,48. B. 17,8 v 2,24.C. 10,8 v 4,48. D. 10,8 v 2,24.

n Cu(NO3)2 = 0,16 n Cu2+ = 0,16 , n NO3- = 0,32 moln H2SO4 = 0,2 n H+ = 0,4V thu c hn hp kim loi nn Ch c mui Fe2+ to thnh3Fe + 2NO3- + 8H+ 3Fe2+ + 2NO + 4H2O (1)0,15---------- 0,4---------------0,1Fe + Cu2+ Fe2+ + Cu (2)0,16-----0,16-----------0,16

Khi lng ng trong 0,6m gam hn hp sau phn ng l : 64.0,16 molBo ton st : m = 0,15.56p(1) + 0,16.56p(2) + (0,6m 0,16.64 ) d m = 17,8Mt khc VNO = 0,1.22,4 = 2,24 Chn p n BCu 4: ho tan hon ton 2,32 gam hn hp gm FeO, Fe3O4 v Fe2O3 (trong s mol FeO bng smol Fe2O3), cn dng va V lt dung dch HCl 1M. Gi tr ca V lA. 0,23. B. 0,18. C. 0,08. D. 0,16.

V s mol FeO , Fe2O3 bng nhau nn ta c thquy i chng thnh Fe3O4 . Vy hn hp trn ch gm Fe3O4 .n Fe3O4 = 2,32 : 232 = 0,01 molFe3O4+ 8HCl 2FeCl3 + FeCl2 + 4H2O0,01 0,08 moln HCl= 0,08 molVHCl= 0,08/1 = 0,08 lt

Chn C .Cu 13: 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 thuc m gam mui khan. Gi tr ca m lA. 38,72. B. 35,50. C. 49,09. D. 34,36.

Quy i hn hp : Fe , FeO , Fe2O3 , Fe3O4 thnh Fe ,OS cho nhn e :Fe 3e Fe3+ O + 2e O-2 N+5 + 3e N+2

x---3x y--2y 1,8---0,6Bo ton mol e : 3x = 2y + 0,18

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CC DNG TON HA V C

56x + 16y = 11,36 x = 0,16 ; y = 0,15 molFe Fe(NO3)30,16 -------0,16 Khi lng mui l : 0,16.242 = 38,72 Chn A

Cu 14: Cho hn hp bt gm 2,7 gam Al v 5,6 gam Fe vo 550 ml dung dch AgNO3 1M. Sau khi ccphn ng xy ra hon ton, thu c m gam cht rn. Gi tr ca m l (bit th t trong dy th in ho:Fe3+/Fe2+ ng trc Ag+/Ag)A. 59,4. B. 64,8. C. 32,4. D. 54,0.

n Al = 2,7/27 = 0,1 mol , n Fe = 5,6/56 = 0,1 mol , n AgNO3 = 0,55.1 = 0,55 molKhi cho hn hp kim loi gm Al , Fe vo dung dch AgNO3 th Al s phn ng trc nu Al ht s n Fephn ng , nu AgNO3 d sau phn ng vi Fe th c phn ngAg+ + Fe2+ Ag + Fe3+

Al + 3AgNO3 Al(NO3)3 + 3Ag (1)Ban u 0,1 0,55Phn ng 0,1 0,3 0,3Kt thc 0 0,25 0,3

Al htTnh theo Al , n AgNO3 = 3.n AlAgNO3 d : 0,55 3.0,1 = 0,25 molFe + 2AgNO3 Fe(NO3)2 + 2Ag (2)

Ban u 0,1 0,25Phn ng 0,1 0,2 0,1 0,2Kt thc 0 0,15 0,1 0,2

Sau phn ng AgNO3 d : 0,15 mol tip tc c phn ngFe(NO3)2 + AgNO3 Fe(NO3)3 + Ag (3)

Ban u 0,1 0,15Phn ng 0,1 0,1 0,1Kt thc 0 0,05 0,1T (1) , (2) , (3) tng s mol Ag = 0,3 + 0,2 + 0,1 = 0,6m Ag= 0,6.108 = 64,8 gam Chn B .

Cu 15 : Cho 9,12 gam hn hp gm FeO, Fe 2O3 , Fe 3O4 tc dng vi dung dch HCl(d). Sau khi cc phn ng xy ra hon ton, c dung dch Y; c cn Y thu c7,62 gam FeCl 2 v m gam FeCl 3 . Gi tr ca m l :A. 9,75 B . 8,75 C . 7,80 D . 6,50

Ta c th tch Fe 3O 4 = FeO + Fe 2 O 3 Lc ny hn hp cht rn ch cn FeO , Fe 2 O 3

FeO + 2HCl FeCl2 + H2 Ox xFe 2 O 3 + 6HCl 2FeCl3 + 3H2 OY 2y

Gi x , y l s mol ca cht FeO , Fe 2O 3 . m cht rn = 72x + 160y = 9,12 gamKhi lng mui FeCl 2 l : 127x = 7,62Giai h : x = 0,06 mol , y = 0,03 mol Khi lng mui FeCl 3 = 2.0,03.162,5 = 9,75 gamChn p n A .

Cu 16 : Th t ch dung dch HNO3 1M (long) t nht cn dng ho tan hon tonmt hn hp gm 0,15 mol Fe v 0,15 mol Cu l (bit phn ng to cht kh duy nhtl NO)A . 1,0 l t B . 0,6 l t C. 0,8 l t D . 1,2 l t

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48/64

CC DNG TON HA V C

Nhn xt : Ln g HNO 3 ti thiu cn dng khi: Fe Fe 2 + , Cu Cu 2 + S cho nhn e :Fe 2e Fe 2 + Cu 2e Cu 2 + N+ 5 + 3e N+ 2

0,15 0 ,3 0,15 0,3 3x x Theo nh lut bo ton mol e : 0 ,3 + 0,3 = 3x x = 0,2 mol Fe , Cu + HNO3 Fe(NO 3)2 + Cu(NO 3) 2 + NO + H2 O0,15 0,15 0,15 0,15 0,2 Bo ton nguyn t N : S mol HNO 3 = 0,3 + 0,3 + 0,2 = 0,8 mol Chn C .

Cu 17: Cho 6,72 gam Fe vo 400 ml dung dch HNO3 1M, n khi phn ng xy ra hon ton, thuc 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. B. 3,20. C. 0,64. D. 3,84.

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2OBan u 0,12 0,4 mol HNO3 ht , sau phn ng c : Fe(NO3)3 0,1 mol ; Fe d : 0,02 mol

Lng Cu ti a uc ho tan ht l khi n tham gia c hai phn ng :Fe + 2Fe(NO3)3 3Fe(NO3)2

Ban u 0,02 0,1 Sau phn ng Fe(NO3)3 d : 0,1 0,04 = 0,06 mol

Cu + 2Fe(NO3)3 Cu(NO3)2 + 2Fe(NO3)2 Ban u 0,06 S mol Cu ti a c ho tan l : 0,03 mol m Cu = 0,03.64 = 1,92 gam Chn p n A

Cu 18 : Cho m gam bt Fe vo 800 ml dung dch hn hp gm Cu(NO 3)2 0,2M v H2SO40,25M. Sau khi cc phn ng xy ra hon ton, thu c 0,6m gam hn hp bt kim loi v Vlt 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.n Cu(NO3)2 = 0,16 n Cu2+ = 0,16 , n NO3- = 0,32 moln H2SO4 = 0,2 n H+ = 0,4V thu c hn hp kim loi nn Ch c mui Fe2+ to thnh3Fe + 2NO3- + 8H+ 3Fe2+ + 2NO + 4H2O (1)0,15---------- 0,4---------------0,1Fe + Cu2+ Fe2+ + Cu (2)0,16-----0,16--------------0,16 Khi lng ng trong 0,6m gam hn hp sau phn ng l : 64.0,16 molBo ton st : m = 0,15.56p(1) + 0,16.56p(2) + (0,6m 0,16.64 ) d m = 17,8Mt khc VNO = 0,1.22,4 = 2,24 Chn p n B

Cu 19 : Ho tan hon ton 24,4 gam hn hp gm FeCl2 v NaCl (c t l s mol tng ng l1 : 2) vo mt lng nc (d), thu c dung dch X. Cho dung dch AgNO 3 (d) vo dungdch X, sau khi phn ng xy ra hon ton sinh ra m gam cht rn. Gi tr ca m l

A. 68,2 B. 28,7 C. 10,8 D. 57,4Gi s mol ca FeCl2 l x 127x + 58,5.2.x= 24,4 x = 0,1.FeCl2 + 2AgNO32AgCl+ Fe(NO3)20,1-------0,2--------0,2-------0,1 molNaCl + AgNO3 AgCl + NaNO30,2-------0,2------0,2Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag (*)

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CC DNG TON HA V C

0,1-------------------------------------0,1m = (0,02 + 0,02)143,5 + 108.0,1 = 68,2 (g);p n A

Dng ton in phn

Cc em thn mn , hu ht cc em u yu gii cc bi ton v in phn , gipcc em khng cn e ngi dng ton ny na , thy Ngc Quang c gng su tp

y cc dng gm cc CU HI L THUYT & DNG BI TP INPHN. c bit cc bi tp c gii cn thn , chi tit , d hiu .

Chc cc em hc tt !

L thuyt :

- Qa trnh in phn din ra ca tt: Qa trnh kh cc ion dng kim loi v H2O

in phn nng chy : Th t kh tng dn t tri qua phi theo chiu tng dn tnh oxiha ca ion kim loi

in phn dung dch : Cc ion kim loi t : Li+ Al3+ khng b in phn (nc s b

phn ).- Qa trnh in phn anot (c dng ) : Th t gim dn nh sau : I- > Br- > I- > H2O (F-khng b in phn )

Mt vi v d v in phn :

VD1 : in phn dd NaCl : NaCl Na+ + Cl-

Catot (-) Anot (+)Na+ khng b in phn 2Cl- - 2e Cl22H2O + e H2 + 2OH-

Phng trnh : 2Cl- + 2H2O Cl2 + H2 + 2OH-

2NaCl + 2H2O 2NaOH + Cl2 + H2

Tng t vi cc phng trnh in phn cc cht : NaCl , CaCl , MgCl2 , BaCl2 , AlCl3 Khng th iu ch kim loi t : Na Al bng phng php in phn dung dch .

VD2 : in phn dung dch : Cu(NO3)2 :

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50/64

CC DNG TON HA V C

Cu(NO3)2 Cu2+ + 2NO3-

Catot(-) Anot (+)

NO3- khng b in phn .Cu2+ + 2e Cu 2H2O - 4e 4H+ + O2 Phng trnh : Cu2+ + H2O Cu + 2H+ + O2

Cu(NO3)2 + H2O Cu + 2HNO3 + O2

Tng t vi trng hp in phn cc mui ca kim loi yu t Zn Hg vi cc gc axit NO3- ,SO42- .

FeSO4 + H2O Fe + H2SO4 + O2

VD3 : in phn dung dch Na2CO3 :Na2CO3 2Na+ + CO32-

Catot (-) Anot (+)

Na+ khng b in phn CO32- khng b in phnH2O + 2e H2 + OH- H2O e H+ + O2

Phng trnh : 2H2O H2 + OH- + H+ + O2H2O H2 + O2

Tng t in phn cc dung dich NaNO3 , Ca(NO3)2 , K2SO4 (Mui ca kim loi t Na + Al3+ vi cc gc axit c cha Oxi ) cng in phn to ra O2 + H2

Cng thc Faraday : S mol e trao i mi in cc :n = It/96500I l cng dng in , t l thi gian tnh bng s

Hoc : n = It/96500.nen e : ha tr ca kim loi

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51/64

CC DNG TON HA V C

BI TON IN PHN

Cu 1:in phn 100ml dung dch cha AgNO3 0.1M v Cu(NO3)2 0.1M vi cng dngin I l 1.93A.Tnh thi gian in phn (vi hiu xut l 100%).

1) kt ta ht Ag (t1)2) kt ta ht Ag v Cu (t2)

a)t1 = 500s, t2 = 1000s b) t1 = 1000s, t2 = 1500sc)t1 = 500s, t2 = 1200s d) t1 = 500s, t2 = 1500s

n AgNO3 = 0,01 mol ; n Cu(NO3)2 = 0,01 mol in phn ht AgNO3 :n 1= It/96500.1 0,01 = 1,93.t1 / 96500 t1 = 500 s

in phn ht 0,01 mol Cu(NO3)2 :n 2 = It2/96500.2 0,01 = 1,93.t2/96500.2 t2 = 1000 s Tng thi gian in phn ht c hn hp trn l : t = t1 + t2 = 500 + 1000 = 1500 s Chn p n d .

Cu 2:in phn 100ml dung dch CuSO4 0.2M vi cng I = 9.65 A.Tnh khi lngCu bm bn catot khi thi gian in phn t1 = 200s v t2 = 500s(vi hiu sut l 100%).

a) 0.32g ; 0.64g b) 0.64g ; 1.28gc) 0.64g ; 1.32g d) 0.32g ; 1.28g

n CuSO4 = 0,2.0,1 = 0,02 molTrc tin ta cn tnh thi gian in phn ht 0,02 mol CuSO4 l :n = It/96500.2 0,02 = 9,65.t / 96500.2 t = 400 sPhng trnh in phn :CuSO4 + H2O Cu + H2SO4 + O2

Khi in phn trong thi gian t1 = 200 s :n = It/96500.2 = 9,65.200/96500.2 = 0,01 mol Khi lng Cu = 0,01.64 = 0,64 gam

Khi in phn trong 500 s : V in phn ht 0,02 mol CuSO4 ht 400s , nn 100s cn li s in phn H2Otheo phng trnh :H2O H2 + O2Khi lng kim loi Cu thu c : 0,02.64 = 1,28 gam Chn p n b .

Cu 3:in phn 100ml dung dch CuSO4 0.1M cho n khi va bt u si bt bn catot th

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CC DNG TON HA V C

ngng in phn. Tnh pH dung dch ngay khi y vi hiu sut l 100%.Th tch dung dchc xem nh khng i. Ly lg2 = 0.30.

a) pH = 0.1 b) pH = 0.7 c) pH = 2.0 d) pH = 1.3

n khi va bt u si bt kh bn catot th Cu2+ va ht .

in phn dung dch : CuSO4 :CuSO4 Cu2+ + SO42-

Catot(-) Anot (+)

SO42- khng b in phn .Cu2+ + 2e Cu 2H2O - 4e 4H+ + O2

0,020,01 0,02 -0,02 S mol e cho anot = s mol e cho catot n H+ = 0,01 mol [H+] = 0,02/0,1 = 0,2 pH = -lg0,2 = 0,7 Chn p n B

Cu 4:in phn 100ml dung dch cha NaCl vi in cc tr ,c mng ngn, cng dng in I l 1.93A. Tnh thi gian in phn c dung dch pH = 12, th tch dungdch c xim nh khng thay i,hiu sut in phn l 100%.

a) 100s b) 50s c) 150s d) 200s

V dung dch c PH = 12 Mi trng kim .p H = 12 [H+] = 10-12 [OH-] = 0,01 S mol OH- = 0,001 mol

NaCl Na+ + Cl-

Catot (-) Anot (+)Na+ khng b in phn Cl- + 2e Cl22H2O + 2e H2 + 2OH-

0,001


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Tng thi gian : 250 + 500 = 750 s Chn d .

Cu 6:in phn 100ml dung dch CuCl2 0.08M. Cho dung dch thu c sau khi in phntc dng vi dung dch AgNO3 d th thu c 0.861g kt ta. Tnh khi lng Cu bm bn

catot v th tch thu c bn anot. Cho Cu = 64.a) 0.16g Cu ; 0.056 l Cl2 b) 0.64g Cu ; 0.112 l Cl2c) 0.32g Cu ; 0.112 l Cl2 d) 0.64g Cu ; 0.224 l Cl2

n CuCl2 = 0,008 molCuCl2 Cu + Cl 2 (1)0,005 ----0,005 0,005 molDung dch sau phn ng c phn ng vi AgNO3 to ra kt ta CuCl2 dCuCl2 + 2AgNO3 Cu(NO3)2 + 2AgCl 0,003---------------------------------------0,006 CuCl2 tham gia phn ng (1) = 0,008 0,003 = 0,005 mol Khi lng Cu = 0,005.64 = 0,32 gam , th tch kh Cl2 thu c : 0,005.22,4 = 0,112 lt

Chn p n C .

Cu 7:in phn 100ml dung dch CuSO4 0.1M vi cng I = 9,65A.Tnh th tch khthu c bn catot v bn anot lc t1 = 200s v t2 = 300s.a) catot: 0 ; 112ml v anot: 112 ; 168mlc) catot: 0 ; 112ml v anot: 56 ; 112ml

b) catot: 112 ; 168ml v anot: 56 ; 84mld) catot: 56 ; 112ml v anot: 28 ; 56ml

in phn dung dch : CuSO4 :

CuSO4 Cu2+ + SO42-

Catot(-) Anot (+)

SO42- khng b in phn .Cu2+ + 2e Cu 2H2O - 4e 4H+ + O2

Thi gian in phn ht : 0,01 mol CuSO4 :0,01 = 9,65.t / 96500.2 t = 200 s

Xt thi im t1 = 200 s n CuSO4 in phn ht = 0,01 molPhng trnh in phn CuSO4 :CuSO4 + H2O Cu + H2SO4 + O20,01 0,01 0,005 mol Th tch kh thu c anot : 0,005.22,4 = 0,112 l = 112 mlKhng c kh thot ra catot .

Xt thi im t2 = 300s , in phn CuSO4 ht 200s , cn in phn H2O ht 100sin phn dung dch : CuSO4 :

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CuSO4 Cu2+ + SO42-

Catot(-) Anot (+)

SO42- khng b in phn .

Cu2+ + 2e Cu 2H2O - 4e 4H+ + O20,01--- 0,02 0,03 ------ 0,0075H2O + 2e H2 + OH-

0,010,005Trong 300 s s mol e trao i ti hai in cc l :n = It/96500 = 300.9,65/96500 = 0,03 molTrong khi ti catot Cu2+ nhn 0,01.2 = 0,02 mol e H2O s nhn 0,03 0,02 = 0,01 mol e cnli Kh thot ra ti catot : H2 , V = 0,005.22,4 = 0,112 ltKh thot ra ti anot O2 : V = 0,0075.22,4 = 0,168 lt Chn p n A .

Cu 8:in phn 100ml dung dch AgNO3 0.2M. Tnh cng I bit rng phi in phntrong thi gian 1000s th bt u si bt bn catot v tnh pH ca dung dch ngay khi y. Thtch dung dch c xem nh khng thay i trong qu trnh in phn. Ly lg2 = 0.30.

a) I = 1.93A,pH = 0,7 b) I = 2.86A,pH= 2.0

c) I = 1.93A,pH = 1.3 d) I = 2.86A,pH= 1.7

AgNO3 Ag+ + NO3-

Catot(-) Anot (+)

NO3- khng b in phn .Ag+ + 1e Ag 2H2O - 4e 4H+ + O2

0,02- 0,02----------------------------- 0,02-- 0,02Khi bt u si bt kh bn catot c ngha l Ag+ va htn Ag+ = n AgNO3 = 0,2.0,1 = 0,02 mol ,Theo cng thc faraday : n = It/965000 0,02 = I.1000/96500 I = 1,93 ATheo s in phn : n H+ = 0,02 mol [H+] = 0,02/0,1 = 0,2 pH = -lg0,2 = 0,7

Chn p n A .Cu 9:in phn 200ml dung dch CuSO4 0.1M v MgSO4 cho n khi bt u si bt bncatot th ngng in phn. Tinh khi lng kim loi bm bn catot v th tch(ktc) thot ra

bn anot.Cho Cu = 64, Mg = 24.a) 1.28g; 2.24 lt b) 0.64; 1.12ltc) 1.28g; 1.12 lt d) 0.64; 2.24 lt

n CuSO4 = 0,02 molV ion Mg2+ khng b in phn nn khi catot bt u si bt kh c ngha l Cu2+ va b

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in phn ht .CuSO4 Cu2+ + SO42-

Ca

C+íc dß¦íng to+ín h+¦a hß+ìc v+¦ c¦í - NGß+îC QUANG

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