CAC DANG TOAN VO CO

GV : Nguyn V MinhCc Dng Thng Gp Dng 1 :

Cng thc gii nhanh Ha V C

Kim Lai + axit loi 1 ( H2SO4 long hoc HCl) mui (sunfat hoc clorua) +

H2

mmui sunfat = mkim loi + 96 n H 2 hay mmui clorua = mkim loi + 71 n H 2 Bo ton e : ncho = nnhn vi ncho= mol kim loi . ha tr kim loi

nnhn= 2. n

H

2

Cu 1: Cho 5,2 g hn hp Al, Fe v Mg vo dd HCl d thu c 2,24 lt kh H2(ktc). C cn dung dch thu c bao nhiu gam mui khan ? A. 10,8 B. 11,5 C. 12,3 D,14,6 2,24 = 0,1 mmui = mkim loai + 71.0,1=5,2+7,1=12,3 . Ta chn C Gii : nH2 = 22,4 Cu 2: Cho 10,8 g hn hp Zn, Cd v Al vo dd H2SO4long, d thu c 0,5 g kh H2. C cn dung dch thu c m gam mui khan. Gi tr m l : A. 40,4 B. 37,2 C. 36,4 D. 34,8 0,5 = 0,25 mmui = mkim lai + 96.0,25=10,8 + 24 = 34,8 . Ta chn D Gii : nH2 = 2 Cu 3 (H khi B 2010): Ho tan hon ton 2,45 gam hn hp X gm hai kim loi kim th vo 200 ml dung dch HCl 1,25M, thu c dung dch Y cha cc cht tan c nng mol bng nhau. Hai kim loi trong X l A. Mg v Ca B. Be v Mg C. Mg v Sr D. Be v Ca Gii : V dung dch Y cha cc cht tan c nng bng nhau nn s mol 2 kim loi kim th bng nhau v bng s mol HCl d (nu c) , nHCl = 0,25 mol M + 2HCl MCl2 + H2 nn , a 2a a 9 + 40 nHCl (d) = a/2 nn 0, 25 2a = 0,5a a = 0,1 M = 24,5 = nn c Be v Ca l hp l, chn D 2 Cu 4 (H khi A 2010): Cho 7,1 gam hn hp gm mt kim loi kim X v mt kim loi kim th Y tc dng ht vi lng d dung dch HCl long, thu c 5,6 lt kh (ktc). Kim loi X, Y l A. natri v magie. B. liti v beri. C. kali v canxi. D. kali v bari. Gii: Gi M i din 2 kim loi, n l ha tr 2M + 2nHCl 2MCln + nH2 n = 1 M = 14,2 7,1 5, 6 Bo ton elctron : M = 14,2n .n = 2. M 22, 4 n = 2 M = 28, 4 m 1 < n < 2 nn 14,2 < M < 28,4 ta chn Na v Mg , p n A Cu 5: Ha tan 9,144g hn hp Cu, Mg, Al bng mt lng va dung dch HCl thu c 7,84 lit kh X (ktc), 2,54g cht rn Y v dung dch Z. Lc b cht rn Y, c cn cn thn dung dch Z thu c lng mui khan l A. 33,99g. B. 19,025g. C. 31,45g. D. 56,3g. Gii: Cht rn Y khng tan l Cu nn ch c Mg v Al phn ng v m(Mg, Al) = 9,144 m(Cu) = 6,604 gam

mmui clorua = m(Mg, Al) + 71 n H 2 = 6,604 + (7,84 : 22,4).71 = 31,45 gam , chn CCu 6: Cho 1,53 gam hh Mg, Fe, Zn vo dd HCl d thy thot ra 448ml kh (ktc). C cn dd sau phn ng th thu c m gam cht rn c khi lng A. 2,95 gam B.2,24 gam C. 3,9 gam D. 1,85 gam 0, 448 = 2,95 gam , chn A Gii: Kh l H2 v mui thu c s l mui clorua : mmui clorua = 1,53 + 71 22, 4 Cu 7: Ha tan hon ton 14 gam mt kim loi vo H2SO4 long d thu c 5,6 lt kh (ktc). Kim loi l : A. Al B. Fe C. Zn D. Mg n = 2 14 5,6 M = 28n Fe vi n l ha tr kim loi Gii: bo ton electron : .n = 2. M 22,5 M = 56

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Ch : Fe tc dng vi axit loi 1 ch ra hoa1 tr II Cu 8 (C 2007): Ha tan han ton 3,22g hn hp X gm Fe, Mg, Zn bng mt lng va dung dch H2SO4 long thu c 1,344 lt hidro(ktc) v dung dch cha m gam mui. Ga tr ca m l? A. 10,27 B. 8,98 C. 7,25 D. 9,52 1, 3 4 4 Gii: Kh l H2 v mui thu c s l mui sunfat : m mui sunfat = 3,22 + 96 = 8,98 gam , chn B 22, 4 Cu 9 (C 2007): Ho tan hon ton 2,9 gam hn hp gm kim loi M v oxit ca n vo nc, thu c 500 ml dung dch cha mt cht tan c nng 0,04M v 0,224 lt kh H2 ( ktc). Kim loi M l A. Na. B. Ca. C. Ba. D. K. Gii: Phn tch: Da vo p n ta thy KL l ha tr II hoc ha tr I ,ta ln lt xt hai trng hp: Nu l KL ha tr II: MO + H2O M(OH)2 M + H2O M(OH)2 + H2 ; 0,01 0,01 mol 0,01 0,01 0,01mol 2,9 = 0.01(M+16) + M.0,01 M =137 Ba, chn C Dng 2 :

Mui cacbonat + axit loi 1 ( H2SO4 long hoc HCl) mui (sunfat hoc clorua) CO2 mmui sunfat = mmui cacbonat + 36 n mmui clorua = mmui cacbonat + 11 nCO CO2

22do CO3 +H 2SO4 SO4 +CO2 +H 2O

2

do CO3 +2HCl 2Cl +CO 2 +H 2 O22

v nmui cacbonat = n mui hidr cacbonat = n CO2Cu 10: Cho 12 g hn hp mui cacbonat ca kim lai kim v kim th vo dung dch cha HCl d thu c 2,24 lt kh ktc. C cn dung dch sau phn ng thu c m gam mui khan. Gi tr m l : A. 13,1 B. 12,1 C. 9,1 D. 11,1

Gii: nCO2 =

2,24 = 0,1 mmui clorua = mmui cacbonat + 11.0,1=12+1,1=13,1 . Ta chn A 22,4

Cu 11: Cho m g hn hp 3 mui cacbonat ca kim nhm IA, IIA v IIIA vo dung dch H2SO4long, d thu c 2,8 lt kh ktc. C cn dung dch sau phn ng thu c 22,65 gam mui khan. Gi tr m l : A. 19,25 B. 20,05 C. 18,15 D. 17,86

Gii: nCO2 =

2,8 = 0,125 m= mmui cacbonat = mmui sunfat 36.n CO =22,65 0,125.36 = 18,15 .Chn C 2 22,4

mmuoi clorua mmuoi cacbonat 8, 75 7, 65 = = 0,1 = nmuoi cacbonat 11 11 7, 65 Gi cng thc chung hai mui cabonat l MCO3 c M + 60 = = 76,5 M = 16,5 nn ta chn C 0,1 Cu 14: Cho 3,6 gam hn hp A. gm 2 mui cacbonat ca 2 kim loi k tip nhau trong phn nhm chnh nhm II.Cho A. tan ht trong dung dch H2SO4 long thu c kh B.. Cho B. sc vo dung dch dung dch Ca(OH)2 d thy to thnh 5 gam kt ta. Hai kim loi l g? A. Ca v Mg B. Ca v Sr C. Mg v Be D.Khng xc nh cGii: nCO2 =

Cu 12: Ha tan 3,06g hn hp 2 mui Cacbonat kim loi ha tr I v II bng dd HCl d thu c 672 ml CO2 (kc) . Nu c cn dd th thu c bao nhiu gam mui khan ? A. 3,39g B. 6,78g C. 9,33g D. Khng xc nh c 0, 672 Gii: Khi lng mui khan mmui clorua = 3,06 + 11. =3,39 . Ta chn A 22, 4 Cu 13: Hn hp X gm hai mui cacbonat ca 2 kim loi kim th hai chu k lin tip. Cho 7,65 gam X vo dung dch HCl d. Kt thc phn ng, c cn dung dch th thu c 8,75 gam mui khan. Hai kim loi l: A. Mg v Ca B. Ca v Sr C. Be v Mg D. Sr v Ba


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GV : Nguyn V MinhGii: Do dung dch Ca(OH)2 d nn n CO2 = n = n CaCO3 =


5 = 0, 05 mol = n muoi cacbonat 100 3, 6 Gi cng thc chung hai mui cabonat l MCO3 c M + 60 = = 72 M = 12 nn ta chn C 0, 05 Cu 15: Ho tan ht 2,25 gam hn hp hai mui cacbonat ca hai kim loi A, B ( k tip nhau trong phn nhm chnh nhm II) bng dung dch HCl thu c 0,56 lt kh CO2 (ktc). Hai kim loi A, B l: A. Be v Mg B. Ca v Sr C. Sr v Ba D. Mg v Ca 0,56 = 0, 025 mol = n muoi cacbonat Gii: n CO2 = 22, 4 2, 25 Gi cng thc chung hai mui cabonat l MCO3 c M + 60 = = 90 M = 30 nn ta chn D 0, 025 Cu 16: 18,4 gam hh 2 mui cacbonat ca 2 kim lai nhm IIA hai chu k k tip nhau trong bng HTTH, khi tc dng ht vi dung dch HCl thu c 0,2 mol CO2 . Hai kim loi l A. Ca v Sr B. Sr v Ba C. Mg v Ca D. Be v Mg Gii: n CO2 = 0, 2 mol = n muoi cacbonat 18, 4 Gi cng thc chung hai mui cabonat l MCO3 c M + 60 = = 92 M = 32 nn ta chn C 0, 2 Cu 17: Cho 115g hn hp gm ACO3, B2CO3, R2CO3 tc dng ht vi dd HCl thy tht ra 0,448l CO2 (ktc). Khi lng mui clorua to ra trong dung dch l: A. 115,22g B.151,22g C. 116,22g D. 161,22g 0, 448 Gii: mmui clorua = mmui cacbonat + 11.nCO2 = 115 + .11 = 115,22 gam . Ta chn A 22, 4 Cu 18: Ho tan hon ton 4,68g hn hp mui cacbonat ca hai kim loi A v B k tip trong nhm IIA vo dd HCl thu c 1,12 lit CO2 ktc. Xc nh kim loi A v B l: (Mg = 24 ; Ca = 40 ; Sr = 88 ; Ba = 137) A. Be v Mg B. Mg v Ca. C. Ca v Sr. D. Sr v Ba. 1,12 = 0, 05 mol = n muoi cacbonat Gii: n CO2 = 22, 4 4, 68 Gi cng thc chung hai mui cabonat l MCO3 c M + 60 = = 93, 6 M = 33, 6 nn ta chn B 0, 05

Dng 3 :

Bo Ton IN TCHCho : dung dch

M m+ :a (mol) X n+ v N :b (mol)

X x- :c (mol) z Z :d (mol)

Bo ton in tch : m.a + n.b = x.c + z.d mmui= khi lng tt c ion = M.a + N.b + X.c + Z.d Cu 19: Mt dung dch cha 0,2 mol Ca 2+ ; 0,1 mol Mg 2+ ;0,1 mol HCO3 v x mol Cl . Tm x ? A. 0,5 B. 0,6 C. 0,7 D. 0,8 Gii: 0,2.2 + 0,1.2 = 0,1.1 + x.1 suy ra x = 0,5 chn A Cu 20: Mt dung dch cha 0,1 mol M 2+ ; 0,05 mol Al3+ ; 0,1 mol Cl v x mol SO 4 2 . C cn dung dch thu c 19,3 mui khan. Tm kim lai M. A. Mg B. Ca C. Fe D. Cu Gii: 0,1.2 + 0,05.3 = 0,1.1 + x.2 suy ra x = 0,125 mmui = M.0,1 + 27.0,05 + 35,5.0,1 + 96.0,125 =19,3 suy ra M = 24 (Mg), chn ACu 21 (H Khi A 2010): Cho dung dch X gm: 0,007 mol Na+; 0,003 mol Ca2+; 0,006 mol Cl-; 0,006 2+ HCO3 v 0,001 mol NO3 . loi b ht Ca trong X cn mt lng va dung dch cha a gam Ca(OH)2


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Ga tr ca a l A. 0,222 B. 0,120 C. 0,444 D. 0,180 Gii: nCa(OH)2 = x. nOH- = 2x v nCa2+ = x. Theo bi: OH- + HCO3- CO32- + H2O 2x 0,006 --------- 0,006 Ca2+ + CO32-- CaCO3. x + 0,003 0,006 Ch c x = 0,003 tha mn. Vy a = 0,003.74 = 0,222 (g) , ta chn A Cu 22 (H Khi A 2010): Dung dch X c cha: 0,07 mol Na+; 0,02 mol SO42- v x mol OH-. Dung dch Y c cha ClO4-, NO3- v y mol H+; tng s mol ClO4- v NO3- l 0,04. Trn X v Y c 100 ml dung dch Z. Dung dch Z c pH (b qua s in li ca H2O) l A. 1 B. 2 C. 12 D. 13 Gii: LBTT: 0,07 = 0,02.2 + x x = 0,03 (mol); y = 0,04 (mol) . Vy nH+ d = 0,01 (mol). [H+] = 0,01: 0.1 = 0,1 (M) pH = 1 , ta chn A Cu 23 (C 2007): Dung dch A cha cc ion Al3+=0,6 mol, Fe2+=0,3mol, Cl- = a mol, SO42- = b mol. C cn dung dch A thu c 140,7gam. Gi tr ca a v b ln lt l? A. 0,6 v 0,3 B. 0,9 v 0,6 C. 0,3 v 0,5 D. 0,2 v 0,3 Gii: bo ton in tch : 0, 6.3 + 0,3.2 = 1.a + 2.b = 2,4 Khi lng mui m = 27.0, 6 + 0,3.56 + 35,5.a + 96.b = 140, 7 35,5a + 96b = 107, 7 Nn ta c a = 0,6 v b = 0,3, chn A Cu 24 (H Khi A 2010): Ha tan hon ton 8,94 gam hn hp gm Na, K v Ba vo nc, thu c dung dch X v 2,688 lt kh H2 (ktc). Dung dch Y gm HCl v H2SO4, t l mol tng ng l 4 : 1. Trung ha dung dch X bi dung dch Y, tng khi lng cc mui c to ra l A. 13,70 gam. B. 18,46 gam. C. 12,78 gam. D. 14,62 gam Gii: Ta c: H2O OH + H2. nOH- = 0,24 (mol). HCl (4x mol) H2SO4 (x mol) th nCl- = 4x ; nSO4 = x; nH+ = 6x = 0,24 x = 0,04. mmui = mKL + mCl- + mSO4 = 8,94 + 4.0,04.35,5 + 0,04.96 = 18,46 (g), chn B Cu 25 (H Khi A 2007) : Ha tan hon ton hn hp gm 0,12 mol FeS2 v mol Cu2S bng dung dch HNO3 va , thu c dung dch X (ch cha 2 mui sunfat) v kh duy nht NO. Gi tr ca a l :

A. 1,8 mol B. 1,08 mol C. 0,18 mol D. 0,06 3+ 2+ : Gii: Dung dch thu c ch cha mui sunfat duy nht nn ch cha cc ion : Fe ; Cu ; SO42-

n Fe = n 3+ = 0,12.1 = 0,12 Fe 0,12 mol FeS2 bao toan nguyen to n Cu = n Cu 2+ = 0, 2.a = 0, 2a a mol Cu 2S n = n SO42- = 0,12.2 + a.1 = 0, 24 + a SBo ton in tch : 0,12 . 3 + 2.2a = (0,24 + a ).2 a = 0,06 , chn D Cu 26: Dung dch Y cha 0,1 mol Ca2+, 0,2 mol Mg2+, 0,2 mol Cl- v x mol HCO3-. C cn dung dch Y thu c bao nhiu gam mui khan ? A. 27,9 gam B. 59,7 gam C.30,4 gam D. 22,0 gam Gii: Bo ton in tch : 0,1 . 2 + 0,2 . 2 = 0,2 . 1 + x . 1 x = 0,4 Nn ch khi b nhit phn th s c phng trnh : 2HCO3 CO 32 + CO 2 + H 2 O 0,4 -------->0,2 m muoi = m Ca 2+ + m Mg 2+ + m Cl- + m CO 2- = 0,1.40 + 0, 2.24 + 35,5.0, 2 + 0, 2.60 = 27,9 , chn A3

Dng 4: xit kim loi + Axit mui + H2O

M 2 O n + HCl ( hay H 2SO 4 ) mui + nc

O ( trong oxit )+ 2H + = H 2 O4

2-


t : 0914449230

GV : Nguyn V Minh


mmui = mkim loi + mgc axit vi mkim loi = m xit mO Hoc c th dng cng th tnh nhanh cho trc nghim :v

n H + =2.n O =n H 2O

+ i vi H2SO4 (long) : m mui sunfat = m xit + 80. n H 2SO4+ i vi HCl

: m mui clorua = m xit + 27,5. n HCl

Cu 27: Ha tan hon ton 19,8g hn hp FeO, MgO, Al2O3 cn va 500ml dung dch HCl 1,6M. Sau khi phn ng hon ton, c cn dung dch thu c m(g) mui khan. Tm m A. 13,1 B. 40,2 C. 39,4 D. 41,8

Gii: nHCl = 0,5.1, 6 = 0,8 nH + = 0,8 nO =

nH + 2

= 0, 4 mO = 16.0, 4 = 6, 4( g )

mkl = 19,8 6, 4 = 13, 4( g ), nCl = 0,8 mCl = 0,8.35, 5 = 28, 4Vy mmui = mkim loi + mgc axit = 13,4 + 28,4 = 41,8 (g) . Chn D Hoc dng cng thc gii nhanh : m mui clorua = m xit + 27,5. n HCl = 19,8 + 27,5. 0,5 . 1,6 = 41,8 (g) Cu 28 (H Khi A 2007): Ha tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500ml dung dch acid H2SO4 0,1M (va ). Sau phn ng c cn dung dch mui thu c bao nhiu gam mui khan? A. 6,81g B. 4,81g C. 3,81g D. 5,81g Gii: nH2SO4 = 0,05 = n SO42 nH+ = 0,1 2H+ + O2 = H2O 0,1 0,05 mol mmui = moxit mO(trong oxit) + mgc axit = 2,81 0,05.16 +0,05.96 = 6,81 gam, chn A Cu 29: Ho tan hon ton m gam hn hp X gm Al v Sn bng dung dch HCl (d), thu c 5,6 lt kh H2 ( ktc). Th tch kh O2 ( ktc) cn phn ng hon ton vi m gam hn hp X l: A. 2,80 lt B. 1,68 lt C. 4,48 lt D. 3,92 lt 5, 6 = 0,5 mol Gii: TH1 : X + HCl : n e-cho = 2. 22, 4 VO n e-cho = 4.n O2 = 4. 2 mol TH2 : X + O2 : 2O 2 O 2 + 4e 22, 4 Do ha tr 2 kim loi khng i nn s mol e cho ca 2 phng trnh bng nhau VO 0,5 = 4. 2 mol VO2 = 2,8 lit , chn A 22, 4 Cu 30: Cho 50 gam hn hp gm ZnO, FeO, Fe2O3, MgO tc dng ht vi 200 ml dung dch HCl 4M (va ) thu c dung dch X. Lng mui c trong dung dch X l : A. 79,2 g B. 78,4 gam C. 72 gam D. 72,9 gamGii: m mui clorua = m xit + 27,5. n HCl = 50 + 27,5. 0,2 . 4 = 72 (g), chn C Cu 31: tc dng va vi 7,68 gam hn hp FeO, Fe2O3, Fe3O4 cn dng 260 ml dung dch HCl 1M. Dung dch thu c cho tc dng vi NaOH d, kt ta thu c nung trong khng kh n khi lng khng i thu c m gam cht rn. Gi tr m l : A. 6 gam B. 7 gam C. 8 gam D. 9 gam 1 Gii: n H+ = n HCl = 0, 26 mol , n O (trong oxit) = n H+ = 0,13 mol m Fe = 7, 68 0,13.16 = 5, 6 g 2 n Fe = 0,1 mol , s hp thc : 2Fe Fe2O3 0,1 0,05 , m Fe2O3 = 160.0,05 = 8 gam, chn C Cu 32.: Ha tan hon ton 15 g hn hp CuO, MgO, Al2O3 cn va V ml dung dch HCl 1,6M. Sau khi phn ng hon ton, c cn dung dch thu c m = 28,2 (g) mui khan. Tm V A. 300 B. 400 C. 500 D. 600


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GV : Nguyn V Minh


Gii: m mui clorua = m xit + 27,5. n HCl 28, 2 = 15 + 27,5.1, 6.V V = 0,3 l = 300 ml , chn A m mui clorua = m xit + 27,5. n HCl 28, 2 = 15 + 27,5.1, 6.V V = 0,3 l = 300 ml , chn ACu 33: Ha tan hon ton m g hn hp CuO, MgO, Al2O3 cn va 400 ml dung dch HCl 1,6M. Sau khi phn ng hon ton, c cn dung dch thu c 40,6 (g) mui khan. Tm m A. 30 B. 40 C. 23 D. 32

m mui clorua = m xit + 27,5. n HCl moxit = 40, 6 27,5.1, 6.0,4 = 23 gam , chn CCu 34: Ha tan hon ton 281 gam hn hp gm Fe2O3, MgO, ZnO trong V ml dung dch acid H2SO4 3 M (va ). Sau phn ng c cn dung dch mui thu c 401 gam mui sunfat khan. Tm V A. 300 B. 400 C. 500 D. 600

m mui sunfat = m xit + 80. n H 2SO4 401 = 281 + 80.3.V V = 0,5 l = 500 ml , chn CCu 35 (H Khi A 2008): 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 l A. 0,23. B. 0,18. C. 0,08. D. 0,16. n FeO = n Fe2O3 coi hn hp ch gm 1 xit duy nht l Fe3O4 v

n Fe3O4 = 0, 01 mol n O = 0, 04 mol n H + = 2n O = 0, 08 mol = n HCl suy ra V = 0,08 lt, chn CCu 36 (H Khi B 2008): Cho 9,12 gam hn hp gm FeO, Fe2O3, Fe3O4 tc dng vi dung dch HCl (d). Sau khi cc phn ng xy ra hon ton, c dung dch Y; c cn Y thu c 7,62 gam FeCl2 v m gam FeCl3. Gi tr ca m l A. 9,75. B. 8,75. C. 6,50. D. 7,80.

Coi hn hp ch gm FeO, Fe2O3 : n FeO = n FeCl2 =

n Fe2O3 =

9,12 72.0, 06 = 0, 03 mol n FeCl3 = 2n Fe2O3 = 0, 06 mol 160

7, 62 = 0, 06 mol 127

m (FeCl3) = 0,06 . 162,5 = 9,75, chn A Dng 5:HNO3 + O2 Fe ( Fedu , FeO, Fe2O3 , Fe3O4 ) mui + sn phm kh + hay H SO dac,t 02 4

H2 O

Bo ton e :

m Fe m m Fe .3 = oxit .2 + n NO2 + 3n NO + 8n N 2O + 10n N 2 + 2n SO2 56 16 m m m Fe(NO3 )3 = Fe .242 v m Fe2 (SO4 )3 = Fe .400 56 2.56mhh = mFe + mO = 56 x + 16 y (1)

Cch khc : Quy i hn hp gm Fe : x mol v O : y mol Qu trnh cho nhn e : Fe 3e Fe v y 2y x 3x Suy ra phng trnh sau :o +3

O + 2e O

o

2

3x = 2y + n NO2 + 3n NO + 8n N 2O + 10n N 2 + 2n SO2 (2) x, y

Nu c cho Cu th ta c phng trnh tng qut :

3n Fe + 2n Cu = 2n O + n NO2 + 3n NO + 8n N2O + 10n N2 + 2n SO2 (2 ') x, yVn cn mt cch khc :

mFe = 0,7. m (hh xit st) + 5,6 . n cho/ nhn6

n cho/ nhn = mol kim loi .ha tr = gim s xi ha . s mol sp khEmail : [email protected] t : 0914449230

GV : Nguyn V Minh


Cu 37: m gam bt st ngoi khng kh, sau mt thi gian bin thnh hn hp (A) c khi lng 12 g gm Fe , FeO , Fe3O4 , Fe2O3 . Cho (A) td hon ton vi dd HNO3 thy sinh ra 2,24 l kh NO duy nht ktc. Tnh m . A. 38,72 B. 35,5 C. 49,09 D,10,08

Gii: S mol e do Fe nhng phi bng s mol e do oxi thu ( O2 thu 4e ) v N ca HNO3 thu ( N thu 3e ) :Qu trnh oxi ha :

+5

+5

Fe m mol 56

+3

Fe

+ 3e

3

m mol 562

Qu trnh kh : O 2

0

12 m 12 m 4 mol 32 32 m 12 m Ta c: 3 = 4 + 0,3 56 32

+

4e

2O

;

N +

+5

3e

N

+2

0,3mol Gii ra : m = 10,08g , chn D

0,1mol

(c th dng cng thc cho nhanh nhng vit qu trnh cho nhn ra s tt hn cho cc em) + Cch gii khc nh sau da theo (1) v (2) :

56 x + 16 y = 12 x = 0,18 = n Fe mFe = 0,18.56 = 10, 08 gam 2, 24 3x 2 y = 3. = 0,3 y = 0,12 22, 4 K t bi ny s c bi gii theo cch 1 hoc 2 hoc cch 3 s trnh by sau y : + Cch 3 : n nhn = 3.nNO = 3. 0,1 nn mFe = 0,7.m xit + 5,6. n nhn = 0,7.12 + 5,6 . 0,3 = 10,08 gam Cu 38 (H Khi B 2008): Cho 11,36 gam hn hp gm Fe2O3, Fe3O4, FeO, Fe phn ng ht vi dd HNO3 long d thu c 1,344 lt kh NO duy nht (ktc) v dung dch X. C cn dung dch X thu c m gam mui khan. Gi tr m l : A. 38,72 B. 35,5 C. 49,09 D,34,36 m hh = mOxit = 11, 36 m Fe m oxit -m Fe m Fe 11,36-m Fe .3= +3n NO vi .3= +3.0,06 suy ra Gii: Cch 1 : 1,344 nNO = = 0, 06 56 16 56 8 22, 4 Vy mFe = 8,96 suy ra m Fe(NO3 )3 =

m Fe 8,96 .242= .242=38,72 , chn A (cch 2 hc sinh t gii ) 56 56

Cu 39: Ha tan han ton 46,4 gam mt oxit kim loi bng dung dch H2SO4 c nng (va ) thu c V lt kh SO2 (ktc) v 120 gam mui. Xc nh cng thc oxit kim loi v V A. FeO; 1,12 B. Fe2O3; 2,24 C. Fe3O4;1,12 D. Fe3O4; 2,24 Gii: quy i xit thnh Fe (x mol) v O (y mol) suy ra m (xit) = 56x + 16y = 46,4 (1)

120 = 0,3 nFe = 0,3.2 = 0, 6 = x ;(1) y = 0,8 = nO 400 V nFe 0, 6 3 V = 2, 24(l ) , chn D = = (C,D) , 3 x = y.2 + 2nSO2 3.0, 6 = 0,8.2 + 2. 22, 4 nO 0,8 4 Cu 40: Cho m gam Fe chy trong oxi mt thi gian thu c 36 gam cht rn A gm 4 cht. Ha tan A bng HNO3 d thu c 6,72 lt NO (ktc). Tnh m? A. 30,24 B. 32,40 C. 24,34 D. 43,20 Gii: quy i xit thnh Fe (x mol) v O (y mol) v s dng (1) v (2) 56 x + 16 y = 36 x = 0,54 = n Fe mFe = 0,54.56 = 30, 24 gam , chn A 6, 72 3x 2 y = 3. = 0,9 y = 0,36 22, 4 + Cch khc : n nhn = 3.nNO = 3. 0,3 nn mFe = 0,7.m xit + 5,6. n nhn = 0,7.36 + 5,6 . 0,9 = 30,24 gam nFe2 ( SO4 )3 =


7

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GV : Nguyn V Minh


Cu 41 (H Khi B 2007): Nung m gam bt Fe trong oxi thu c 3 gam hn hp cht rn X. Ha tan ht hn hp X trong dung dch HNO3(d) thot ra 0,56 lt (ktc) NO (l ssn phm kh duy nht). Ga tr ca m l? A. 2,52 B. 2,22 C. 2,62 D. 2,32 Gii: quy i xit thnh Fe (x mol) v O (y mol) v s dng (1) v (2) 56 x + 16 y = 3 x = 0, 045 = n Fe mFe = 0, 045.56 = 2,52 gam , chn A 0,56 3x 2 y = 3. 22, 4 = 0, 075 y = 0, 03 Cu 42: Nung nng 16,8 gam bt st ngoi khng kh, sau mt thi gian thu c m gam hn hp X gm oxit st v st d. Ha tan ht hn hp X bng H2SO4 c nng thu c 5,6 lt SO2 (ktc). Ga tr ca m l? A. 24g B. 26g C. 20g D. 22g Gii: quy i xit thnh Fe (x mol) v O (y mol) v s dng (1) v (2)

nFe =

3 x 2.n SO2 16,8 = = 0,3 = x ; 3x = 2y + 2n SO2 (2) y = 2 56 mhh = mFe + mO = 56 x + 16 y = 56.0,3 + 0, 2.16 = 20g , chn C + Cch khc : n nhn = 2. n SO = 0,5 mol2

3.03 2 2

5, 6 22, 4

= 0, 2

mFe = 0,7.m xit + 5,6. n nhn suy ra m oxit =

m Fe 5, 6.n nhan 16,8 5, 6.0,5 = = 20 gam 0, 7 0, 7

Cu 43: Ha tan 13,92 g Fe3O4 bng dd HNO3 thu c 448 ml kh NxOy (ktc).Xc nh NxOy? A. NO B. N2O C.NO2 D. N2O5 n Fe = 0, 06.3 13,92 0, 448 = 0, 06 mol = 0, 02 mol Gii: n Fe3O4 = , n = n o = 0, 06.4 22, 4 232 Gi k l gim s xi ha ca kh

3n Fe = 2n O + k .n 3.0,18 = 2.0, 24 + k .0, 02 k = 3 N O , Chn ACu 44 (H Khi A 2009): Ha tan hon ton 20,88 gam mt xit st bng dung dch H2SO4 c nng thu c dung dch X v 3,248 lt kh SO2 (sn phm kh duy nht). C cn dung dch X thu c m gam mui sunfat khan. Gi tr m l : A. 52,2 B. 54,0 C. 58,0 D. 48,4 Gii: quy i xit thnh Fe (x mol) v O (y mol) v s dng (1) v (2) 56 x + 16 y = 20,88 x = 0, 29 = n Fe n m Fe2 (SO4 )3 = Fe .400 = 58 gam , chn C 3, 248 2 y = 0, 29 3x 2 y = 2. 22, 4 = 0, 29 Cu 45 (HQGHN 2000): m (g) phoi bo Fe ngoi khng kh, sau mt thi gian c 12 g cht rn X gm Fe, FeO, Fe3O4, Fe2O3. Ha tan ht X trong dung dch H2SO4 , nng c 2,24 lt SO2 (ktc). Gi tr ca m l: A. 9,52 B. 9,62 C. 9,42 D. 9,72 Gii: quy i xit thnh Fe (x mol) v O (y mol) v s dng (1) v (2) 56 x + 16 y = 12 x = 0,17 = n Fe m Fe = 0,17.56 = 9,52 gam , chn A 2, 24 y = 0,155 3x 2 y = 2. 22, 4 = 0, 2 Cu 46: Nung m gam bt Cu trong oxi thu c 37,6 gam hn hp rn X gm Cu, CuO v Cu2O. Ha tan hon ton X trong dung dch H2SO4 c, nng (d) thy thot ra 3,36 lt kh ( ktc). Gi tr ca m l: A. 25,6 gam B. 32 gam C. 19,2 gam D. 22,4 gam Gii: quy i xit thnh Cu (x mol) v O (y mol) v s dng (1) v (2)

+2


8

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GV : Nguyn V Minh


64 x + 16 y = 37, 6 x = 0,5 = n Cu m Cu = 0,5.64 = 32 gam , chn B 3,36 2 x 2 y = 2. 22, 4 = 0,3 y = 0,35 Cu 47 (H Khi A 2007):Nung m g st trong khng kh, sau mt thi gian ngi ta thu c 104,8 g hh rn A gm Fe,FeO,Fe2O3 v Fe3O4. Ha tan hon ton A trong HNO3 d thu c dung dch B v 12,096 lt hn hp kh NO v NO2 (ktc) c t khi so vi He l 10,167. Gi tr m l: A.72 B.78,4 C.91,28 D, p s khc Gii: Gi a l s mol NO, b l s mol NO2 30a + 46b S mol hh kh l : n = a + b = 0,54 mol , M = = 10,167.4 30a + 46b = 21,96 a+b Ta c : a = 0,18 , b = 0,36 , n nhn = 3.n NO + 1.n NO2 = 0,18.3 + 0,36.1 = 0,9 mol mFe = 0,7.m xit + 5,6. n nhn = 0,7 . 104,8 + 0,9 . 5,6 = 78,4, chn B Cu 48: Ha tan hon ton 8,64 gam FeO bng dung dch HNO3 th thu c 336 ml kh duy nht (ktc). Cng thc ca cht kh l: A. N2 B. NH3 C. N2O D. NO2 Gii: Cn nh r gim s xi ha tng sn phm kh n FeO = n Fe = n O = 0,12 mol , gi X l gim s xi ha ca sn phm kh

3n Fe = 2n O + X.n spk X =

3n Fe 2n NO 3.012 2.0,12 = = 8 N 2 O , Chn C 0,336 n spk 22, 4

Cu 49: m gam bt st ngoi khng kh, sau mt thi gian s chuyn thnh hn hp B gm 4 cht rn c khi lng 12 gam. Cho hn hp B phn ng ht vi dung dch HNO3 d thy thot ra 2,24 lt NO (ktc). Tnh m v khi lng HNO3 phn ng ? A. 10,08 g v 34,02 g A. 10,8 g v 34,02 g C. 10,8 g v 40,32 g D. 10,08 g v 40,32 g Gii: n nhn = 3.nNO = 3. 2, 24 = 0,3 mol mFe = 0,7.m xit + 5,6. n nhn = 0,7.12 + 5,6 . 0,3 = 10,08 gam 22, 4 Fe(NO3)3 Fe 0,18 0,18 mol , bo ton nguyn t N : n N/HNO3 = n N/Fe(NO3 )3 + n N/NO = 3.0,18 + 0,1 = 0, 64 mol m HNO3 = 0, 64.63 = 40,32 g

Ta chn D Cu 50: Cho hn hp gm FeO, CuO, Fe3O4 c s mol 3 cht u bng nhau tc dng ht vi dung dch HNO3 thu hn hp kh gm 0,09 mol NO2 v 0,05 mol NO . S mol ca mi cht l: A. 0,12 B. 0,24 C. 0,21 D. 0,36 n Fe = 4 x Gii: t x = n FeO = n CuO = n Fe3O 4 n O = 6 x n = x Cu

3n Fe + 2n Cu = 2n O + n NO2 + 3n NO 3.4 x + 2.x = 2.6 x + 0, 09 + 3.0, 05 x = 0,12 , chn ACu 51: Cho 22,72 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung dch HNO3 long d thu c V lt kh NO duy nht (ktc) v dung dch X. C cn dung dch X thu c 77,44 gam mui khan. Gi tr ca V l A. 4,48. B. 2,688. C. 5,6. D. 2,24. 77, 44 = 0,32 mol m Fe = 0,32.56 = 17,92 gam Gii: mui chnh l Fe(NO3)3 : n Fe = n Fe(NO3)3 = 242

m O = m hh m O = 22, 72 17,92 = 4,8 gam n O = 0,3 mol , 3n Fe = 2n O + 3

VNO VNO = 2, 668 l , chn B 22, 4


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GV : Nguyn V MinhSn phm kh N O 2 , N 2 , N O, N 2 O, N H 4 NO 3 i vi HNO3+4

Cng thc gii nhanh Ha V C+2 +1 -3

Dng 6: Kim loi + Axit (H2SO4c, HNO3) mui + sn phm kh + H2O+4 o-2 o

S O 2 , H 2 S,S i vi H2SO4cv nnhan

Mui (kim lai phi ha tr cao nht) v Al, Fe, Cr khng tc dng vi H2SO4 v HNO3 c ngui.

ncho = mol kimloai . hoa triSp kh gim s xi ha

= do giam so OXH .nsan pham khuH2 S6 (-2) = 8-2

+ i vi H2SO4 c :+4

S O2

S60=6

o

64=2

nH 2 SO4 = nS = nSO 2 + nS (trong sp khu ) =4

n( cho / nhan ) 24

+ nS (trong sp khu )

mmuoi = mkimloai + mSO 2 = mkimloai + 96.nSO 2 = mkimloai + 96.4

n( cho / nhan ) 2N H 4 NO 3 (mui)5 (-3) = 8-3

+ i vi HNO3 : Sp kh gim s xi ha

N O254=1

+4

N2(5-0).2 = 10

o

NO52=3

+2

N2 O(5 1).2 = 8

+1

nHNO3 = nN = nNO + nN (trong sp khu ) = n( cho / nhan ) + nS (trong sp khu )3

mmuoi = mkimloai + mNO = mkimloai + 62.nNO = mkimloai + 62.3 3

n( cho / nhan ) 1

Ch : Nu sp kh c NH4NO3 th khi lng mui sau phn ng phi cng thm khi lng ca NH4NO3 Cu 52 (C 2011): Dy gm cc kim loi u tc dng c vi dung dch HCl nhng khng tc dng vi dung dch HNO3 c , ngui l: A. Fe, Al, Cr B. Cu, Fe, Al C. Fe, Mg, Al D. Cu, Pb, Ag Gii: Chn A , HNO3 c ngui khng tc dng Al, Fe, Cr Cu 53 : Mt hn hp gm hai bt kim loi Mg v Al c chia thnh hai phn bng nhau: - Phn 1: cho tc dng vi HCl d thu c 3,36 lt H2. - Phn 2: ho tan ht trong HNO3 long d thu c V lt mt kh khng mu, ho nu trong khng kh (cc th tch kh u o ktc). Gi tr ca V l A. 2,24 lt. B. 3,36 lt. C. 4,48 lt. D. 5,6 lt. Gii: S mol e kim loi nhng khi tc dng vi HCl v HNO3 nh nhau ( Do 2 kim loi c ha tr khng i ).

Nn s mol e H v N nhn bng nhau .

+

+5

N + 3e 3x V = 2,24 lt . chn A 10 Email : [email protected] H + 2e 0,3 3x = 0,3

+

H2 0,15 mol x = 0,1

+5

N x mol

+2

t : 0914449230

GV : Nguyn V Minh


Cu 54 : Cho 1,35 gam hn hp Cu , Mg , Al tc dng vi HNO3 d c 896 ml ( ktc) hn hp gm NO v NO2 c M = 42 . Tnh tng khi lng mui nitrat sinh ra (kh ktc). A. 9,41 gam. B. 10,08 gam. C. 5,07 gam. D. 8,15 gam. Gii: nhh kh = 0,04 . a + b = 0,04 v 30a + 46b = 42 . 0,04 = 1,68 a = 0,01 = nNO ; b = 0,03 = nNO2 n nhan = 3.n NO + 1.n NO = 3.0, 01 + 0, 03.1 = 0, 06 mol 2 m hh mui

= mhh kim loi + mNO = mhh kim loi + 62.nnhn = 1,35 + 62.0,06 = 5,07 gam. Chn C 3

Cu 55 : Ho tan hon ton 1,805 g hn hp gm kim loi A c ho tr khng i duy nht v Fe bng dung dch HCl thu c 1,064 lt kh H2 . Khi ho tan 1,805 gam hn hp trn bng dd HNO3 long d thu c 0,896 lt kh NO duy nht(ktc) . Cc kh o cng iu kin . Kim loi A l: A. Cu B. Cr C. Al D. Mn.Gii: Hn hp kim loi b ha tan hon ton trong HCl A phi tc dng vi HCl .

A x

An+ + ne nx

Fe Fe2+ + 2e y 2y

2 H

+

+ 2e 0.095

H2 0.0475+2

nx + 2y = 0,095 (1) v Ax + 56y = 1,805 (2)

A x

An+ + ne nx

Fe Fe3+ + 3e y 3y

N + 3e 0,12

+5

N 0.04

A = 9n n 1 2 3 Chn A = 27 ( Al ) , chn C A 9 18 27 Cu 56: Ha tan hon ton 3,6 gam hn hp Al, Fe, Mg bng dd HNO3 thu c 0,01 mol NO; 0,01 mol N2O v khng c sp kh no khc. C cn dung dch sau phn ng thu c m (g) mui khan. Tnh m. A. 10,42 B. 11,42 C. 9,84 D. 12,04 Gii: nnhn=3.0,01 + 8.0,01 =0,11 mmui = mkim loi + mgc axit = 3,6 +0,11.62= 10,42 gam, chn A

nx + 3y = 0,12 (3) . T (1) , (2) y = 0.025 . T (1) , (2) nx = 0,045 v Ax = 0,405

Cu 57: Ha tan hon ton hn hp Al, Fe, Mg vo 800ml dung dch HNO3(va ) thu c 0,08 mol NO; 0,06 mol N2O v 0,01 mol N2. Vy nng mol ca dung dch HNO3 l A. 2M B. 1,5M C.1,3M D.1,8M Gii: nnhn=3.0,08 + 8.0,06+10.0,01 =0,82 nN(trong sp kh) = 0,08.1 + 0,06.2 + 0,01.2=0,22 1, 04 = 1,3M , chn C suy ra nHNO3 = nnhn + nN (trong sp kh) = 0,82 + 0,22 = 1,04 suy ra CM(HNO3)= 0,8 Cu 58: Ha tan hon ton 2,7 gam mt kim loi cha r ha tr vo dd HNO3 d thy thot ra 0,672 lt kh (ktc) khng mu khng mi khng chy ( sp kh duy nht ). Tm kim loi A. Al B. Fe C. Zn D. Mg 0, 672 = 0, 03 , gim s OXH l 10 Gii: Kh khng mu khng mi khng chy l N2 , nN 2 = 22, 4 Bo ton e : mol kim loi . ha tr (tm t l n) = gim s OXH . s mol sp kh

n=1 M =9 2,7 .n=10.0,03 M =9.n, n=2 M =18 , chn A M n=3 M =27 (Al) Cu 59: Chia m gam Al thnh 2 phn bng nhau : Phn mt tc dng vi lng d dd NaOH sinh ra x mol kh H2 Phn hai tc dng vi lng d dd HNO3 long sinh ra y mol kh N2O (sp kh duy nht). Quan h gi x v y l : A. y = 2x B. x = y C. x = 4y D. x = 2y 11 Email : [email protected] t : 0914449230

GV : Nguyn V MinhGii: s mol Al hai phn bng nhau Al + NaOH d : n Al .3=2.n H 2 =2.x


v

Al + dd HNO3 d : n Al .3 = 8.n N 2 O = 8y

Ta c c 2.x = 8.y x = 4y , chn C Cu 60: Cho 3,6 gam Mg tc dng vi dd HNO3 d sinh ra 2,24 lt kh X (sp kh duy nht ktc). Kh X l : A. NO B. N2O C. NO2 D. N2 Gii: gi X l gim s xi ha ca kh cn tmBo ton e :

3,6 2,24 .2 = X. X = 3 NO . Chn A 24 22,4

Cu 61: Cho m gam Al tan hon ton trong dd HNO3, thu c 44,8 lt (ktc) hn hp kh NO, N2O, N2 theo t l mol 1 : 2 : 2. Gi tr m l A. 35,1 B. 16,8 C. 140,4 D.2,7 Gii: nhh kh = 44,8 : 22,4 = 2 mol , t t l cho ta t s mol : NO : x mol , N2O : 2x mol , N2 : 2x mol Nn ta c : x + 2x + 2x = 2 x = 0, 4 mol

n nhan = 3n NO + 8n N2O + 10n N2 = 3.x + 8.2 x + 10.2 x = 39 x = 39.0, 4 = 15, 6 molm .3 = 15, 6 m = 140, 4 gam , Chn C 27 Cc em hy c gng thuc gim s xi ha mi kh lm bi tt hn Cu 62: Khi cho 1,92 gam hn hp X gm Mg v Fe c t l mol 1:3 tc dng hon ton vi HNO3 to ra hn hp kh gm NO v NO2 c th tch 1,736 lt (ktc). Tnh khi lng mui to thnh v s mol HNO3 phn ng. A. 8,074gam v 0,018mol B. D. 8,4gam v 0,8mol C. 8,7gam v 0,1mol D. 8,74 gam v 0,1875mol Gii: t s mol Mg : x , Fe : 3x mhh = 24.x + 56.3x = 1,92 x = 0,01 t s mol NO : a v NO2 : b nhh kh = 0,0775 = a + b Bo ton e : 2.x + 3.3x = 3.a + 1.b = 0,11 mol a = 0,01625 ; b = 0,06125 nHNO3 = nnhn + n(N /NO) + n(N/NO2) = 0,11 + 0,01625.1 + 0,06125.1 = 0,1875 mol mmui = mkim loi + 62.nnhn = 1,92 + 62.0,11 = 8,74 gam , chn D Cu 63: Ha tan hon ton 11,2 gam Fe vo HNO3 d thu c dung dch A v 6,72 lt (ktc) hn hp kh B gm NO v mt kh X, vi t l th tch l 1:1. Xc nh kh X? A. NO B. N2O C. NO2 D. N2 Gii: NO v kh X, vi t l th tch l 1 : 1 nn nNO = nX = 0,15 mol . Gi X l gim s xi ha ca kh 11, 2 Bo ton electron : .3 = 3.0,15 + X.0,15 X = 1 NO 2 , Chn C 56 Cu 64: Ha tan hon ton 19,2g kim loi M trong dung dch HNO3 d thu c 8,96 lt (ktc) hn hp kh gm NO2 v NO c t l th tch 3:1. Xc nh kim loi M. A. Fe (56) B. Cu (64) C. Al (27) D. Zn (65) Gii: NO2 v NO c t l th tch 3:1 nn ta t s mol NO2 : 3x v NO : 1x nhh kh = 4x = 8,96 /22,4 x = 0,1 mol 19,2 Bo ton electron : .n = 3.0,1 + 1.0,3 M = 32n M = 64, n = 2 (Cu) , chn B M Cu 65: Cho 13,5 gam hn hp gm Al v Ag tan trong HNO3 d thu c dung dch A v 4,48 lt hn hp kh gm (NO,NO2) c khi lng 7,6 gam. Tnh % khi lng mi kim loi. A. 30 v 70 B. 44 v 56 C. 20 v 80 D. 60 v 40 Gii: t s mol NO l a , NO2 l b mhh = 30.a + 46.b = 7,6 , nhh = a + b = 0,2 a = b = 0,1 mol Ta c 13,5 gam hn hp gm Al : x mol v Ag : y mol 27x + 108y = 13,5 gam Bo ton electron : 3x + 1y = 3.a + 1.b = 0,4 x = y = 0,1 mol 12 Email : [email protected] t : 0914449230

Bo ton e :

GV : Nguyn V MinhNn %m Al =


27.0,1 .100 = 20% %m Ag = 80% , chn C 13,5 Cu 66: Cho 3 gam hn hp gm Cu , Ag tan ht trong dung dch gm HNO3 v H2SO4 thu 2,94 gam hn hp 2 kh NO2 v SO2 c th tch 1,344 lt (ktc). Tnh % khi lng mi kim loi? A. 30 v 70 B. 44 v 56 C. 20 v 80 D. 64 v 36 46a + 64b = 2,94 a = 0, 05 Gii: NO2 : a mol v SO2 : b mol 1,344 a + b = 22, 4 = 0, 06 b = 0, 01 Cho 3 gam hn hp gm Cu : x mol , Ag : y mol 64x + 108y = 3 gam Bo ton electron : 2x + 1y = 1.a + 2.b = 0,07 x = 0,03 v y = 0,01 mol 64.0, 03 Nn %m Cu = .100 = 64% %m Ag = 36% , chn D 3 Cu 67: Trn 60g bt Fe vi 30g lu hunh ri un nng (khng c kkh ) thu c cht rn A. Ho tan A bng dd axit HCl d c dd B v kh C. t chy C cn V lt O2 (ktc). Tnh V, bit cc phn ng xy ra hon ton. A. 32,928 lt B. 33 lt C. 34 lt D. 35 lt 30 Gii: nFe > nS = . nn Fe d v S ht 32 Kh C l hh H2 v H2S . t chy C thu c SO2 v H2O . H+ nhn e to H2 , sau H-2 nhng e to li H+ . Do : Trong phn ng c th coi ch c Fe v S nhng e , cn O2 nhn e . 30 30 mol 4 mol xmol 32 32 60 30 Theo nh lut bo ton electron : 2 + 4 = 4x x = 1,47 56 32 VO2 = 32,928 lt, chn A

Fe 60 mol 56

Fe

+2

+ 2e 60 2 mol 56

S

+4

S + 4e

O2

+ 4e

2O

2

4x mol

Cu 68: Th tch dd FeSO4 0,5M cn thit phn ng va vi 100ml dd cha KMnO4 0,2M v K2Cr2O7 0,1M mi trng axit l : C. 80 ml D. 640 ml A. 160 ml B. 320 ml n K 2Cr2O7 = 0,01 Gii : Ta c : n KMnO4 = 0,02

Fe Fe3+ + 1e x mol x mol

+2

Mn + 5e Mn 0,02 0,1

+7

+2

2 Cr + 6e 2 Cr 0,02 0,06

+6

+3

x = 0,1 + 0,06 = 0,16

V FeSO4 =

0,32 lt = 320 ml, chn B

Cu 69: Ha tan m gam Al trong dung dch HNO3 long sau phn ng thu c 0,896 lt hn hp kh NO v N2O (ktc) c t khi so vi hiro bng 16,75. Khi lng mui khan thu c sau phn ng l : A. 12,07 gam B. 12,78 gam C. 10,65 gam D. 14,91 gam. a + b = 0, 04 a = 0, 03 Gii: NO : a mol v N2O : b mol 30a + 44b = 16, 75.2.0, 04 b = 0, 01 m 17 Bo ton e : mol .3 = 3.0, 03 + 8.0, 01 m = 1, 53 gam n Al = n Al ( NO3 )3 = 27 300 m mui = 17:300 . 213 = 12,07, chn A Cu 70: Ha tan hon ton 14,8g hh (Fe, Cu) vo lng d dung dch hn hp HNO3 v H2SO4 c, nng. Sau phn ng thu c 10,08 lt NO2 v 2,24(l) SO2 (ktc). Khi lng ca Fe trong hn hp ban u l: A. 5,6 B. 8,4 C. 18,0 D. 18,2


13

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56 x + 64 y = 14,8 x = 0,15 Gii: Fe : x mol , Cu : y mol 10, 08 2, 24 3x + 2 y = 1. 22, 4 + 2 22, 4 y = 0,1 mFe = 56 x = 8, 4 gam , chn B Cu 71: Ha tan 5,6g Fe bng dung dch H2SO4 long d th thu c dd X . Dung dch X phn ng va v i Vml dd KMnO4 0,5M . Gi tr ca V l : A. 20ml B. 40ml C. 60ml D. 80ml Gii: nFe = 0,1 molFe2+ + 2e Fe2+ Fe2+ + 1e Mn + 5e 0,1 mol 0,1 mol 0,1 mol 0,1 mol x mol 5x mol Theo nh lut bo ton electron : 5x = 0,1 x = 0,02 mol V = 40 ml , chn B

Fe

+7

Mn

+2

Cu 72: Ho tan hon ton 1,2 gam kim loi M vo dung dch HNO3 d thu c 0,224 lt kh N2 ktc (sn phm kh duy nht). M l kim loi no di y? A. Zn B. Al C. Ca D. Mg Gii: gi n l ha tr kim loi 1,2 0, 224 Bo ton elctron : M = 12n M = 24, n = 2 Mg , ta chn D n = 10. M 22, 4 Cu 73: Cho 9,72 gam kim loi M phn ng ht vi dung dch HNO3 long thu c 0,672 lt kh NO (ktc). Kim loi M dng l : A. Cu B. Mg C. Fe D. Ag. Gii: gi n l ha tr kim loi 9, 72 0, 672 Bo ton elctron : M = 108n M = 108, n = 1 Ag , ta chn D n = 3. M 22, 4 Cu 74: Ha tan hon ton 16,2g mt kim loi ha tr III bng dung dch HNO3,thu c 5,6 lt (kc) hn hp X gm NO v N2. Bit t khi hi ca X so vi kh oxi bng 0,9. Xc nh tn kim loi em dng? A. Al B. Fe C.Cu D.Na Gii: NO : x mol , N2 : y mol 0,56 = 0, 25 x = 0,1 nhh = x + y = 22, 4 m = 30 x + 28 y = 0,9.32.0, 25 y = 0,15 hh

16, 2 3 = 3.0,1 + 10.0,15 M = 27 Al , chn A M Cu 75: Ho tan 8,1 gam kim loi M bng dung dch HNO3 long thy c 6,72 lt kh NO duy nht ( ktc)thot ra. M l kim loi: A. Al B. Cu C. Fe D. Mg Gii: gi n l ha tr kim loi 8,1 6, 72 M = 9n M = 27, n = 3 Al , ta chn A Bo ton elctron : n = 3. M 22, 4 Dng : To mui NH4NO3 (du hiu nhn bit : tao bo ton electron 2 v khng bng nhau nn phi c thm mui NH4NO3 v gii li bi ton vi x l s mol NH4NO3 ) (cu 76 v cu 77 ) Cu 76 (H Khi B 2008) : Cho 2,16 gam Mg tc dng vi dung dch HNO3 (d). Sau khi phn ng xy ra hon ton thu c 0,896 lt kh NO ( ktc) v dung dch X. Khi lng mui khan thu c khi lm bay hi dung dch X l A. 13,32 gam. B. 6,52 gam. C. 8,88 gam. D. 13,92 gam. Gii: nMg = 0,09 mol , nNO = 0,04 mol Bo ton e : 0,09. 2 0,04.3 nn c to mui NH4NO3 : x mol Bo ton e khi c mui NH4NO3 : 0,09.2 = 0,04.3 + 8.x x = 0,0075 mol Bo ton electron : Mg Mg(NO3)2 0,09 0,09 mol


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m mui = mMg ( NO3 )2 + mNH 4 NO3 = 0, 09.148 + 0, 0075.80 = 13,92 gam , chn DCu 77 (H Khi A 2009): Ho tan hon ton 12,42 gam Al bng dung dch HNO3 long (d), thu c dung dch X v 1,344 lt ( ktc) hn hp kh Y gm hai kh l N2O v N2. T khi ca hn hp kh Y so vi kh H2 l 18. C cn dung dch X, thu c m gam cht rn khan. Gi tr ca m l A. 38,34. B. 34,08. C. 106,38. D. 97,98. Gii: nAl = 0,46 mol, N2O : a mol , N2 : b mol 1,344 = 0, 06 nhh = a + b = 22, 4 a = b = 0, 03 mol m = 44a + 28b = 2.18.0, 06 hh Bo ton e : 0,46. 3 8.0,03 + 10.0,03 nn c to mui NH4NO3 : x mol Bo ton e khi c mui NH4NO3 : 0,46. 3 = 8.0,03 + 10.0,03 + 8.x x = 0,105 mol Al Al(NO3)3 0,46 0,46 mol

m mui = mAl ( NO3 )3 + mNH 4 NO3 = 0, 46.213 + 0,105.80 = 106,38 gam , chn CCu 78: Cho 1,35 g hn hp gm Cu , Mg , Al tc dng ht vi dung dch HNO3 thu c hn hp kh gm 0,01 mol NO vo 0,04 mol NO2. Khi lng mui to ra trong dung dch l : A. 3,45g B. 4,35g C. 5,69g D. 6,59g

Gii : N + 3e N N + 1e N 0,03 mol 0,01 mol 0,04 mol 0,04 mol n nhn = 0,03 + 0,04 = 0,07 = s mol gc NO3 trong mui Khi lng hh mui = m kim loi + m NO3 trong mui = 1,35 + 62 . 0,07 = 5,69g , chn CCu 79 : Cho hn hp gm 0,15 mol CuFeS2 v 0,09 mol Cu2FeS2 tc dng vi dung dch HNO3 d thu c dung dch X v hn hp kh Y gm NO v NO2. Thm BaCl2 d vo dung dch X thu c m gam kt ta. Mt khc, nu thm Ba(OH)2 d vo dung dch X, ly kt ta nung trong khng kh n khi lng khng i thu c a gam cht rn. Gi tr ca m v a l:

+5

+2

+5

+4

A. 111,84g v 157,44g C. 112,84g v 157,44g Gii: Ta c bn phn ng: CuFeS2 0,15 Cu2FeS2 0,09 n SO2 = 0, 48 mol;4

B. 111,84g v 167,44g D. 112,84g v 167,44g

Cu2+ + Fe3+ + 2SO42 0,152+

0,153+

0,3 0,18

2Cu + Fe + 2SO42 0,18 0,09 Ba2+ + SO42 BaSO4 0,48 0,48

m = 0,48 233 = 111,84 gam.2Fe Fe2O3 0,24 0,12

nCu = 0,33 mol; nFe = 0,24 mol. Cu CuO 0,33 0,33

a = 0,33 80 + 0,12 160 + 111,84 = 157,44 gam. Chn ADng 7: Phn ng Nhit Luyn


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GV : Nguyn V Minh

Cng thc gii nhanh Ha V C Al 2 O 3 CO 2 , iu kin : kim loi CO hay CO 2 H 2O

Al CO t cao xit kim loi + kim loi + C H 2 o

> Al

Quy i xit = kim loi + O Cng thc

n O (trong Oxit) = n CO = n CO2 = n H2O = n H2 = n (H2 ,CO)

Cu 80 (H Khi A 2009): Cho lung kh CO (d) i qua 9,1 gam hn hp gm CuO v Al2O3 nung nng n khi phn ng hon ton, thu c 8,3 gam cht rn. Khi lng CuO c trong hn hp ban u l A. 0,8 gam. B. 8,3 gam. C. 2,0 gam. D. 4,0 gam. Gii: mO = 9,1 8,3 = 0,8 (g) nO = nCuO = 0,05(mol) mCuO = 0,05.80 = 4 (g) , Chn D Cu 81 : Dy cc xit b CO kh nhit cao l : A. CuO, FeO, ZnO, MgO B. CuO, Fe3O4, Fe2O3, Al2O3 C. Na2O, CaO, MgO, Al2O3 D. ZnO, PbO, CuO, Fe2O3 Gii: xit kim loi tham gia p nhit luyn phi ng sau Al nn ta loi cc xit ca kim loi Na, Ca, Al, Mg. Chn D Cu 82 : Kh hon ton 6,64 gam hn hp gm Fe, FeO, Fe2O3, Fe3O4 cn dng va 2,24 lt CO (ktc). Khi lng Fe thu c l bao nhiu ? A. 5,4 gam B. 5,04 gam C.2,24 gam D. 3,84 gam

Gii: xit kim loi = Fe + O , n O (trong oxit) = n CO =

2, 24 = 0,1 m O = 0,1.16 = 1,6 (gam) 22, 4

Suy ra mFe = mxit - mO = 6,64 1,6 = 5,04 gam. Ta chn B Cu 83 : Kh hon ton 6,4 gam hn hp gm CuO v Fe2O3 bng kh H2 thy to ra 1,8 gam H2O. Khi lng hn hp kim loi thu c sau phn ng l : A. 4,5 gam B. 4,8 gam C. 4,9 gam D. 5,2 gam Gii: xit kim loi = hn hp kim loi + O

n O =n H 2O =

1,8 = 0,1 m O =0,1.16=1,6 (gam) m kim loai = 6, 4 1, 6 = 4,8g . Ta chn B 18

Cu 84 : Cho V lt hn hp kh (ktc) gm CO v H2 phn ng vi mt lng d hn hp rn gm CuO v Fe3O4 nung nng. Sau khi phn ng xy ra hon ton thy khi lng hn hp rn gim 0,32 gam. Tnh V. A. 0,448 B. 0,112 C. 0,224 D. 0,560 Gii: Khi lng hn hp rn gim 0,32 gam = khi lng xi trong xit tham gia phn ng

nO =

0,32 =0,02 = n (H2 ,CO) V(H2 ,CO) =V=0, 02.22, 4 = 0, 448 . Ta chn A 16

Cu 85 : Kh hon ton a gam mt xit st bng CO nhit cao ngi ta thu c 14,56 gam Fe v 8,736 lt CO2 (ktc). Vy cng thc xit st l : A. FeO B. Fe3O4 C.Fe2O3 D. Fe2O3 hoc Fe3O4

n 14,56 8,736 0, 26 2 =0,26, n O (oxit)=n CO2 = =0,39, Fe = = . Ta chn C 56 22,4 n O 0,39 3 Cu 86 (H Khi A 2010) : Cho m gam hn hp bt X gm ba kim loi Zn, Cr, Sn c s mol bng nhau tc dng ht vi lng d dung dch HCl long, nng thu c dung dch Y v kh H2. C cn dung dch Y thu c 8,98 gam mui khan. Nu cho m gam hn hp X tc dng hon ton vi O2 (d) to hn hp 3 oxit th th tch kh O2 (ktc) phn ng l A. 2,016 lt. B. 0,672 lt. C. 1,344 lt. D. 1,008 lt. Gii : 3 kim loi trn khi phn ng vi HCl long nng u b oxi ha thnh s oxi ha +2. Cn khi tc dng O2, Zn tGii: n Fe =


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GV : Nguyn V Minh+2, Cr to +3, Sn to +4. - Gi s mol mi kim loi l a (mol) th: m M Cl = 3a.(2


65 + 52 + 119 + 71) = 8,98 => a = 0,02 . 3 2.0, 02 + 3.0, 02 + 4.0, 02 - Bo ton (e) cho qu trnh tc dng O2: nO2 = = 0, 045(mol ) 1, 008(lit ) 4 Chn D Cu 87 (C 2009) : Kh hon ton mt xit st X nhit cao cn va V lt CO ( ktc), sau phn ng thu c 0,84 gam Fe v 0,02 mol kh CO2. Cng thc X v gi tr V ln lt l : A. FeO v 0,224 B. Fe2O3 v 0,448 C. Fe3O4 v 0,448 D. Fe3O4 v 0,224 Gii: nO (trong xit) = nCO = nCO2 = 0,02 mol ; nFe = 0,015 mol n Fe 0, 015 3 = = Fe3O 4 , V = 0,02.22,4 = 0,448 lt, chn C nO 0, 02 4 Cu 88 (C 2011): t chy hon ton 17,4 gam hn hp Mg v Al trong kh oxi (d) thu c 30,2 gam hn hp oxit. Th tch kh oxi (ktc) tham gia phn ng l: A. 17,92 lt B. 4,48 lt C. 11,20 lt D. 8,96 lt m mkl 30, 2 17, 4 1 Gii: n O 2 = n O = oxit = = 0,8 mol VO 2 = 0,8.22, 4 = 17,92 lit , chn A 2 16 16 Cu 89 : Thi lng kh CO d i qua ng ng hn hp hai xit Fe3O4 v CuO nung nng n khi phn ng xy ra hon ton thu c 2,32 gam hn hp kim loi. Kh thot ra c a vo bnh ng dung dch Ca(OH)2 d thy c 5 gam kt ta trng. Khi lng hn hp hai xit kim loi ban u l : A. 3,12 gam B. 3,21 gam C. 4 gam D. 4,2 gam Gii: Do Ca(OH)2 d nn ta lun c : nCO2 = nCaCO3 = nO = nCO = 0, 05 molm xit = m kim loi + mO = 2,32 + 16.0,05 = 3,12 gam , chn A Cu 90 (s dng quy i xit = kim loi + O): Cho 2,13 gam hn hp X gm ba kim loi Mg, Cu v Al dng bt tc dng hon ton vi oxi thu c hn hp Y gm cc oxit c khi lng 3,33 gam. Th tch dung dch HCl 2M va phn ng ht vi Y l A. 50 ml. B. 57 ml. C. 75 ml. D. 90 ml. Gii: n O = VHCl =

moxit mkim loai 3,33 2,13 = = 0,8 mol , nH = n 16 16+

HCl

= 2.n O = 0,15 mol

0,15 = 0, 075 lit = 75 ml , chn C 2 Cu 91 (s dng quy i xit = kim loi + O): t chy hon ton 26,8 g hn hp 3 kim loi Fe, Al, Cu thu c 41,4 g hn hp 3 oxit. Th tch dung dch H2SO4 1M cn dng ha tan va hn hp oxit trn l A. 1,8250 lt. B. 0,9125 lt. C. 3,6500 lt. D. 2,7375 lt Gii: n O =

moxit mkim loai 41, 4 26,8 0,9125 = = 0,9125 mol = n H 2SO4 VH 2SO4 = = 0,9125 lt 16 16 1

Ta chn B Cu 92 : Cho 31,9 gam hn hp Al2O3, ZnO, FeO, CaO tc dng ht vi CO d nung nng thu c 28,7 gam hn hp Y. Cho Y tc dng vi dung dch HCl d thu c V lt H2(ktc). Th tch H2 l: A . 5,6 lt B. 6,72 lt C. 4,48 lt D. 11,2 lt Gii: Khi lng nguyn t xi = gim khi lng cht rn mO = 31,9 28,7 = 3,2 gam n O = n H2 = n CO = 0, 2 mol V = 0, 2.22, 4 = 4, 48 lit , chn CCu 93 : Cho V lt (ktc) kh H2 i qua bt CuO un nng c 32 gam Cu. Nu cho V lt H2 i qua bt FeO un nng th lng Fe thu c l : A. 24 gam B. 26 gam C. 28 gam D. 30 gam 32 = 0,5 mol mFe = 56.0,5 = 28 gam , chn C Gii: n H2 = n Cu = n Fe = 64


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Cu 94 : Cho 34,8 gam hn hp gm xit v mui cacbonat ca kim loi km R. Ha tan ht hn hp trn bng mt lng va dung dch cha 0,6 mol HCl. Tn kim loi R l : A. Na B. K C. Li D. Cs Gii: R 2 O + 2HCl 2RCl + H 2 O ; R 2 CO3 + 2HCl 2RCl + H 2O + CO 2 Qua hai phng trnh ta thy s mol HCl = 2 ln tng s mol hn hp nn thay hn hp bng 1 cht c s 34,8 mol l 0,3 M = = 116 2 R + 16 < M < 2 R + 60 28 < R < 50 l K (39), chn B 0,3 Dng 8: Ha tan hon ton (K, Na, Ca, Ba) + H2O dd kim ( cha ion OH ) +-

H2

Ta c pt ion sau :

+e H 2O OH +

nOH = 2.nH 2 1 H2 2 mran = mkimloai + mOH

Dung dch sau phn ng trung ha bi axit th

n H+ =n OH-

Cu 95 (H khi B 2007): Cho mt mu hp kim Na-Ba tc dng vi nc (d) thu c dung dch X v 3,36 lt H2 ktc. Th tch dung dch H2SO4 2M cn dung trung ha dd X l A. 60ml B. 30ml C. 75ml D. 150ml+e Gii: H 2 O OH +

1 3, 36 H 2 nOH = 2.nH = 2. = 0, 3 2 2 22, 4 Gi th tch dung dch H2SO4 2M l V n H + =2.n H 2SO 4 = 2.2.V=4Vn H+ = n OH- 4V = 0,3 V = 0,075 lit = 75 ml , chn C

Dung dch sau phn ng trung ha bi axit thCu 96 : Cho a (g) hh Na, K, Ca tc dng vi nc (d) thu c dung dch X v 0,224 lt H2 ktc. Th tch dung dch H2SO4 0,1M cn dung trung ha dd X l A. 0,15 lt B. 0,1 lt C. 0,12 lt D. 0,20 lt Gii: gi V l th tch cn tm 3,36 V = 0,1 lt, chn B n H+ = n OH = 2.n H2 V.2.2 = 2. 22, 4 Cu 97 : Ha tan ht mu hp kim K-Ba vo nc thu c dung dch X v 0,224 lt H2 ktc. trung ha hon ton 1/10 dung dch X trn cn bao nhiu lt dd HCl pH = 2 ? A. 0,2 B. 0,19 C. 0,18 D. 0,16 0, 224 = 0, 02 mol, [ H + ] = 102 M Gii: gi V l th tch cn tm n OH = 2.n H2 = 2. 22, 4 1/10 dung dch X n OH= = 0, 002 mol n H+ = n OH 102.V = 0, 002 V = 0, 2 lt, chn A Cu 98 : Cho hh Na, K, Ca vo nc thu c dung dch A v V (lt) kh H2 ktc. Trung ha 1/3 dung dch A cn 200ml dung dch hn hp HNO3 0,1M v H2SO4 0,5M. Tm V. A. 7,25 B. 7,392 C. 7,27 D. 7,28 V V = Gii: dung dch X : nOH = 2nH 2 = 2 22, 4 11, 2 V 1 / 3 dung dch X : nOH = = nH + = 0,1.1.0, 2 + 0,5.0, 2.2 V = 7,392 , chn B 33, 6 Cu 99 : Ha tan mt mu hp kim Ba Na vo nc c dung dch X v 7,392 lt kh (27,30C, 1 atm). Th tch dung dch HCl 0,2 M cn dng trung ha ht dung dch X l : A. 2 lt B. 1,5 lt C. 3 lt D. 2,5 lt PV 1.1, 792 Gii: n H2 = = = 0,3 mol , RT 0, 082.(273 + 27,3)


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GV : Nguyn V Minhn H+ = n HCl = n OH = 2.n H2 = 0, 6 mol VHCl = 0, 6 : 0, 2 = 3 lt

Cng thc gii nhanh Ha V CTa chn C

Cu 100 : Ha tan hon ton mt lng gm 2 kim loi kim vo nc thu c 200ml dung dch A v 1,12 lt H2 (ktc). Tnh pH ca dung dch A. A. 12 B. 11,2 C. 13,1 D. 13,7 14 0,1 10 = 0,5M [H + ] = = 2.1014 Gii: dung dch A : nOH = 2nH 2 = 0,1 mol; [OH ] = 0, 2 0,5 Nn pH = log[ H + ] = 13, 7 , chn D Cu 101 : Cho m gam hn hp X gm Na Ba tc dng vi nc thu c dung dch Y v 3,36 lt H2 (ktc). Th tch dung dch axit HNO3 2M cn dng trung ha lng dung dch Y l A. 0,15 lt B. 0,3 lt C. 0,075 lt D. 0,1 lt 3,36 = 0,3 mol Gii: n OH = 2.n H2 = 2. 22, 4 0,3 dung dch Y n OH = =0,15 mol n H+ = n HNO3 = n OH V .1.2 = 0,15 V = 0, 075 l = 75 ml , 2 chn C+ Dng 9: Cho Cu vo dung dch hn hp cha cc ion ( H , NO3 ) NO

hoc v d nh phn ng ca Cu kim loi vi hn hp dung dch NaNO3 v dung dch H2SO4 l3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

Dung dch Cu phn ng c th cha nhiu axit nhng nu c ion

Fe3+ th cn ch ti phn ng gia Cu v

n Cu n n + n 3+ nH + sau lp t s Cu , H , NO3 v sau ta s mol ca cht hay ion c t s nh Fe Cn tnh 3 8 2 nNO 3nht vo v tnh V(NO) Cu 102(C 2011): nhn ra ion NO3- trong dung dch Ba(NO3)2, ngi ta un nng nh dung dch vi: A. dung dch H2SO4 long B. kim loi Cu v dung dch Na2SO4 C. kim loi Cu v dung dch H2SO4 long D. kim loi Cu Gii: Chn C (xem trong phng php)Cu 103 (H Khi B 2007): Khi cho Cu tc dng vi dung dch cha H2SO4 long v NaNO3. Vai tr ca NaNO3 trong phn ng l? A. Cht xc tc B. Cht oxi ha C. Mi trng D. Cht kh Gii: Vai tr ca NaNO3 trong phn ng l cht xi ha , chn B Cu 104 : Xem phn ng: a Cu + b NO3- + c H+ d Cu2+ + e NO + f H2O Tng s cc h s (a + b + c + d + e + f) nguyn, nh nht, phn ng trn cn bng, l: A. 18 B. 20 C. 22 D. 24 + 2+ Gii: 3Cu + 8H + 2NO3 3Cu + 2NO + 4H2O, tng h s = 22 chn C Cu 105 (H Khi A 2008): Cho 3,2 gam bt ng tc dng vi 100ml dung dch hn hp gm HNO3 0,8M v H2SO4 0,2 M. Sau khi phn ng xy ra hon ton thu c V lt NO duy nht (ktc). Gi tr V l A. 0,448ml B. 1,792 C. 0,672 D. 0,746

Gii: nCu =

nH + = 0,12 3, 2 = 0, 05; nHNO3 = 0,8.0,1 = 0, 08; nH 2 SO4 = 0, 2.0,1 = 0, 02 64 nNO3 = 0, 08 19


t : 0914449230

GV : Nguyn V Minh3C u + 8H + + 2N O 3 - 3C u 2+ + 2N O 2 + 4H 2 O


Vy V(NO) = V = 0,03.22,4 = 0,672 . Chn C 0,12 0, 03 0, 05 0,12 0, 08 tinh theo H + 3 8 2 Cu 106 (H Khi B 2007): Thc hin hai th nghim : Th nghim 1 : Cho 3,84 gam Cu phn ng vi 80ml dung dch HNO3 1M thot ra V1 lt kh NO duy nht Th nghim 2 : Cho 3,84 gam Cu phn ng vi 80ml dung dch cha HNO3 1M v H2SO4 0,5M thy thot ra V2 lt kh NO duy nht (ktc). Mi quan h gia V1 v V2 l : A. V1 = V2 B. V2 = 2V1 C. V2 = 2,5V1 D. V2 = 1,5V1 3,84 n H + = 0,08 mol = 0,06 mol n Cu = 64 Gii: TN1: n NO3 = 0,08 mol n HNO = 0,08 mol 33Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Ban u :0,06 0,08 0,08 mol H+ phn ng ht 0,02 mol Phn ng: 0,03 0,08 0,02 V1 tng ng vi 0,02 mol NO.TN2: nCu = 0,06 mol ; n HNO3 = 0,08 mol ; n H2SO4 = 0,04 mol. Tng: n H + = 0,16 mol ; n NO = 0,08 mol.3

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Ban u: Phn ng: 0,06 0,16 0,08 mol Cu v H+ phn ng ht 0,04 mol 0,06 0,16 0,04

V2 tng ng vi 0,04 mol NO. Nh vy V2 = 2V1. Chn B Cu 107 (H Khi B 2010): Cho 0,3 mol bt Cu v 0,6 mol Fe(NO3)2 vo dung dch cha 0,9 mol H2SO4 (long). Sau khi cc phn ng xy ra hon ton, thu c V lt kh NO (sn phm kh duy nht, ktc). Gi tr ca V l A. 6,72 B. 8,96 C. 4,48 D. 10,08 + 2+ 3Cu + 8H + 2NO3 3Cu + 2NO + 4H 2O0,3 0,8 0,2 0, 2 mol3Fe 2+ + 4H + + NO3 3Fe3+ + NO + 2H 2O

VNO = (0,2 + 0,2).22,4 = 8,96 lt ,chn B

0,6 0,8 0,2 0, 2 mol Cu 108 : Ha tan ht 3,6 gam FeO bng HNO3 long, va . Thm H2SO4 long d vo dung dch sau phn ng thu c mt dng dch c th ha tan ti a m gam bt Cu v to ra V lt NO (ktc). Gi tr m v V ln lt l : A. 16 gam v 3,36 lt B. 14,4 gam v 3,36 lt C. 1,6 gam v 3,36 lt D. 16 gam v 4,48 lt HNO3 Gii : FeO Fe(NO 3 )3 0,05 0,05 mol Thm H2SO4 long d. Dung dch cha cc tc nhn ca phnng Cu l Fe3+ , H + , NO3

2Fe3+ + Cu 2Fe 2+ + Cu 2+ 0,05 0,025 mol

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O 0,15 0,015 mol 0,225mCu = 0,25. 56 = 16 gam , V = 0,15.22,4 = 3,36, chn A Cu 109 (C 2010): Cho a gam Fe vo 100 ml dung dch hn hp gm HNO3 0,8M v Cu(NO3)2 1M. Sau 20 Email : [email protected] t : 0914449230

GV : Nguyn V Minh+5


khi phn ng xong c kh NO l sn phm kh duy nht ca N v 0,92a gam hn hp kim loi. Gi tr a l : A. 11,0 B. 11,2 C. 8,4 D. 5,6 n 2+ = 0,1 mol Cu Gii : n H+ = 0, 08 mol n NO3- = 0, 28 mol Kim loi cn d nn Fe+ b chuyn v Fe2+ : 0,03 0,08 0,02 2+ + Cu Fe + Cu2+ Fe 0,1 0,1 0,1 Nn : a 56.(0,03 + 0,1) + 64.0,1 = 0,92a a = 11, chn A Cu 110 : Ha tan 1,28 gam Cu vo 50ml dung dch hn hp H2SO4 0,1M v NaNO3 0,5M thu c a mol kh NO dung nht. Tnh a ? A. 0,0025 B.0,0133 C. 0,025 D. 0,032 n = 0, 02 mol Cu Gii : n H + = 0, 01 mol n NO3- = 0, 025 mol Ban u : 0,02 0,01 0,025 mol 0,0025 mol , chn A Phn ng 0,01 Cu 111 : Cho 0,09 mol Cu vo 400ml dung dch cha HNO3 0,3M v H2SO4 0,1M. n khi phn ng kt thc, th tch kh NO duy nht thot ra (ktc) l A. 0,672 lt B. 0,896 lt C. 1,344 lt D. 1,12 lt n = 0, 09 mol Cu Gii : n H + = 0, 2 mol n NO3- = 0,12 mol Ban u : 0,09 0,2 0,12 mol Phn ng 0,2 V = 0,05.22,4 = 1,12 lt , chn D

3Fe + 8H+ + 2NO3 3Fe2+ + 2NO + 4H2O

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O

0,05 mol

Cu 112 : Ha tan 12,8 gam Cu vo 200ml dung dch hn hp H2SO4 1M v KNO3 0,5M. Th tch kh NO duy nht ktc l : A. 2,24 lt B. 2,99 lt C. 4,48 lt D.11,2 lt n = 0, 2 mol Cu Gii : n H + = 0, 4 mol n NO3- = 0,1 mol

Ban u : 0,2 0,4 0,1 mol 0,1 mol Phn ng 0,4 V = 0,1.22,4 = 2,24 lt , chn A Cu 113 : Cho 1,92 gam Cu vo 100 ml dung dch cha ng thi KNO3 0,16M v H2SO4 0,4M thy sinh ra mt cht kh c t khi hi so vi H2 l 15. Th tch kh ktc l : A. 0,672 lt B. 1,446 lt 21 C. 0,3584 lt D. 0,4568 lt

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O


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n = 0, 03 mol Cu Gii : n H + = 0, 08 mol v Kh c M = 15.2 = 30 l NO n NO3- = 0, 016 mol 3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Ban u : 0,03 0,08 0,016 mol 0,016 mol Phn ng 0,016 V = 0,016.22,4 = 0,3584 lt , chn C Cu 114 : Cho 1,12 gam Cu vo 50 ml dung dch cha ng thi KNO3 0,16M v H2SO4 0,1M thy c kh NO ( sn phm kh duy nht ca s kh N+5 ) bay ra. kt ta ton b Cu2+ trong dung dch sau phn ng cn ti thiu bao nhiu lt dung dch NaOH 0,5M. A. 0,07 lt B. 0,015 lt C. 0,064 lt D. 0,048 lt n = 0, 0175 mol Cu Gii : n H + = 0, 01 mol n NO3- = 0, 008 mol 3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O 0,00375 0,01 0,0025 mol 0,00375 n OH = 2 n Cu 2+ = 0, 0075 mol nn V = 0,0075 : 0,5 = 0,015 lt , Chn B2 Dng 10: Cho dd chc ion H+ vo dung dch cha CO3 , HCO3

Yu cu : tnh s mol ca cc ion sau H + , CO32 , HCO3 ( nu c thm HCO3- t gi thuyt )+ 2+ u tin s c phn ng H + CO3 HCO3 + + Sau nu H+ cn d mi xy ra tip phn ng sau : H + HCO3 CO 2 + H 2 O

-

2Vy : nu c kh thot ra th CO3 ht v nu khng c kh thot ra th H + ht

-

2Nu bi ton hi ngc li : cho CO3 vo dung dch cha H + th ch c 1 phn ng duy nht 22H + + CO3 CO 2 + H 2 O

Cu 115 (H Khi A 2009):Dung dch X cha hn hp gm Na2CO3 1,5M v KHCO3 1M. Nh t t tng git cho n ht 200 ml dung dch HCl 1M vo 100 ml dung dch X, sinh ra V lt kh ( ktc). Gi tr ca V l A. 4,48. B. 1,12. C. 2,24. D. 3,36. 2Gii: n (Na2CO3) = 0,1.1,5 = 0,15 mol = n (CO3 ) n (KHCO3) = 0,1.1 = 0,1 mol = n ( HCO3-) n (HCl) = 0,2.1 = 0,2 mol = n (H+) 2H + + CO3 HCO3 + u tin s c phn ng 0,15 0,15 0,15vy tng s mol HCO3 l 0,15 + 0,1 = 0,25 mol + Sau do H+ cn d ( 0,02 0,15 = 0,05 mol ) nn xy ra tip phn ng sau : H + du + HCO3 CO 2 + H 2 O vy VCO2 =0,05.22,4=1,12 l chn B 0,05 0,05 0,05

Cu 116 (H Khi A 2007):Cho t t dung dch chc a mol HCl vo dung dch cha b mol Na2CO3 ng thi khuy u, thu c V lt kh (ktc) v dung dch X. Khi cho d nc vi trong vo dung dch X thy c xut hin kt ta. Biu thc lin h gia V, a v b l : A. V = 22,4(a + b) B. V = 11,2 (a b) C. V = 11,2(a + b) D. V = 22,4(a b)


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Gii: Khi cho d nc vi trong vo dung dch X thy c xut hin kt ta chng t c mui NaHCO3 : Ca(OH) 2 + 2NaHCO3 CaCO3 +Na 2 CO3 + 2H 2O Vy bi ton c ngha l cho a mol H + vo b mol CO 3 cho kh CO2 v mui NaHCO32H + + CO3 HCO3

2-

b btnh theo s mol CO3 do H + phi cn d mi to phn ng th hai c , ta chn D ( a b) ( a b) ( a b) Cu 117 (H Khi A 2010) : Nh t t tng git n ht 30 ml dung dch HCl 1M vo 100ml dung dch cha Na2CO3 0,2 M v NaHCO3 0,2M. Sau phn ng thu c s mol CO2 l : A. 0,01 B. 0,015 C. 0,020 D. 0,030 Gii: n CO2 = 0, 02 mol, n HCO = 0, 02 mol, n H + = 0, 03 mol3 3

2-

H + du

+

HCO3 CO 2 + H 2 O

V(CO 2 )=(a-b).22,4

0, 02 0, 02 0, 02 mol vy tng s mol HCO3 l 0,02 + 0,02 = 0,04 mol + Sau do H+ cn d ( 0,03 0,02 = 0,01 mol ) nn xy ra tip phn ng sau : H + du + HCO3 CO 2 + H 2 O 0,01 0,01 0,01 Vy ta chn A Cu 118 : Cho t t dung dch HCl vo dung dch Na2CO3 n d th A. C si bt kh ngay lp tc. B. Ban u khng c si bt, mt thi gian sau si bt. C. Khng hin tng. D. C kt ta mu trng. Gii : Chn B Cu 119 : Cho t t 100ml dung dch HCl 2M vo 200ml dung dch Y cha KHCO3 1M v K2CO3 1M. Th tch kh CO2 thot ra ktc l : A. 0,0 lt B. 4,48 lt C. 2,24 lt D. 3,36 lt n CO2 = 0, 2 mol, n HCO = 0, 2 mol, n H + = 0, 2 mol Gii:3 3

+ u tin s c phn ng

H

+

2+ CO3 HCO3

0, 2 mol 0, 2 0, 2 Do H ht phn ng trn nn s khng c kh CO2 thot ra, chn A Cu 120 : Nh rt t t n ht 500ml dung dch HCl nng 1,2M vo 480ml dung dch Na2CO3 1M ang c khuy nh, u. Sau phn ng thu c mt dung dch v V lt kh (ktc). Tnh V ? A. 2,688 B. 13,44 C. 10,752 D. 6,288 n CO2 = 0, 48 mol, n H + = 0, 6 mol Gii:+3

u tin s c phn ng

H+

2 + CO3 HCO3

+ u tin s c phn ng 3

H+

2 + CO3 HCO3

0, 48 0, 48

0, 2 mol

s mol HCO l 0,2 mol + Sau do H+ cn d ( 0,6 0,48 = 0,12 mol ) nn xy ra tip phn ng sau : H + du + HCO3 CO 2 + H 2 O V (CO2) = 0,12.22,4 = 2,688 mol , chn A 0,12 0,12 0,12 mol Dng 11 : PHNG PHP QUY I TRONG HA HC Phm vi p dng : nhng bi ton hn hp cho qua Axit c tnh xi ha mnh ra sn phm kh ri em dung dch cho vo kim thu kt ta, sau nung kt ta


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Cng thc gii nhanh Ha V C Mg , S MgS , S MgS , Mg

Mg 1/ MgS S Cu CuS 2/ S Cu2 S Fe FeS 3/ S FeS 2 gii

quy i v

quy i v

Cu , S CuS , S CuS , Cu CuS , Cu2 S .............. Fe, S FeS , S FeS 2 , Fe FeS , FeS 2 ..............

quy i v

Ch : ch a a v 2 cht trong nhiu cht sau ta thng dng bo tan e- v pt hn hp Tnh : V kh sinh ra, khi lng rn thu c, khi lng hn hp u Cu 121 : Ha tan han ton 20,8 gam hn hp gm Fe, FeS2 v S bng HNO3 d, thot ra sn phm kh duy nht l 53,76 lt kh NO2 (ktc) v dung dch A. Cho NaOH d vo dung dch A thu kt ta, nung kt ta trong khng kh n khi lng khng i thu c m (g) cht rn. Tnh gi tr m. A. 16 gam B. 17 gam C. 18 gam D. 19 gam Gii: NO2 : 2,4 mol Ta quy i hn hp v Fe (x mol)v FeS2 (y mol) => mhh = 56x + 120y = 20,8 (1) (c th quy i cch khc, v nu gii ra s mol S m th bi tan vn ng )

FeS 2 15e Fe +3 + 2 S +6 Fe 3e Fe +3 y 15 y y x 3x x Vy tng s mol e- nhng l ncho = 3x + 15y v nnhn =( gim s oxh) . (s mol kh ) = 1.2,4 x = 0, 05 = n Fe Fe3+ : x + y = 0, 2mol 3.x + 15.y =2,4 (2) => ddA 2 y = 0,15 = n FeS2 SO4 = nS = 2.0,15 = 0,3

0

0

Fe 3 + OH Fe (OH ) 3 nung Fe2 O3 : a ( mol ) , O2nFe = nFe3+ = nFe( Fe2O3 ) 0, 2 = 2a a = 0,1 mFe2O3 = 0,1 (56 2 + 16 3) = 16 g , chn ACu 122 : Ha tan hon ton 30,4g rn X gm Cu, CuS, Cu2S v S bng HNO3 d, thot ra 20,16 lt kh NO (sp kh duy nht ktc) v dung dch Y. Thm Ba(OH)2 d vo dung dch Y thu c m gam kt ta. Tm gi tr m. A. 81,55 B. 104,20 C. 110,95 D. 115,85 Gii: NO : 0,9 mol Ta quy i hn hp v Cu (x mol) v S (y mol) mhh = 64x + 32y = 30,4 gam (1)

Cu 2e Cu +2 x 2x x

0

S 6e S y 6y

0

+6


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v nnhn = ( gim s oxh) . (s mol kh ) = 3.0,9 = 2,7 mol64 x + 32 y = 30, 4 Ta c h : x = 0,3; y = 0,35 2 x + 6 y = 2, 7 2 OH Cu 2+ Cu(OH) 2 Nn m Cu(OH)2 = 0,3.98 = 29, 4 gam 0, 3 0,3 mol Ba 2+ + SO 2 BaSO 4 4 Nn m BaSO4 = 0,35.233 = 81,55 gam , 0, 35 0,35 mol Khi lng kt ta = 29,4 + 81,55 = 110,95 gam, chn C

Vy tng s mol e- nhng l ncho = 2.x + 6.y

Cu 123 : Hn hp X gm Mg, MgS v S. Ha tan han ton m gam X trong HNO3 d thu c 2,912 lt kh N2 duy nht (ktc) v dung dch Y. Thm Ba(OH)2 d vo Y thu c 46,55 g kt ta. Tm m. A. 4,8 B. 7,2 C.9,6 D. 12,0 Gii: N2 : 0,13 mol , Ta quy i hn hp v Mg (x mol) v S (y mol) mhh = 24x + 32y

Mg 2e Mg +2 x 2x x

0

S 6e S y 6y

0

+6

Bo ton electron : 2x + 6y = 10.0,13 (1) Mg 2+ + 2OH Mg(OH)2 Ba 2+ + SO2 BaSO 4 4 x x y y m = 58 x + 233 y = 46,55 gam (2) Suy ra x = 0,2 mol , y = 0,15 mol , mhh = 24x + 32y = 9,6 gam , chn CCu 124 : Ha tan hon ton 25,6 g hn hp X gm Fe, FeS, FeS2 v S vo dung dch HNO d thu c V lt kh NO duy nht ktc v dung dch Y. Thm Ba(OH)2 d vo Y thu c 126,25 g kt ta. Tm V. A. 17,92 B. 19,04 C. 24,64 D. 27,58 Gii: Ta quy i hn hp v Fe (x mol) v S (y mol) mhh = 56x + 32y = 25,6 gam (1) m = m Fe(OH)3 + m BaSO4 = 107 x + 233 y = 126, 25 (2)

y = 0,45 mol V Bo ton electron : 3x + 6y = 3. V = 24,64 lt, chn C 22,4Cu 125 : Ha tan hon ton 3,76 gam hn hp: S, FeS, FeS2 trong HNO3 c d c 10,752 lt NO2 ( 27,30C v 1,1 atm) l sn phm kh duy nht v dung dch X. Cho dung dch Ba(OH)2 d vo X, lc kt ta nung n khi lng khng i th khi lng cht rn thu c l: A. 17,545 gam B. 18,355 gam C. 15,145 gam D. 2,4 gam Gii: NO2 : 0, 48 mol Ta quy i hn hp v Fe (x mol) v S (y mol) mhh = 56x + 32y = 3,76 gam (1) Bo ton electron : 3x + 6y = 1.0,48 (2) Ta c : x = 0,03 , y = 0,065 mol m BaSO4 = 233 y = 15,145 gam

(1) v (2) ta c : x = 0,2 mol ;

2Fe(OH)3 Fe 2O3 0,03 0,015 mol mFe2O3 = 0, 015.160 = 2, 4 gam Khi lng cht rn = 15,145 + 2,4 = 17,545 gam , chn A Dng 12 :


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HNO 3 FeS, FeS2 , CuS, Cu 2S SO 4 2- ,Fe3+ ,Cu 2+ + NO

( d.d ch cha mui sunfat duy nht )Cu 126 ( H Khi A 2007) : Ha tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo acid HNO3 (va ), thu c dung dch ch cha 2 mui sunfat) v kh duy nht NO. Gi tr ca a l ? A. 0,04 B. 0,075 C. 0,12 D.0,06 3+ Gii: Dung dch thu c ch cha mui sunfat duy nht nn ch cha cc ion : Fe ; Cu2+ ; SO42-

Bo ton in tch : 0,12 . 3 + 2.2a = (0,24 + a ).2 a = 0,06 , chn D

n Fe = n 3+ = 0,12.1 = 0,12 Fe 0,12 mol FeS2 bao toan nguyen to n Cu = n Cu 2+ = 0, 2.a = 0, 2a a mol Cu 2S n S = n SO42- = 0,12.2 + a.1 = 0, 24 + a

Cu 127 : Ha tan hon ton hh gm a mol FeS2 ; 0,5a mol FeS v 0,06 mol Cu2S vo axit HNO3 va thu c dung dch X ch cha 2 mui sunfat v kh NO duy nht . Gi tr ca a l : A.0,24 B.0,2 C.0,12 D.0,06 3+ 2+ Gii: Dung dch thu c ch cha mui sunfat duy nht nn ch cha cc ion : Fe ; Cu ; SO42-

Bo ton in tch : 1,5a.3 + 0,12.2 = (0,06 + 2,5.a ).2 a = 0,24 , chn A Cu 128: Khi cho hn hp FeS v Cu2S phn ng vi dung dch HNO3 d s thu c dung dch cha cc ion B. Cu2+ , Fe3+ , H+ , NO3- . A. Cu2+, S2-, Fe2+ , H+, NO3 -. 2+ 3+ + 32D. Cu2+ , SO42- , Fe2+ , H+ , NO3- . C. Cu , SO4 , Fe , H , NO . Gii: Chn C v cn c HNO3 dCu 129 : Ha tan hon ton hn hp gm x mol FeS2 v 0,05 mol Cu2S vo acid HNO3 (va ), thu c dung dch ch cha 2 mui sunfat) v kh duy nht NO. Gi tr ca x l ? A. 0,15 B. 0,2 C. 0,1 D.0,05

n = n 3+ = a + 0,5a = 1,5a a mol FeS2 Fe Fe bao toan nguyen to 0,5a mol FeS n Cu = n Cu 2+ = 0, 06.2 = 0,12 0, 06 mol Cu S 2 n S = n SO42- = 2,5a + 0, 06

Bo ton in tch : x.3 + 0,1.2 = (0,05 + 2.x).2 x = 0,1 , chn C Cu 130 : Cho hn hp A gm x mol FeS2 v y mol Cu2S tc dng vi HNO3 long, d un nng ch thu c mui sunfat duy nht ca cc kim loi v gii phng kh NO duy nht. T l x/y c gi tr l : A. 0,5 B. 1 C. 1,5 D. 2

n Fe = n 3+ = x Fe x mol FeS2 bao toan nguyen to n Cu = n Cu 2+ = 0, 05.2 = 0,1 Gii: 0, 05 mol Cu 2S n = n SO42- = 2 x + 0, 05 S

n Fe = n 3+ = x Fe x mol FeS2 bao toan nguyen to n Cu = n Cu 2+ = 2 y Gii: y mol Cu 2S n = n SO42- = 2 x + y S

Bo ton in tch : x.3 + 2y.2 = (y + 2.x).2 x : y = 2 , chn D

Cu 131 : Ha tan hon ton a gam hn hp X ( FeS2 v Cu2S ) vo dung dch HNO3 (va ) thu c dung dch Y ch cha 2 mui sunfat duy nht v 17,92 lt kh NO l sn phm kh duy nht. Tm a : A. 12 B. 16 C. 24 D. 25


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Bo ton in tch : x.3 + 2y.2 = (y + 2.x).2 x : y = 2 hay x = 2y N + 3e N 2,4 0,8 molFeS2 x Cu 2S 2Cu0 +2 0+5 +2

n Fe = n 3+ = x Fe x mol FeS2 bao toan nguyen to n Cu = n Cu 2+ = 2 y y mol Cu 2S n S = n SO42- = 2 x + y

Fe

+3

+ +

2S

+6

++6

15e 15x +

1S

10e

y 10y Bo ton electron : 15x + 10y = 2,4 suy ra x = 0,06 mol , y = 0,12 mol Vy a = 120.0,12 + 160.0,06 = 24 gam, chn C Dng 13 : TON MUI ( TON THUN) Cho CO2 hocSO2 hoc H2S hoc P2O5 hoc H3PO4 vo dung dch cha mt trong cc dung dch NaOH, KOH , Ca(OH)2 , Ba(OH)2 Cho NH3 vo dung dich cha H3PO4 V nhiu dng tng t , ta cn lp s mol ca 2 nguyn t quan trng s thy sn phm v s dng bo ton nguyn t s tm ra s mol sn phm d dng m khng cn phi vit phng trnh phn ng

Cu 132 : 0,1 mol CO2 hp th vo dd cha 0,2 mol NaOH. Tm cc cht tan c trong dung dch. A. Na2CO3 B. NaHCO3 D. NaHCO3 v Na2CO3 C. Na2CO3 v NaOH d Gii: Nhn xt : ch c th c 2 mui l Na2CO3 ( Na : C = 2 : 1= 2 ) hoc NaHCO3 ( Na : C = 1 : 1 = 1) nC = nCO2 = 0,1 mol , nNa = n NaOH = 0,2 mol n 0, 2 = 2 to mui Na2CO3 ( Na : C = 2 : 1 ), chn A lp t l Na = nC 0,1 Cu 133 : 0,1 mol CO2 hp th vo dung dch cha 0,18 mol KOH. Tm cc sn phm A. K2CO3 B. KHCO3 D. KHCO3 v K2CO3 C. K2CO3 v KOH d Gii: Nhn xt : ch c th c 2 mui l K2CO3 ( K : C = 2 : 1 =2 ) hoc KHCO3 ( K : C = 1 : 1 = 1) ng dng nC = nCO2 = 0,1 , nK = n KOH = 0,18 K 2 CO3 n 0,18 lp t l 1 < K = = 1,8 < 2 c c 2 mui nC 0,1 KHCO3 Ta chn D Cu 134 : 0,5 mol P2O5 cho vo dd cha 0,8 mol Ca(OH)2. Tm cc sn phm ? B. Ca(H2PO4)2 v CaHPO4 A. Ca3(PO4)2 D. Ca(H2PO4)2 B. Ca(H2PO4)2 v Ca3(PO4)2 Ca3(PO4)2 ( Ca : P = 3 : 2 = 1,5 ) Gii: Nhn xt : ch c th c 3 mui l Ca(H2PO4)2 ( Ca : P = 1 : 2 = 0,5 ) ( Ca : P = 1 : 1 = 1 ) v CaHPO4 n P = 2.0,5 = 1 mol , n Ca = 0,8.1 = 0,8 mol


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CAC DANG TOAN VO CO

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