If you can't read please download the document
Upload
phong-pham
View
4.349
Download
6
Embed Size (px)
Citation preview
2. Nguoithay.vnlch nhau 23 gam . Ga tr ca m l ?Fe + HCl FeCl2a aFe + HNO3 Fe(NO3)3aa 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 gamCu 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 molDng phng php bo ton nguyn t :FeS2 Fe2(SO4)3 (1)0,120,06 (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 DCu 6 :Cho kh CO i qua ng s cha 16 gam Fe2O3 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 + H2ONh vy : Fe2O3 Fe2(SO4)3 0,10,1 ( 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 : 16Cu Cu(NO3)2 Cu(OH)2 CuO0,1 0,1 ( Bo ton nguyn t Cu )Fe Fe(NO3)3 Fe(OH)3 Fe2O3Nguoithay.vn -2- 3. Nguoithay.vn0,1 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 gamCu 8 :Ho tan hon ton hn hp X gm 0,4 mol FeO v 0,1 mol Fe2O3 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 gamFeO Fe(NO3)3 Fe(OH)3 Fe2O30,4 0,2 (bo ton Fe )Fe2O3 Fe(NO3)3 Fe(OH)3 Fe2O30,10,1 Cht rn sau khi nung l : Fe2O3 : 0,3 mol Khi lng 0,3.16 = 48 gamCu 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 Fe2O3aa 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 = 0,1m = 0,1.56 + 0,1.160 = 21,6 BI TON CO2Dn kh CO2 vo dung dch kim NaOH , KOH th t xy ra phn ng :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)Nguoithay.vn-3- 4. Nguoithay.vnCu 1:Cho 1,568 lt CO2 ktc li chm qua dung dch c ha tan 3,2 gam NaOH . Hy xcnh 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)xxx ---------- CO2 + 2NaOH Na2CO3 (2) y 2yyGi 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 )Cu 2 :Cho 8 lt hn hp kh CO v CO2 trong CO2 chim 39,2 % i qua dung dch ccha 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) xxx 2----------2CO2 + Ca(OH)2 Ca(HCO3)2(2)yy/2Gi 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 gamNguoithay.vn -4- 5. Nguoithay.vnCu 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 X un k dung dch Xthu thm c 100 gam kt ta . Ga tr ca m l ?(C6H10O5)n + n H2O n C6H12O6 3,75/n 3,75Phn ng ln men : C6H12O6 2CO2 + 2H2O 3,75---------7,5Dn 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 + H2O1- ----------------------1 molCa(OH)2 + CO2 CaCO3 + Ca(HCO3)2 + H2O 5,51Bo ton nguyn t C S mol CO2 = 7,5 Khi lng tinh bt : 162.3,75.100/81 = 750Cu 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 + H2O0,120,08Bo 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 MCu 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 ?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 gamCu 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 ?Nguoithay.vn -5- 6. Nguoithay.vnGii 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 + H2O0,1 -----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 MCu 9. Cho ton 0,448 lt kh CO2 (ktc) hp th hon ton bi 200 ml dung dch Ba(OH)2 thu -c1,97 gam kt ta. Hy la chn nng mol/l ca dung dch Ba(OH)2.A. 0,05MB. 0,1MC. 0,15M D. p n khc.Gii tng t Cu 4Cu 10. Cho 3,36lt kh CO2 (ktc) hp th bng 100 ml dung dch NaOH 1M v Ba(OH)2 1M ththu -c bao bao nhiu gam kt ta.A. 19,7 gamB. 24,625 gam C. 14,775 gamD. c A, B u ng.Gii tng t cu 5Cu 11. Hn hp X gm 2 mui cacbonat ca 2 kim loi thuc 2 chu k lin tip ca phn nhmchnh nhm II. Ha tan ht 16,18 gam hn hp A trong dung dch HCl thu -c kh B. Cho ton bkh B hp th vo 500 ml dung dch Ba(OH)2 1M. Lc b kt ta , ly dung dch n-c lc tc dngvi l-ng d- dung dch Na2SO4 th thu -c 11,65 gam kt ta. Xc nh cng thc ca 2 mui.A. BeCO3 v MgCO3 B. MgCO3 v CaCO3C. CaCO3 v SrCO3 D. c A, B u ngn Ba(OH)2 = 0,5mol , n BaSO4 = 0,05 molXt 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 mnTrng hp 2 : C 2 phn ng : Vit tm tt liBa(OH)2 + CO2 BaCO3 + Ba(HCO3)2 + H2O (1)Nguoithay.vn-6- 7. Nguoithay.vn 0,50,05 Ba(HCO3)2 + Na2SO4 BaSO4 + 2NaHCO3 0,05 -----------------------0,05 Bo ton s mol Ba (1) S mol BaCO3 = 0,045 Bo ton C (1) S mol CO2 = 0,45 + 0,1 = 0,55 MCO3 + 2HCl MCl2 + CO2 + H2O 0,55-------------------------0,55 MCO3 = 16,18/0,55 = 29,4 Mg , Ca tha mn Chn D .Cu 12 : Mt bnh cha 15 lt dd Ba(OH)2 0,01M. Sc vo dd V lt kh 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 ltn 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 .Trng hp (1) ch c 1 phn ng :CO2 + Ba(OH)2 BaCO3 + H2O0,150,1 S mol CO2 tnh theo BaCO3 : n CO2 = 0,1 mol V CO2 = 2,24 molTrng hp 2 : C c 2 phn ng : Vit tm tt liCO2 + Ba(OH)2 BaCO3 + Ba(HCO3)2 + H2O 0,15-----v----0,1 ----------0,05Bo 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 V CO2 = 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 XTheo gi thit : S mol O2 = 0,3 mol ; n BaCO3 lc u = 0,1 mol ; n BaCO3 sau khi un nng dd = 0,05molCch tm khi lng nguyn t C nhanh nht cho bi ton ny :un nng dd :Ba(HCO3)2 BaCO3 + CO2 + H2O0,050,05 CO2 + Ba(OH)2 BaCO3 + Ba(HCO3)2 + H2O 0,1 0,05 Bo ton nguyn t C S mol CO2 = 0,1 + 0,05.2 = 0,2 molKhi lng dd gim = m BaCO3 ( m CO2 + m H2O ) = 5,5 m H2O = 5,4 n H2O = 0,3 mol (> n CO2 Hp cht no )T s mol H2O v CO2 n H = 0,6 ; n C = 0,2 ,n O trong CO2 + n O trong H2O = 0,4 + 0,3 = 0,7 n O trong X = 0,7 0,6 = 0,1 ( 0,6 l s mol O trongO2 ) Nguoithay.vn -7- 8. Nguoithay.vnGi ctpt ca X l CxHyOz x : y : z = 2 : 6 : 1 C2H6OCu 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)yy/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 .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:19,7B:17,72C:9,85D: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- + H2OBan u0,2 0,05 Tnh theo HCO3- : S mol CO32- = 0,05 molTip 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-0,050,05 molBa(OH)2 Ba2+ + 2OH-0,1 0,10,2 mol n OH- = 0,05 + 0,2 = 0,25 molCO2 + OH- HCO3-Ban u 0,20,25 CO2 ht , n OH- d : 0,25 0,2 = 0,025 mol , n HCO3- = 0,2 molOH- + HCO3- CO32- + H2OBan u0,05 0,2 0,05 HCO3- d : 0,2 0,05 = 0,15 mol , n CO32- : 0,05 molBa2+ + CO32- BaCO3Ban u 0,10,05 Ba2+ d : 0,1 0,05 = 0,05 mol , n BaCO3 = 0,05 mol Khi lng kt ta : 197.0.05 = 9,85 gam Nguoithay.vn -8- 9. Nguoithay.vn Chn CPHNG PHP QUY ICc em thn mn , gii nhng bi ton hn hp gm nhiu cht nu gii bng cch thngthng , 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 I hn 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 th quy i chng thnh Fe3O4 . Vy hn hp trn ch gm Fe3O4 .n Fe3O4 = 2,32 : 232 = 0,01 molFe3O4 + 8HCl 2FeCl3 + FeCl2 + 4H2O0,010,08 mol n HCl = 0,08 mol V HCl = 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 lA. 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 + H2Ox 0,06 y 11,362y.63242x 0,06.30 18.yTrong : x mol Fe(NO3)3 c : 3x mol N , Trong 0,06 mol NO c 0,06 mol N S mol N v tri : 3x + 0,06 molTheo nh lut bo ton nguyn t : n N v tri = 3x + 0,06 n HNO3 = 3x + 0,06 (v 1 mol HNO3 tng ng vi 1 mol N ) .Mt khc trong y mol H2O c 2y mol H S H HNO3 cng l 2y ( Bo ton H ) n HNO3 = 2y mol 3x + 0,06 = 2y 3x 2y = -0,06 (1)p dng nh lut bo ton khi lng : 11,36 + 126y = 242x + 1,8 + 18y 242x 108y = 9,56 (2)Nguoithay.vn -9- 10. Nguoithay.vnGiai (1) , (2) x = 0,16 mol , y = 0,27 m Fe(NO3)3 = 242.0,16 = 38,72 gam Chn A Cch 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 3e Fe+3 x 3x O + 2e O-2 y 2y N+5 + 3e N+20,18 0,06 Theo nh lut bo ton e : 3x 2y = 0,18 Tng khi lng hn hp l : 56x + 16y = 11,36 Gii h x = 0,16 mol , y = 0,15 mol Vit phn ng ca Fe vi HNO3 Fe + HNO3 Fe(NO3)3 + NO + H2O 0,160,16 mol khi lng mui l : 0,16.242 = 38,72 gamCu 3 : Cho 9,12 gam h n hp gm FeO, Fe 2 O 3 , Fe 3 O 4 tc dng vi dung dch HCl (d) . Saukhi cc phn ng xy ra hon ton, c dung dch Y; c cn Y t hu c 7,62 gam FeCl 2 vm gam FeCl 3 . Gi tr ca m l :A. 9,75B. 8,75 C. 7,80 D. 6,50Ta c th q uy i F e 3 O 4 = Fe O + Fe 2 O 3 L c n y h n h p ch t r n ch c n F eO , Fe 2 O 3Fe O + 2 H Cl F e Cl 2 + H 2 Ox xFe 2 O 3 + 6 H Cl 2 F e Cl 3 + 3 H 2 OY2yG i x , y l s mo l c a c h t Fe O , F e 2 O 3 . m ch t r n = 7 2 x + 1 6 0 y = 9 ,1 2 g a mKh i l n g mu i F e Cl 2 l : 1 2 7 x = 7 ,6 2Gia i h : x = 0 ,0 6 mo l , y = 0 ,0 3 mo l Kh i l n g mu i Fe C l 3 = 2 .0 ,0 3 .1 6 2 ,5 = 9 ,7 5 g a mCh n 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 lA. CaB. BaC. KD. NaDng phng php quy i nguyn t : M , M2On quy i thnh M , OTa c : Mx + 16y = 2,9x = 0,5.0,04 = 0,02M + nH2O M(OH)n + n/2H2x xM Mn+ + ne;O + 2e O2-;2H+ + 2e H2x----------nxy--2y0,02 0,01Bo ton s mol e nx = 2y + 0,02Gii h ta c : n = 2 ; M = 137 Bap n BCu 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. Nguoithay.vn- 10 - 11. Nguoithay.vnn NO = 0,15 mol ; kim loi d l Cu Ch to thnh mui st IIQuy i hn 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+2x--2xy--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 = 0,6Mui khan : Cu(NO3)2 = 0,375.188 = 70,5 ; Fe(NO3)2 = 0,45.180 = 81 Tng khi lng : 151,5 gam .p n ACu 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 mQuy i Cu , CuO , Cu2O v hai nguyn t : Cu , OCu 2e Cu +2O + 2e O-2S + 2e S +6 +4Gi x , y l s mol ca Cu , O 2x = 2y + 0,4 x y = 0,264x + 16y = 24,8 x = 0,35 , y = 0,15 m = 64.0,35 = 22,4Cu 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 mQuy i hn hp Fe , FeCl2 , FeCl3 thnh Fe , ClS cho nhn e :Fe 3e Fe+3Cl- + e Cl-1S+6 + 2e S+4Gi x , y l s mol ca Fe , Cln SO2 = 0,2 mol 3x = y + 0,4Fe Fe+3 Fe(OH)3 Kt ta l Fe(OH)3 c n = 0,3 x = 0,3 y = 0,5Do m = 0,3.56 + 0,5.35,5 = 34,55Cu 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 gamkim loi .Tnh nng mol/lit ca dung dch HNO3Dng 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 IIFe - 2e Fe+2x2xO + 2e O-2y2yN+5 + 3e N+20,30,1Nguoithay.vn- 11 - 12. Nguoithay.vnTheo 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 0,270,1Bo ton nguyn t N (1) s mol HNO3 = 2.0,27 + 0,1 = 0,64 Nng mol ca HNO3 : 0,64 / 0,2 = 3,2Cu 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 mLi 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 -22H + 2e H2+1 2a + 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 = 0,11 Tng s mol Fe : 0,06 + 0,04 = 0,1 Khi lng Fe : 0,1 . 56 = 5,6 gamCu 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 gamB. 9 gamC. 8,2 gam D. 10,7 gamHn hp FeS , FeS2 , S c to ra t cc nguyn t : Fe , SS mol NO2 = 53,76/22,4 = 2,4 molS bin i e :Fe 3e Fe+3x 3xS - 6e S+6y 6yN+5 + 1e N+42,4 2,4 molTng s mol e cho = tng s mol e nhn 3x + 6y = 2,4Tng khi lng hn hp ban u : 56x + 32y = 20,8Gii h ta c : x = 0,2 , y = 0,3 mol Theo di s bin i cc nguyn t :Fe Fe3+ Fe(OH)3 Fe2O30,2 0,1 ( Bo ton nguyn t Fe )S SO42- Na2SO4 tan Khi lng cht rn l : khi lng Fe2O3 : 0,1.160 = 16 gam Chn ANguoithay.vn- 12 - 13. Nguoithay.vnPHNG 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 khV 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- H2O2.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 4.Ho tan hn hp kim loi gm Na, Ba vo dung dch gm HCl , H2SO4Nguoithay.vn - 13 - 14. Nguoithay.vnHCl H+ + Cl-H2SO4 2H+ + SO42-Ba + H+ Ba2+ + H2Na + H+ Na+ + H2Ba2+ + 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 + H2NaOH 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)2CO3Na + H2O NaOH + H2Ca + H2O Ca(OH)2 + H2NaOH 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 + H2NaOH 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+ + H210.Trn NaOH , KOH vi NaHCO3 v Ca(HCO3)2KOH K+ + OH-NaOH Na+ + OH-NaHCO3 Na + HCO3-Ca(HCO3)2 Ca2+ + 2HCO3-Nguoithay.vn- 14 - 15. Nguoithay.vnOH- + 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 tinPhn II : Bi tpCu 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 thnhNaOH Na+ + OH-0,20,2Ba(OH)2 Ba2+ + 2OH-0,1 0,10,2HCl H+ + Cl-0,15 0,15H2SO4 2H+ + SO42-0,3 0,60,3 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 0,750,4Phn ng0,4 0,4Kt thc0,350 Ba2+ + SO42- BaSO4 Ban u 0,10,3Phn ng 0,10,10,1Kt thc0 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 = 0,6Khi lng kt ta : 233.0,1 = 23,3 gamCu 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-xxBa(OH)2 Ba2+ + 2OH-Nguoithay.vn- 15 - 16. Nguoithay.vny y2yNaHSO4 Na+ + HSO4-0,015 0,015Ba(HSO4)2 Ba + 2HSO4- 2+0,030,030,06Tng 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 u0,055 0,075Phn ng 0,055 0,055 0,055Kt thc0 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 taY ln nht th gi tr ca m lA.1,17 B.1,71 C.1,95 D.1,59Gi 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 xKOH K+ + OH-2x xAl2(SO4)3 2Al3+ + 3SO42-0,020,040,06 Tng s mol OH- : 0,09 + x ; Al3+ : 0,04 molAl3++ 3OH- Al(OH)3 (1)0,04 0,09 + xBa2+ + 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 gamCu 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 molOH- : 0,06 mol Na+ : 0,02 mol 2a + 0,02 = 0,06 a = 0,02 molNguoithay.vn- 16 - 17. Nguoithay.vnDung dch (2)HCO3- : 0,04 molCO3- : 0,03 molNa+ : b mol 0,04 + 2.0,03= b b = 0,1Phn ng ha hc ca cc ion OH- + HCO3- CO32- + H2OBan u 0,06 0,04Phn ng 0,04 0,040,04Kt thc 0,02 0 0,04Tng s mol CO32- : 0,04 + 0,03 = 0,07 molBa2+ + CO32- BaCO3Ban u 0,020,04Phn ng0,020,02 0,02Kt thc0 0,020,02 Khi lng kt ta thu c l : 0,02.197 = 3,94 gamCu 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 ) .HCl H+ + Cl-0,25 0,25H2SO4 2H+ + 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+ + 3H2 H+ phn ng = 2.n H2 = 0,475 mol H+ d : 0,5 0,475 = 0,025 mol [H+] = 0,025/0,25 = 0,1 PH = 1s : PH = 1Cu 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 , BaCl2Na2O + H2O 2NaOHa 2aNaOH Na+ OH-2a2aNH4Cl NH4+ + Cl-a aNaHCO3 Na+ + HCO3-aaBaCl2 Ba2+ + 2Cl-a aCc phng trnh ion :OH- + NH4+ NH3 + H2Oaa OH- d a mol , NH4+ htOH- + HCO3- CO32- + H2Oaa c OH- , HCO3- u htCO3 + Ba2+ BaCO3 2-a a C CO32- , Ba2+ u htVy dung dch ch cn ion Cl- , Na+ Mui NaClNguoithay.vn - 17 - 18. Nguoithay.vnCu 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)2 Ba2+ + 2OH-0,010,01 0,02NaOH Na+ + OH-0,010,01H2SO4 2H+ + SO42-0,0150,03 0,03HCl H+ + Cl-0,005 0,005 Tng s mol OH- : 0,03 mol , H+ 0,035 molMc d c phn ng : Ba2+ + SO42- BaSO4 nhng n khng nh hng n PH Khng xt n . OH-+ H+ H2OBan u 0,03 0,035Phn ng 0,03 0,03Kt thc 0 0,005Tng th tch ca dung dch sau phn ng : 100 + 400 = 500 ml = 0,5 lt [H+] = 0,01 PH =2Cu 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 khB. vn cC. Si bt kh v vn c D. Vn c, sau trong sut tr li.Ca(HCO3)2 Ca2+ + 2HCO3-aa2aCa(HSO4)2 Ca2+ + 2HSO4-a a 2aHCO3- + HSO4- CO2 + SO42- + H2Oaa aaCa2+ + SO42- CaSO4 C kt ta to thnh v si bt khCu 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.5/4C.3/4D.4/3NaOH Na+ + OH-0,1V10,1V1Ba(OH)2 Ba2+ + 2OH-0,2V1 0,4V1 Tng s mol OH- : 0,5V1H2SO4 2H+ + SO42-0,1V20,2V2HCl H+ + Cl-0,2V20,2V2 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 + V2 n OH- = 0,1V1 + 0,1V2 0,5V1 0,4V2 = 0,1V1 + 0,1V2 0,4V1 = 0,5V2 V1 : V 2 = 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 ?Nguoithay.vn- 18 - 19. Nguoithay.vnA. 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 + CaCl2 CaCO3 + 2NaClBan u a a+bPhn ng aa aKt thc 0a aKt ta thu c l : a molTrng hp 2 Ca(OH)2 Ca2+ + 2OH- a+ba + b 2(a+b) NaHCO3 Na+ + HCO3- A aa Na2CO3 2Na+ + CO32- B2bbPhng trnh ion : OH- + HCO3- CO32- + H2OBan u a + b aPhn ng a a aKt thc b0a Tng s mol CO32- sau : a + bTng s mol Ca2+ : a + bPhng trnh ion khc : Ca2+ + CO32- CaCO3 Kt ta thu c l : a + b molCu 13 :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 V2Trng hp 1 :n Cu = 3,84/64 = 0,06 molHNO3 H+ + NO3-0,080,08 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + H2OBan u 0,06 0,08 0,08Phn ng0,08 0,02Nhn thy : 0,06/3 > 0,08>8 > 0,08/2 H+ ht Tnh theo H+ Th tch kh NO : 0,02.22,4Trng hp 2 :n Cu = 0,06 molH2SO4 2H+ + SO42-0,040,08HNO3 H+ + NO3-0,08 0,08 0,08 Tng s mol ca H+ : 0,16 mol 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + H2OBan u0,06 0,16 0,08Nguoithay.vn- 19 - 20. Nguoithay.vnPhn ng0,160,04Nhn thy : 0,06/3 = 0,16>8 > 0,08/2 Cu v H+ ht Tnh theo H+ Th tch ca NO : 0,04.22,4 Th tch V2 = 2V1Cu 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 lnB.C kh bay ln v c kt ta xanhC.C kt taD.C kh bay ln v c kt ta mu xanh sau kt ta li tan .CuSO4 Cu2+ + SO42-0,10,1H2SO4 2H+ + SO42-0,2 0,2Khi cho Na vo dung dch Na phn ng vi H+Na + H+ Na+ + H2Ban u 0,3 0,2Phn ng0,2 0,2 0,2Kt thc 0,1 0D Na : 0,1 molNa + H2O NaOH + H20,1 0,1NaOH Na+ + OH-0,1 0,1Cu2+ + 2OH- Cu(OH)2Ban u 0,10,1Phn ng0,05 0,050,05 mol Kt ta Cu(OH)2 : 0,05 molCu 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 molCl- : 0,1 mol NO3- : 0,2 mol 2a + 2b + 2c = 0,1 + 0,2 = 0,3 molK2CO3 2K+ + CO32-V.1 1.VMg2+ + CO32- MgCO3 A aBa + CO32- BaCO3 2+B bCa2+ + CO32- CaCO3 C c Tng s mol CO32- phn ng : a + b + c = 0,15 mol V = 150 mlCu 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:19,7B:17,72C:9,85D: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-Nguoithay.vn - 20 - 21. Nguoithay.vnCO2 + 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- + H2OBan u0,20,05 Tnh theo HCO3- : S mol CO32- = 0,05 molTip tc c phn ng : CO32- + Ba2+ BaCO3 Ban u0,05 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 = 1Cu 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 dungdch 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 0,125Do H+ htNa v K ( Gi chung l R ) phn ng vi H2O thu c 0,15 0,125 = 0,025 mol :2R + H2O 2ROH + H20,025 S mol OH- = 0,05 mol Khi 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 gamCu 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 HClGi 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)0,05 phn ng (2) : HCO3 l 0,05 mol , H l 0,05 mol - +V dung dch c phn ng vi nc vi trong d to kt ta nn HCO3- dHCO3- + 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 molVy H tham gi (1) , (2) l : 015 + 0,05 = 0,2 CM = 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 BaCl20,3M 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 + H2 S mol OH- = 0,25 molNguoithay.vn - 21 - 22. Nguoithay.vnBa2+ : 0,3.0,2 + 0,8.0,2 = 0,22 molHCO3- : 0,32 molPhn ng : OH- + HCO3- CO32- + H2O0,25 0,25 0,25 Kt ta BaCO3 tnh theo Ba2+ : 0,22.197 = 43,34Cu 22 : Cho 200 ml gm HNO3 0,5M v H2SO4 0,25M tc dng vi Cu d c V lit NO (ktc) c cndung 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 ; 12,4D.1,12 ; 12,7HNO3 H+ + NO3-0,10,1 0,1H2SO4 2H+ + SO42-0,050,1 Tng s mol ca H+ : 0,2 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O 0,20,1 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 = 12,7Th tch kh NO l : 0,05.22,4 = 11,2 ltCu 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 gamXt trng hp kim loi M ha tr II c phn ng vi axitHCl H+ + Cl-3x3x 3x( x l s mol ca H2SO4 )H2SO4 2H+ + SO42-x2xx Tng s mol H+ l : 5xV kim loi d nn H+ ht2H+ + 2e H25x 2,5x 2,5x = 0,5 mol x = 0,2 molV kim loi d : 3,4 gam nn khi lng kim loi phn ng : 20 3,4 = 16,6 gam Khi 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 gamCu 24: Dung dch X cha cc ion: Fe3+, SO4 2-, 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.NaOH Na+ + OH-BaCl2 Ba2+ + 2Cl-n kh = 0,672/22,4 = 0,03 molOH- + NH4+ NH3 + H2O (1) 0,030,03Fe3+ + 3OH- Fe(OH)3 (2)0,010,01Nguoithay.vn- 22 - 23. Nguoithay.vnss Kt ta l : Fe(OH)3 n 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 ion m mui = 0,03.18 + 0,01.56 + 0,02.96 + 0,02.35,5 = 3,73 gam Chn ACu 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 molHCl H+ + Cl-0,5 0,5H2SO4 2H+ + SO42-0,140,28 Tng s mol ca H+ l : 0,5 + 0,28 = 0,78 molS nhn e :2H+ + 2e H20,39 S mol H+ nhn l : 0,39.2 = 0,78 H+ 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 H2SO4 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 lA. 1,2B. 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- H2OBan 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 - -p n DNguoithay.vn- 23 - 24. Nguoithay.vnBI TON LIN QUAN N PHCc cng thc cn nh :p H = -lg[H+][H+].[OH-] = 10-14p 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 CH3COOH0,4M l .A.0,4B.2,59C.5,14D.3,64CH3COOH CH3COO- + H+B0,4Ply 0,4. 0,4. 0,4. [H+] = 0,4.1,8.10-5 = 0,72. 10-5 p H = 5,14Cu 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.2,5B.2,7 C.5,2 D.3,5Dung 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 = 2.10-3Nguoithay.vn- 24 - 25. Nguoithay.vn pH = 2,7Cu 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.2,0B.4,1C.4,9D.1,4HCl H+ + Cl-NaOH Na+ + OH-0,03 0,03 0,03 0,0280,028+-H+ OH H2O0,030,028 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 .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 = 9V1B.V1 = 1/3 V2 C.V1 = V2 D.V1 = 3V2Ban u th tch V1 : PH = 3 [H+] = 0,001 mol n H+ = 0,001V1Thm 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.0,55MC.0,75MD.0,85Mn Ba(OH)2 = 0,05 mol ; n KOH = 0,05 molBa(OH)2 Ba2+ + 2OH- KOH K+ + OH-0,05 ------------------0,10,05----------0,05 Tng s mol OH- = 0,15 mol [OH-] = 0,15/0,2 = 0,75 MCu 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.0,003MC.0,25M D.0,025MDung dch H2SO4 c pH = 1 [H+] = 0,1 M 50 ml dung dch trn c n H+= 0,005 mol ,n SO42- = 0,0025 molKOH K+ + OH-0,005 --0,0005------0,005 molPhn ng : H++OH- H2OBan 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 mlNguoithay.vn - 25 - 26. Nguoithay.vn CM = 0,0025/0,1 = 0,025 molCu 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.5/4C.3/4D.4/3NaOH Na+ + OH- Ba(OH)2 Ba2+ + 2OH-0,1V1 ------------0,1V1 0,2V1 ---------------0,4V1 Tng s mol OH- = 0,5V1H2SO4 2H+ + SO42-HCl H+ + Cl-0,1V2------0,2V2 0,2V2------0,2V2 Tng s mol H+ = 0,4V2Phng trnh : H+ +OH- H2OBan u0,4V2 0,5V1pH = 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 = 5/4Bi 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+ + SO42-0,1250,25 Tng s mol H+ = 0,25 + 0,25 = 0,5 MPhng trnh phn ng :Al + 6H+ Al3+ + 3H2 Mg + 2H+ Mg2+ + H2 S mol H+ phn ng l : 2.n H2 = 2.0,2375 = 0,475 mol H+ 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 ABi 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 0,02Nguoithay.vn - 26 - 27. Nguoithay.vn NaOH Na+ + OH- 0,01 0,01 Tng s mol OH- : 0,02 + 0,01 = 0,03 mol H2SO4 2H+ + SO42- 0,0150,03 HCl H+ + Cl- 0,005 0,005 Tng s mol ca H+ : 0,035 mol Phn ng :H+ + OH- H2O Ban u 0,035 0,03 Phn ng0,03 0,03 Kt thc 0,005 0 Sau 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 BCu 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 ; V2H2SO4 2H+ + SO42- NaOH Na+ + OH-0,02V1--0,04V1 0,035V2---------0,035V2Phn ng : 2H+ + OH- H2OBan u : 0,04V1 0,035V2+ D H = 0,04V1 0,035V2 = 0,01V1 + 0,01V2 0,03V1 = 0,045V2 V1 = 1,5V2Cu 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 H2OBan 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 Nguoithay.vn - 27 - 28. Nguoithay.vn 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 H2OBan u 11,8V1 OH- d = 1,8V1 1 = 0,1(V1 + 0,5) 1,7V1 = 1,05 V1 =Cu 13 : Trn 250 ml dung dch hn hp HCl 0,08 mol/l v H2SO4 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 dungdch 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 0,01V0,01VHCl H+ + Cl-0,03V 0,03V 0,03VH++ OH- H2OBan u 0,01V 0,03V OH- d : 0,03V 0,01V = 0,02 V , Tng th tch l : 2V [H+] = 0,02V/2V = 0,01 PH = - lg0,01 = 2 Chn C .Cu 16 : Trn 100 mldung d ch c pH = 1 gm HCl v HNO 3 vi 100 ml dung d ch NaOHnng a (mol/l) thu c 200 ml dung d ch c pH = 12. Gi tr ca a l (bi t trong midung dch [ H + ][OH - ] =10 -14)A. 0,15 B.0,30C. 0,03 D. 0,12n H C l = x mo l , n H N O 3 = y mo lHCl H+ + Cl-xxH NO3 H + + N O 3 -yy T n g s mo l H + : x + yV P H = 1 [ H + ] = 0 ,1 n H + = 0 ,1 . 0 , 1 = 0 ,0 1 mo l x + y = 0 ,0 1 m o lN Na OH = 0 ,1 .aNa O H Na + + OH -0 ,1 .a0 ,1 .aC p h n n g : H+ + OH - H 2 O ( 1 )Ba n u 0 ,0 1 mo l 0 ,1 .aV P H =1 2 OH - d ,PH =1 2 [ H + ] = 1 0 - 1 2 [ OH - ] = 1 0 - 2 n O H - = V.[ OH - ] = 1 0 - 2 .0 ,2 = 0 , 0 0 2mo lTh eo (1 ) O H - d : 0 ,1 .a 0 ,0 1 = 0 ,0 0 2 a = 0 , 1 2 MCh n p n DNguoithay.vn - 28 - 29. Nguoithay.vnCu 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 lA. 1,2B. 1,0 C. 12,8D. 13,0 +Tng 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- H2OBan 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 BI TP V HNO3 V MUI NITRAT Cu 1: Nung nng hn hp 27,3 gam hn hp NaNO3 , Cu(NO3)2 . Hn hp kh thot ra c dn vo 89,2 ml nc th cn d 1,12 lt kh ktc khng b hp th . Tnh khi lng ca mi mui trong hn hp u NaNO3 NaNO2 + O2 (1) a a/2 Cu(NO3)2 CuO + 2NO2 + O2 (2) b 2bb 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 H2SO4 1M . NO l kh duy nht thot ra . Xc nh mi quan h gia V1 v V2 bit rng kh thot ra hai th nghim l nh nhau . Th nghim (1) : HNO3 H+ + NO3- 4V14V1 4V1 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O 4V1 4V1 Tnh theo H+V1 Th nghim (2) : HNO3 H+ + NO3- 3V23V2 3V2 H2SO4 2H+ + SO42- V22V2 Tng s mol ca H+ : 5V2 , S mol ca NO3- : 3V2 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O 5V23V2 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 lt NO.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- Nguoithay.vn- 29 - 30. Nguoithay.vn0,080,08 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O0,060,08 0,08 Tnh theo H+0,02 molTh nghim (2) :HNO3 H+ + NO3-0,080,08 0,08H2SO4 2H+ + SO42-0,040,08 Tng s mol ca H+ : 0,16 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O0,060,16 0,08 0,04 Tnh theo H+ v Cu S mol k trong trng hp ny l : 0,04 V2 = 2V1Cu 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 ; 10,8C.1,12 ; 12,4 D.1,12 ; 12,7HNO3 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 0,20,1 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 ltCu 5 : Cho 0,96 gam Cu vo 100ml dung dch cha ng thi KNO3 0,08M v H2SO4 0,2M sinhra 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 litB. 0,3584 litC. 0,448 lit D. 0,336 litKh c t khi so vi H2 l 15 NOKNO3 K++ NO3-0,008 0,008H2SO4 2H+ + SO42-0,020,02 Tng s mol ca H+ : 0,2 , S mol ca NO3- : 0,083Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O0,015 0,02 0,0080,008 mol Tnh theo NO3- --------------------------- Th tch kh NO : 0,008.22,4 = 0,1792 ltCu 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 gamB.0,49 gamC.9,4 gam D.0,94 gamCu(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 gamCu 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 molNguoithay.vn - 30 - 31. Nguoithay.vnNhn thy : Cu - 2e Cu2+0,3 0,6 molNO3- NO NO2 NO3- Nh vy N khng thay i s oxi ha trong c qu trnhO2 4e 2O-2 0,6 mol s mol O2 phn ng : 0,6/4 = 0,15 VO2 = 3,36 ltCu 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 khNO (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 lA. 360. B. 240.C. 400. D. 120.Cu 9 : Cho 2,16 gam Mg tc d ng vi dung dch HNO3 (d). Sau khi ph n ng xy rahon ton thu c 0,896 lt kh NO ( kt c) v dung d ch X. K hi lng mui khanthu c khi lm bay hi dung d ch X l :A. 8,88 gam B. 13,92 gamC. 6,52 gam D. 13,32 gam3 Mg + 8 H N O 3 3 Mg ( NO 3 ) 2 + 2 N O + 4 H 2 O ( 1 )0 ,0 60 ,0 6 mo l0 ,0 4 mo l4 Mg + 1 0 H N O 3 4 Mg ( NO 3 ) 2 + NH 4 NO 3 + 3 H 2 O (2 )0 ,0 30 ,0 30 ,0 0 7 5 mo ln N O = 0 ,8 9 6 /2 2 ,4 = 0 , 0 4 mo l , n M g = 0 ,0 9 mo l , t p h n n g (1 ) n M g (1 ) = 3 /2 n N O =3 /2 .0 ,0 4 = 0 ,0 6 mo l . n M g (2 ) = 0 ,0 9 0 ,0 6 = 0 ,0 3 mo l ,T p (1 ) n M g ( N O 3 ) 2 = 0 ,0 6 mo lT p (2 ) n M g ( N O 3 ) 2 = 0 ,0 3 mo l , n N H 4 N O 3 = 0 , 0 0 7 5 mo l Kh i l n g mu i kh a n kh i l m b a y h i d u n g d ch l : 1 4 8 .0 ,0 9 + 0 ,0 0 7 5 .8 0 = 1 3 ,9 2 g C h n p n BNguoithay.vn - 31 - 32. Nguoithay.vnCu 10 : Th tch dung dch HNO3 1M (long) t nh t cn dng ho tan hon tonmt hn hp gm 0,15 mol Fe v 0,15 mol Cu l (bi t phn ng t o cht kh duy nhtl NO)A. 1,0 lt B. 0,6 lt C. 0,8 lt D. 1,2 ltNhn xt : Lng HNO 3 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 3xx Theo nh lut bo ton mol e : 0,3 + 0,3 = 3x x = 0,2 molFe , Cu + HNO 3 Fe(NO 3 ) 2 + Cu( NO 3 ) 2 + NO + H 2 O0,15 0,150,150,150,2 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(NO3)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 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 = 0,15.56 p(1) + 0,16.56 p(2) + (0,6m 0,16.64 ) d m = 17,8Mt khc V NO = 0,1.22,4 = 2,24 Chn p n BCu 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 mlB.720 mlC.780 mlD.840 ml10Al + 36HNO3 10Al(NO3)3 + 3N2 + 18H2O0,1 0,3610Fe + 36HNO3 10Fe(NO3)3 + 3N2 + 18H2Ox3,6xx ( x l s mol Fe phn ng vi HNO3 )Fe + 2Fe(NO3)3 3Fe(NO3)2y 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 = 1/30 Tng s mol : HNO3 phn ng l : 0,36 + 3,6.4/15 = 1,32 mol V = 1,32/2 = 0,66 lit = 660 mlNguoithay.vn - 32 - 33. Nguoithay.vnBI TP V NHML 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 + 3BaSO4Phn ng nhit nhm : Al + FexOy Al2O3 + FeSau 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 hocFexOy 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)3V 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 = 0,23NaOH + Al(NO3)3 NaNO3 + Al(OH)3Ban u 0,7 0,2Phn ng0,6 0,2 0,2Kt thc0,1 00,2 C tip phn ng :NaOH + Al(OH)3 NaAlO2 +H2OBan u 0,1 0,2Phn ng0,1 0,1Kt thc0,1Vy kt ta Al(OH)3 thu c l 0,1 mol Khi lng 7,8 gamLuyn tp :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 taY ln nht th gi tr ca m l . A.1,17 B.1,71C.1,95D.1,59Nguoithay.vn - 33 - 34. Nguoithay.vnDng 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 Al 2(SO4)3 = 0,02 molK + 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 molAl2(SO4)3 2Al3+ + 3SO42-0,020,04 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 BCu 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 ltTH2 : C c hai phn ng :AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1) 0,30,9 0,3Al(OH)3 + NaOH NaAlO2 + H2O (2) x xx 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 mln HCl = 0,2 mol ; n AlCl3 = 0,1 molHCl + NaOH NaCl + H2O (1)0,20,2AlCl3 + 3NaOH Al(OH)3 + 3NaCl (2)0,10,3Phn 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 ltNguoithay.vn- 34 - 35. Nguoithay.vnCu 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 lMgCl2 +2NaOH Mg(OH)2 + 2NaCl (1)V.12V.1AlCl3 + 3NaOH Al(OH)3 + 3NaCl (2)V.13V.1 kt ta ln nht th (1) , (2) va Tng s mol NaOH phn ng : 5V = 0,5 V = 0,1 ltCu 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.12V.1AlCl3 + 3NaOH Al(OH)3 + 3NaCl (2)V.13V.1V.1Al(OH)3 + NaOH NaAlO2 + H2O (3)V.1V.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 mlCu 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.AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)Al(OH)3 + NaOH NaAlO2 + H2O (2)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 ltTH2 : C c hai phn ng :AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1) 0,150,45 0,15Al(OH)3 + NaOH NaAlO2 + H2O (2) xx 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 V NaOH dng = 0,48/0,2 = 2,4 ltS : 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 nCu 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 HCln cht rn (Al2O3) = 0,075 molC th c hai phn ng :HCl + NaAlO2 + H2O Al(OH)3 + NaCl .0,2 0,20,23HCl + Al(OH)3 AlCl3 + H2O3xxNguoithay.vn - 35 - 36. Nguoithay.vnV kt ta thu c c 0,075 mol Nn chc chn phi c hai phn ng .Al(OH)3 Al2O3 + H2O0,2 x0,2 - 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 molCu 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 + 1/2H2x x1/2xAl + 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 + 3/2H2Th 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 + H2x x x2Al + Ba(OH)2 + 2H2O Ba(AlO2)2 + 3/2 H2x 3/2xTng s mol H2 = x + 3/2x = 2,5x = 0,06Phn II :Ba + H2O Ba(OH)2 + H2x x x2Al + Ba(OH)2 + 2H2O Ba(AlO2)2 + 3/2 H2y3/2y x + 3/2y = 0,465 x = 0,024 ; y = 0,294 Khi lng Al : 0,294.27 = 7,938 gamCu 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.2MB.1,5MC.1MD.1,5MTrng 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 Nguoithay.vn- 36 - 37. Nguoithay.vnChng t rng . Ch c phn ng :AlCl3 + 3NaOH Al(OH)3 + 3NaCl0,08 0,24 mol 0,08 molV AlCl3 d a molAlCl3 + 3NaOH Al(OH)3 + 3NaCl a3a aLc ny Al(OH)3 c to thnh l : 0,08 + aAl(OH)3 + NaOH NaAlO2 + H2OBan u 0,08 + aPhn ng xxKt thc0,08 + 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 = 1MCu 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 l-ng khng i th -c 5,1g cht rna) Nu V = 200 ml th a c gi tr no sau y:A. 2MB. 1,5M hay 3MC. 1M hay 1,5M D. 1,5M hay 7,5Mb) Nu a = 2 mol/l th gi tr ca V l:A. 150 mlB. 650 mlC. 150 ml hay 650 ml D. 150 ml hay 750 mlBi 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 ng AlCl3 + 3NaOH Al(OH)3 + 3NaCl(1)TH2 : C c hai phn ng :AlCl3 + 3NaOH Al(OH)3 + 3NaCl (1)Al(OH)3 + NaOH NaAlO2 + H2O (2)Bi ton nhit nhm :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 + H2OV 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 .Nguoithay.vn- 37 - 38. Nguoithay.vnCu 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. 22,75B. 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 + 3/2H2x 3/2 x S mol kh H2 thu c phn II : 3/2x = 0,84/ 22,4 = 0,0375 mol x = 0,025 molPhn (1) tc dng vi H2SO4 :2Al + 3H2SO4 Al2(SO4)3 + 3H20.0250,0375Fe + H2SO4 long FeSO4 + H2y y y = 0,12Al + Fe2O3 Al2O3 + 2Fe (1)0,1 0,05 0,1Khi 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 ACu 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 +-xxAl2(SO4)3 2Al3+ + 3SO42-0,10,2H2SO4 2H+ + SO42-0,10,2n 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 0,1 Tng OH- = 0,2 + 0,3 = 0,5 mol V = 0,5/2 = 0,25 ltC c hai phn ng :Al3+ + 3OH- Al(OH)3 (1)0,20,6 0,2Al(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 mol Tng s mol OH- l : 0,2 + 0,6 + 0,1 = 0,9 mol V NaOH = 0,9/2 = 0,45 lt Chn ACu 15: Ho tan hon ton 0,3 mol hn hp gm Al v Al4C3 vo dung dch KOH (d), thu c a molhn 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.Nguoithay.vn- 38 - 39. Nguoithay.vnAl4C3 + 12H2O 4Al(OH)3 + 3CH4x4x 3xAl(OH)3 + KOH KAlO2 + H2O4x4xAl + KOH + H2O KAlO2 + 3/2 H2yy3/2y Tng th tch kh l : 3x + 3/2yDung dch gm : KAlO2 : 4x + y mol v KOH dCO2 + KAlO2 + H2O Al(OH)3 + KHCO34x + y 4x + yn kt ta = 0,6 mol 4x + y = 0,6 molx + y = 0,3 x = 0,1 ; y = 0,2 mol Tng s mol ca kh : 3.0,1 + 3/2.0,2 = 0,6 mol Chn p n BCu 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 + H2x x x/2 Al + NaOH + H2O NaAlO2 + 3/2 H2 (2)Ban u 2xx NaOH 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 = 2x 2x = 0,4 x = 0,2 mol Vy Al d : 0,2.27 = 5,4 gam . Chn BCu 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 mol2Al + Fe2O3 Al2O3 + Fe (1)0,20,1 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 + H2O NaAlO2 + 3/2 H2(2)0,1 0,1 mol 0,15 mol T (2) n NaOH (2) = 2/3n H2 = 0,1 molT (1) n NaOH = 2n NaOH = 0,3 mol Tng s mol NaOH = 0,2 + 0,1 = 0,3 mol V NaOH = 0,3/1 = 300 mlChon CCu 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 lNguoithay.vn - 39 - 40. Nguoithay.vn A. 54,0 B. 75,6C. 67,5 D. 108,0Ti catot(-) :Al3+ + 3e AlTi anot (+) : O-2 2e O2Kh oxi sinh ra anot t chy dn C : C + O2 CO2CO2 + C 2COCo 4e C+4 v Co 2e C+2Phng trnh in phn :2Al2O3 + 3C 4Al + 3CO2 (1)0,8---0,6Al2O3 + C 2Al + 3CO (2)1,2 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-Nguoithay.vn - 46 - 47. Nguoithay.vn Phng trnh : 2Cl- + 2H2O Cl2 + H2 + 2OH-2NaCl + 2H2O 2NaOH + Cl2 + H2Tng 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 :Cu(NO3)2 Cu2+ + 2NO3-Catot(-) Anot (+) NO3- khng b in phn .Cu + 2e Cu 2+2H2O - 4e 4H+ + O2 Phng trnh : Cu + H2O Cu + 2H+ + O2 2+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 + O2VD3 : in phn dung dch Na2CO3 : Na2CO3 2Na+ + CO32-Catot (-)Anot (+)Na+ khng b in phn CO32- khng b in phn H2O + 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 + H2Cng thc Faraday : S mol e trao i mi in cc :n = It/96500I l cng dng in , t l thi gian tnh bng sHoc : n = It/96500.nen e : ha tr ca kim loiNguoithay.vn - 47 - 48. Nguoithay.vnBI TON IN PHNCu 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 = 1000sb) t1 = 1000s, t2 = 1500sc)t1 = 500s, t2 = 1200sd) t1 = 500s, t2 = 1500sn 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 ; 0.64gb) 0.64g ; 1.28gc) 0.64g ; 1.32gd) 0.32g ; 1.28gn 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 + O2Khi 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 gamKhi 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 .Nguoithay.vn - 48 - 49. Nguoithay.vnCu 3:in phn 100ml dung dch CuSO4 0.1M cho n khi va bt u si bt bn catotth ngng in phn. Tnh pH dung dch ngay khi y vi hiu sut l 100%.Th tch dungdch c xem nh khng i. Ly lg2 = 0.30. a) pH = 0.1 b) pH = 0.7 c) pH = 2.0 d) pH = 1.3n 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 . Cu + 2e Cu2+ 2H2O - 4e 4H+ + O20,02 0,01 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 BCu 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) 100sb) 50sc) 150sd) 200sV dung dch c PH = 12 Mi trng kim .p H = 12 [H+] = 10-12 [OH-] = 0,01 S mol OH- = 0,001 molNaCl Na+ + Cl- Catot (-) Anot (+) Na+ khng b in phnCl- + 2e Cl2 2H2O + 2e H2 + 2OH- 0,001 2a C.b < 2a D.b < 2a hoc b > 2aCu(NO3)2 Cu2+ + 2NO3-aaNaCl Na+ + Cl- b bCatot(-)Anot (+)Na+ khng b in phn NO3- khng b in phn .Cu2+ + 2e Cu2Cl- - 2e Cl2 Phng trnh : Cu + 2Cl Cu + Cl2 (1)2+-a bNu d Cu2+ sau (1) : a > b/2 ( 2a > b ) th ti Anot c phn ng : Cu2+ + 2H2O Cu + 4H+ + O2 Dung dch thu c c axit nn c phn ng vi Al2O3Nu d Cl- sau (1) : a < b/2 ( b < 2a) Th ti catot c phn ng : 2H2O + 2Cl- 2OH- + H2 + Cl2 Dung dch thu c c mi trng bazo C phn ng vi Al2O3 :NaOH + Al2O3 NaAlO2 + H2O Chn p n D .Cu 13 :in phn 500 ml dung dch A : FeSO4 v KCl vi in cc tr , gia cc in ccc mng ngn xp ngn cch . Sauk hi in phn xong anot thu c 4,48 lt kh B ktc . ca tt thu c kh C v bnh in phn thu c dung dch D . Dung dch D ha tan ti a15,3 gam Al2O3 .1.Tnh nng mol/l cc cht trong A2.Tnh th tch kh C thot ra catot3.Sau khi in phn khi lng dung dch A gim i bao nhiu gam ?FeSO4 Fe2+ + SO42-a aKCl K+ + Cl- b b Catot(-)Anot (+)Nguoithay.vn- 53 - 54. Nguoithay.vnNa+ khng b in phnNO3- khng b in phn .Fe + 2e Fe 2+ 2Cl- - 2e Cl2 Phng trnh : Fe2+ + 2Cl- Fe + Cl2abV dung dch thu c c phn ng vi Al2O3 lng tnh C th mi trng kim (Cl- d ) , C th mitrng axit (Cu2+ d )Nu d Cl- :Fe2+ + 2Cl- Fe + Cl2 (1)ab Cl- d = b 2a mol a---------------------a2Cl- + H2O OH- + Cl2 + H2(2)b 2a---------b 2a-------b/2-a----b-2aTheo gi thit : S mol kh thu c anot : Cl2 : a + (b/2-a) = b/2 = 0,2 mol b = 0,42OH- + Al2O3 2AlO2- + H2O 0,3---0,15 b 2a = 0,3 mol a = 0,05 mol Tha mn .A.Nng mol ca FeSO4 = 0,05/0,5 = 0,1M ; Nng mol ca KCl = 0,4/0,5 = 0,8MB.Kh thot ra catot : theo (2) S mol H2 = b -2a = 0,4 0,1 = 0,3 mol V H2 = 6,72 ltC.Khi lng dung dch gim = khi lng Fe + H2 + Cl2 = 0,05.56 + 0,3.2 + 0,2.35,5 = 10,5 gNu Fe2+ d : Kh thot ra anot l Cl2 v O2Fe2+ + 2Cl- Fe + Cl2ab Fe2+ d = a b/2 mol b-----------b/2Fe2+ + H2O Fe + 1/2O2 + 2H+ (3)a-b/2------------------------- a/2-b/4 ----2a bAl2O3 + 6H+ 2Al3+ + 3H2O0,15--0,9 2a b = 0,9Tng Kh Cl2 v O2 : (a/2 b/4) + b/2 = 0,2 a + b = 0,4 gii h c nghim m loiCu 14 : in phn 200 ml dung dch hn hp gm HCl 0,1M v CuSO4 0,5M bng incc tr . Khi catot c 3,2 gam Cu th th tch kh thot ra Anot lA.0,56 ltB.0,84 ltC.0,672 ltD.0,448 ltCuSO4 Cu2+ + SO42- 0,1 0,1HCl H+ + Cl- 0,020,02 Catot(-)Anot (+)SO42- khng b in phn .Cu2+ + 2e Cu 2Cl- - 2e Cl2 0,1--0,050,02-0,012H2O 4e 4H+ + O20,08 --------0,02 molKhi catot thot ra 3,2 gam Cu tc l 0,05 mol S mol Cu2+ nhn 0,1 mol , m Cl- cho ti a 0,02 mol 0,08 mol cn li l H2O cho T s in phn kh thot ra ti anot l : Cl2 0,01mol ; O2 0,02 mol Tng th tch : 0,03.22,4 = 0,672 lt Nguoithay.vn - 54 - 55. Nguoithay.vn Chn p n C .I HC KHI B 2009 . Cu 15: in phn c mng ngn 500 ml dung dch cha hn hp gm CuCl2 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 ln nht ca m lA. 4,05B. 2,70 C. 1,35 D. 5,40 S mol e trao i khi in phn : n = 5.3860/96500 = 0,2 mol n CuCl2 = 0,1.0,5 = 0,05 mol ; n NaCl = 0,5.0,5 = 0,25 mol n Cu2+ = 0,05 mol , n Cl- = 0,25 + 0,05 = 0,3 mol Vy Cl- d , Cu2+ ht , nn ti catot s c phn ng in phn nc .(sao cho s mol e nhn catot l 0,2) Ti catot :Ti anot : Cu2+ + 2e Cu 2Cl- + 2e Cl2 0,05---0,10,2--0,2 2H2O + 2e H2 + 2OH- 0,1----(0,2 - 0,1)-0,1 Dung dch sau khi in phn c 0,1 mol OH- c kh nng phn ng vi Al theo phng trnh : Al + OH- + H2O AlO2- + 3/2 H2 0,1-0,1 mAlmax = 0,1.27= 2,7 (g) p n B Cu 16 : 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 lA. 54,0 B. 75,6C. 67,5 D. 108,0 Ti catot(-) :Al3+ + 3e AlTi anot (+) : O-2 2e O2 Kh oxi sinh ra anot t chy dn C : C + O2 CO2 CO2 + C 2CO Co 4e C+4 v Co 2e C+2 Phng trnh in phn : 2Al2O3 + 3C 4Al + 3CO2 (1) 0,8---0,6 Al2O3 + C 2Al + 3CO (2) 1,2 2aB. b = 2aC. b < 2aD. 2b = aBi 15: Trong cng nghip natri hiroxit -c sn xut bng ph-ng php A. in phn dung dch NaCl, khng c mng ngn in cc B. in phn dung dch NaNO3, khng c mng ngn in ccNguoithay.vn - 56 - 57. Nguoithay.vn C. in phn dung dch NaCl, c mng ngn in cc D. in phn NaCl nng chyBi 16: iu ch Cu t dung dch Cu(NO3)2 bng ph-ng php no th thu -c Cu tinh khit99,999% ? A. Ph-ng php thy luyn. B. Ph-ng php nhit luyn C. Ph-ng php in phn D. C A, B, Cc - Bi tp t giI SU TP :Bi 1: Tin hnh in phn in cc tr, c mng ngn 1 dung dch cha m(g) hn hp CuSO4, NaClcho ti khi n-c bt u in phn c 2 in cc th dng li. anot thu -c 0,448 lt kh (ktc),dung dch sau phn ng c th ho tan ti a 0,68g Al2O3.1. Tnh m2. Tnh khi l-ng catot tng trong qu trnh in phn3. Tnh khi l-ng dung dch gim I sau qu trnh in phn. (Gs n-c bay hi khng ng k)Bi 2: Ho tan 12,5g CuSO4.5H2O vo dung dch cha a(g) HCl -c 100ml dung dch X. in phndung dch X vi in cc tr, dng in 1 chiu 5A trong 386 giy.1. Vit cc PTHH c th xy ra khi in phn.2. Tnh nng mol/l cc cht tan trong dung dch sau in phn3. Sau in phn ly in cc ra ri cho vo phn dung dch 5,9g 1 kim loi M (ng sau Mg trongdy in ho). Khi phn ng kt thc, ng-i ta thu -c 0,672 lt kh (1,6atm v 54,6 0C) v lc dungdch thu -c 3,26g cht rn. Xc nh m v tnh a.4. Nu khng cho M m tip tc in phn , v nguyn tc phi in phn bao lu mi thy kh thotra K.Bi 3: Nung hon ton 45,6g hn hp 2 mui hirocacbonat ca kim loi R v R thu -c hn hpcht rn A v hn hp kh B. Cho B hp th ht vo 2 lt dung dch Ba(OH)2 0,3M (d=1,2) thu -c102,44g kt ta.Sau phn ng khi l-ng dung dch cn 2325,48g v dung dch vn c tnh baz. Ho tan ht cht rnA cn 500ml dung dch HCl 3,65% th thu -c 2 mui clorua ca R v RNu em in phn nng chy mui clorua ca R trong A th cn thi gian t(giy) vi c-ng I =10A. Trong khi , cng vi thi gian v c-ng nh- trn em in phn nng chy mui cloruaca R trong A th -c 11,04g R.a. Hy xc nh R, Rb. Tnh D ca dung dch HCl dng.Bi 4: Trong 500ml dung dch A cha 0,4925g mt hn hp gm mui clorua v hiroxit ca kimloi kim. Dung dch A c pH =12. Khi in phn 1/10 dung dch A cho n khi ht clo th thu -c11,2ml kh clo (2730C v 1atm).a) Xc nh kim loib) 1/10 A tc dng va vi 25ml dung dch CuCl2. Tm nng mol ca dd CuCl2c) Hi phi in phn 1/10A trong bao lu vi I = 96,5A -c dung dch c 1 cht tan vi pH=13.Bi 5: Tin hnh in phn dung dch hn hp gm HCl 0,01M + CuCl2 0,1M + NaCl 0,1M, in cctr, mng ngn xp. V th biu din s bin thin pH ca dung dch theo qu trnh in phn.Bi 6: in phn 100ml dung dch cha Cu2+, Na+; H+;SO42- c pH = 1, in cc tr. Saumt thi gian in phn, rt in cc ra khi dung dch, thy khi l-ng trong dungdch gim 0,64 gam v dung dch c mu xanh nht, th tch dung dch khng i.1.Vit cc ph-ng trnh phn ng xy ra trong qu trnh in phn.2. Tnh nng H+ c trong dung dch sau khi in phn.Bi 7: M l kim loi c tng s ht c bn l 87, X l halogenua. in phn dung dch MXa bngdng in 5A, in cc tr, sau 21 pht 27 giy ngng in phn, thy trn catot sinh ra 1,9575 gamkimloi M. Xc nh tn kimloi M v nguyn t X bit MXa c khi l-ng phn t l 218,7Bi 8: Nu ph-ng php tch tng cht sau ra khi hn hp: KCl; BaCl2; MgCl2Bi 9: iu ch cc kimloi t hn hp sau:NaCl; BaCl2; AlCl3; CuCl2Bi 10: Nhng qu trnh no xy ra khi in phn dung dch Cu(NO3)2 bng than ch? Sau nu ichiu dng in th iu g s xy ra?Nguoithay.vn - 57 - 58. Nguoithay.vnBi 11: Tnh thi gian in phn dung dch NiSO4 bng dng in 2A ph kn c 2 mt mt lkim loi mng c kch th-c 10 10 cm bng mt lp niken c b dy 0,05mm. Bit rng niken ckhi l-ng ring 8,9g/cm3 v hiu sut in phn l 90%.Bi 12: in phn dung dch NaCl cho n khi ht mui vi dng in 1,61A thy ht 60pht1. Tnh khi l-ng kh thot ra, bit rng in cc tr, mng ngn xp.2.Trn dung dch sau in phn vi dung dch H2SO4 cha 0,04 mol ri c cn dung dch. Tnh khil-ng mui khan thu -c.Bi 13: in phn 2 lt dung dch CuSO4 0,5M vi in cc tr. Sau mt thi gian, ngng in phnv cho i qua dung dch sau in phn mt l-ng d- kh A th thu -c 72gam cht kt ta mu en.Bit rng, khi t chy A trong oxi d- th thu -c hi n-c v kh B , B lm mt mu dung dch n-cbrom1. Xc nh cng thc phn t ca cc kh A, B2. Tnh th tch kh thot ra trn anot ktc3. Tnh th tch dung dch HNO3 60% ( d= 1,37g/ml) cn thit ho tan l-ng kim loi thot ra trncatot.Bi 14. in phn n ht 0,1 mol Cu (NO3)2 trong dung dch vi in tc tr, th sau in phn khil-ng dung dch gim bao nhiu gamA. 1,6g B. 6,4g C. 8,0 gam D. 18,8gBi 15. Tnh th tch kh (ktc) thu -c khi in phn ht 0,1 mol NaCl trong dung dch vi in cctr, mng ngn xp.A. 0,024 litB. 1,120 litC. 2,240 lit D. 4,489 litBi 16: in phn dung dch CuCl2 vi in cc tr , sau mt thi gian thu -c 0,32 gam Cu catotv mt l-ng kh X anot. Hp th hon ton l-ng kh X trn vo 200 ml dung dch NaOH nhit th-ng). Sau phn ng nng NaOH cn li l 0,05M ( gi thit th tch ca dung dch NaOHkhng thay i). Nng ban u ca dung dch NaOH l. (H KHI A 2007)A. 0,15 M B. 0,2M C. 0,1 MD. 0,05MBi 17: in phn 200 ml dung dch CuSO4 vi in cc tr bng dng in mt chiu I = 9,65 A.Khi th tch kh thot ra c hai n cc u l 1,12 lt (ktc) th dng in phn. Khi l-ng kim loisinh ra katt v thi gian in phn l:A. 3,2gam v1000 sB. 2,2 gam v 800 sC. 6,4 gam v 3600 s D. 5,4 gam v 1800 sBi 18. in phn 200ml dd CuSO4 0,5 M v FeSO4 0,5M trong 15 pht vi in cc tr v dng inI= 5A s thu -c catot:A. ch c ng B. Va ng, va stC. ch c stD. va ng va st vi l-ng mi kim loi l ti aBi 19: in phn dung dch CuSO4 bng in cc tr vi dng in c c-ng I = 0,5A trong thigian 1930 giy th khi l-ng ng v th tch kh O2 sinh ra lA: 0, 64g v 0,112 litB: 0, 32g v 0, 056 ltC: 0, 96g v 0, 168 ltD: 1, 28g v 0, 224 ltBi 20: in phn 200ml dung dch hn hp gm HCl 0,1M v CuSO4 0,5M bng in cc tr. Khi katt c 3,2g Cu th th tch kh thot ra ant lA : 0, 56 lt B : 0, 84 lt C : 0, 672 ltD : 0,448 lit Bi 21: in phn dd cha 0,2 mol FeSO4 v 0,06mol HCl vi dng in 1,34 A trong 2 gi (incc tr, c mng ngn). B qua s ho tan ca clo trong n-c v coi hiu sut in phn l 100%.Khi l-ng kim loi thot ra katot v th tch kh thot ra anot (ktc) ln l-t l:A. 1,12 g Fe v 0, 896 lit hn hp kh Cl2 , O2.B. 1,12 g Fe v 1, 12 lit hn hp kh Cl2 v O2.C. 11,2 g Fe v 1, 12 lit hn hp kh Cl2 v O2.D. 1,12 g Fe v 8, 96 lit hn hp kh Cl2 v OBi 22: Dung dch cha ng thi 0,01 mol NaCl; 0,02 mol CuCl2; 0,01 mol FeCl3; 0,06 mol CaCl2.Kim loi u tin thot ra catot khi in phn dung dch trn l :A. FeB. Zn C. CuD. CaBi 23: in phn vi in cc tr dung dch mui clorua ca kim loi ho tr (II) vi c-ng dngin 3A. Sau 1930 giy, thy khi l-ng catot tng1,92 gam. Kim loi trong mui clorua trn l kimloi no d-i y (cho Fe = 56, Ni = 59, Cu = 64, Zn = 65)A. Ni B. ZnC. CuD. FeBi 24: in phn dng in cc tr dung dch mui sunfat kim loi ho tr II vi c-ng dngin 3A. Sau 1930 giy thy khi l-ng catot tng 1,92 gam, Cho bit tn kim loi trong mui sunfat(cho Fe = 56, Ni = 59, Cu = 64, Zn = 65)A. Fe B. CaC. CuD. MgNguoithay.vn- 58 - 59. Nguoithay.vn(coi th tch dung dch khi in phn l khng i, khi c mt NaCl th dng thm mng ngn)Bi 25: Tin hnh in phn hon ton dung dch X cha AgNO3 v Cu(NO3)2 thu -c 56 gam hnhp kim loi catot v 4,48 lt kh anot (ktc). S mol AgNO3 v Cu(NO3)2 trong X ln l-t l (choAg = 108, Cu = 64) A. 0,2 v 0,3 B. 0,3 v 0,4 C. 0,4 v 0,2 D. 0,4 v 0,2Bi 26: in phn 100ml dung dch A cha ng thi HCl 0,1M v NaCl 0,2 M vi in cc tr cmng ngn xp ti khi anot thot ra 0,224 lt kh (ktc) th ngng in phn. Dung dch sau khi inphn c pH (coi th tch dung dch thay i khng ng k) l A. 6B. 7C. 12 D. 13Bi 27: in phn n ht 0,1 mol Cu(NO3)2 trong dung dch vi in cc tr, th sau in phn khil-ng dung dch gim bao nhiu gam ? ( cho Cu = 64; O = 16) A. 1,6 gamB. 6,4 gamC. 8,0 gamD. 18,8 gamBi 28: Khi in phn 25,98 gam iotua ca mt kim loi X nng chy, th thu -c 12,69 gam iot.Cho bit cng thc mui iotua A. KI B. CaI2 C. NaID. CsIBi 29:in phn dung dch CuCl2 vi in cc tr, sau mt thi gian thu -c 0,32 gam Cu catotv mt l-ng kh X anot. Hp th hon ton l-ng kh X trn vo 200 ml dung dch NaOH ( nhit th-ng). Sau phn ng, nng NaOH cn li l 0,05M (gi thit th tch dung dch khng thayi). Nng ban u ca dung dch NaOH l (cho Cu = 64) A. 0,15MB. 0,2M C. 0,1M D. 0,05MBi 20: Ho tan 40 gam mui CdSO4 b m vo n-c. in phn ht caimitrong dung dchcn dng dng in 2,144A v thi gian 4 gi. Phn trm n-c cha trong mui l A. 18,4%B. 16,8%C. 18,6%D. 16%Bi 31: in phn 300ml dung dch CuSO4 0,2M vi c-ng dng in l 3,86A. Khi l-ng kimloi thu -c catot sau khi in phn 20 pht l (cho Cu = 64; S = 32; O = 16) A. 1,28 gam B.1,536 gam C. 1,92 gam D. 3,84 gamBi 32: in phn dung dch MSO4 khi anot thu -c 0,672 lt kh (ktc) th thy khi l-ng catottng 3,84 gam. Kim loi M l (cho Cu = 64; Fe = 56; Ni = 59; Zn = 65) A. Cu B. Fe C. Ni D. ZnBi 33: in phn nng chy mui clorua ca kim loi M, anot thu -c 1,568 lt kh (ktc), khil-ng kim loi thu -c catot l 2,8 gam. Kim loi M l A. Mg B. Na C. KD. CaBi 34: C 200ml dung dch hn hp Cu(NO3)2 v AgNO3. in phn ht ion kim loi trong dungdch cn dng dng in 0,402A, thi gian 4 gi, trn catot thot ra 3,44 gam kim loi. Nng mol/lit ca Cu(NO3)2 v AgNO3 l A. 0,1 v 0,2 B. 0,01 v 0,1C. 0,1 v 0,01D. 0,1 v 0,1Bi 35: Tin hnh in phn (c mng ngn xp) 500 ml dung dch cha hn hp HCl 0,02M v NaCl0,2M. Sau khi anot bay ra 0,448 lt kh ( ktc) th ngng in phn. Cn bao nhiu ml dung dchHNO3 0,1M trung ho dung dch thu -c sau in phn A. 200 ml B. 300 ml C. 250 ml D. 400 mlBi 36: Ho tan 1,28 gam CuSO4 vo n-c ri em in phn ti hon ton, sau mt thi gian thu-c 800 ml dung dch c pH = 2. Hiu sut phn ng in phn l A. 62,5%B. 50%C. 75%D. 80%Bi 37: Ho tan 5 gam mui ngm n-c CuSO4.nH2O ri em in phn ti hon ton, thu -c dungdch A. Trung ho dung dch A cn dung dch cha 1,6 gam NaOH. Gi tr ca n l A. 4B. 5C. 6D. 8Bi 38: in phn dung dch mt mui nitrat kim loi vi hiu sut dng in l 100%, c-ng dng in khng i l 7,72A trong thi gian 9 pht 22,5 giy. Sau khi kt thc khi l-ng catot tngln 4,86 gam do kim loi bm vo. Kim loi l A. Cu B. Ag C. Hg D. PbBi 39: Tin hnh in phn (c mng ngn xp) dung dch X cha hn hp gm 0,02 mol HCl v0,05 mol NaCl vi C-ng dng in l 1,93A trong thi gian 3000 giy, thu -c dung dch Y.Nu cho qu tm vo X v Y th thy (cho H = 1; Cl = 35,5) A. X lm qu tm, Y lm xanh qu tm. B. X lm qu tm, Y lm qu tm. C. X l qu tm, Y khng i mu qu tm. D. X khng i mu qu tm, Y lm xanh qu tm.Nguoithay.vn - 59 - 60. Nguoithay.vnNguoithay.vn - 60 -
