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Cng thc gii nhanh Ha V C

1

Cc Dng Thng GpDng 1 :

Kim Lai + axit loi 1 ( H2SO4long hoc HCl) mui (sunfat hoc clorua) + 2H

mmuisunfat = mkim loi + 96 2Hn hay mmuiclorua = mkim loi + 71 2Hn

Bo ton e : ncho = nnhn vi ncho= mol kim loi . ha tr kim loi

nnhn

= 2.2

Hn 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 bao nhiu gam mui khan ? A. 10,8 B. 11,5 C. 12,3 D,14,6

Gii :2H

2,24n 0,1

22,4= = mmui = mkim loai + 71.0,1=5,2+7,1=12,3 . Ta chn C

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 thuc m gam mui khan. Gi tr m l : A. 40,4 B. 37,2 C. 36,4 D. 34,8

Gii :2

0,50,25

2Hn = = mmui = mkim lai + 96.0,25=10,8 + 24 = 34,8 . Ta chn D

Cu 3 (H khi B 2010): Ho tan hon ton 2,45 gam hn hp X gm hai kim loi kim th vo 200 m

dung dch HCl 1,25M, thu c dung dch Y cha cc cht tan c nng mol bng nhau. Hai kim loi trongX 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 vbng s mol HCl d (nu c) , nHCl = 0,25 molM + 2HCl MCl2 + H2 nn ,a 2a a

nHCl (d) = a/2 nn9 40

0,25 2 0,5 0,1 M 24,52

a a a+

= = = = nn c Be v Ca l hp l, chn D

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 tcdng 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 tr2M + 2nHCl 2MCln + nH2

Bo ton elctron :1 M 14,27,1 5,6

.n 2. M 14,2n2 M 28,4M 22,4

n

n

= == =

= =

m 1 < n < 2 nn 14,2 < M < 28,4ta chn Na v Mg , p n ACu 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 mukhan 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

mmuiclorua = m(Mg, Al) + 71 2Hn = 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 ththu c m gam cht rn c khi lng

A. 2,95 gam B.2,24 gam C. 3,9 gam D. 1,85 gam

Gii: Kh l H2 v mui thu c s l mui clorua : mmuiclorua = 1,53+ 710 , 4 4 8

2 2 , 4= 2,95 gam , chn A

Cu 7: Ha tan hon ton 14 gam mt kim loi vo H2SO4 long d thu c 5,6 lt kh (ktc). Kim loi lA. Al B. Fe C. Zn D. Mg

Gii: bo ton electron :214 5,6

.n 2. M 28n56M 22,5

nFe

M

== =

=vi n l ha tr kim loi

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Ch : Fe tc dng vi axit loi 1 ch ra hoa1 tr IICu 8 (C 2007): Ha tan han ton 3,22g hn hp X gm Fe, Mg, Zn bng mt lng va dung dchH2SO4 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

Gii: Kh l H2 v mui thu c s l mui sunfat :m mui sunfat = 3,22+ 961 , 3 4 4

2 2 , 4= 8,98 gam , chn B

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 lA. 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 0,01mol2,9 = 0.01(M+16) + M.0,01 M =137 Ba, chn C

Dng 2 :

Mui cacbonat + axit loi 1 ( H2SO4long hoc HCl) mui (sunfat hoc clorua) 2CO

mmuisunfat = mmui cacbonat + 36 2C On do2- 2-

3 2 4 4 2 2CO +H SO SO +CO +H O

mmuiclorua = mmui cacbonat + 11 2C On do2- 2

3 2 2CO +2HCl 2Cl +CO +H O v nmui cacbonat = n mui hidr cacbonat = 2COn

Cu 10: Cho 12 g hn hp mui cacbonat ca kim lai kim v kim th vo dung dch cha HCl d thu 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:2CO

2,24n 0,1

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

Cu 11: Cho m g hn hp 3 mui cacbonat ca kim nhm IA, IIA v IIIA vo dung d ch H2SO4long, d thuc 2,8 lt kh ktc. C cn dung dch sau phn ng thu c 22,65 gam mui khan. Gi trm l :

A. 19,25 B. 20,05 C. 18,15 D. 17,86Gii:

2CO

2,8n 0,125

22,4= = m= mmui cacbonat = mmui sunfat

2CO36.n =22,65 0,125.36 18,15 = .Chn C

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 mCO2 (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

Gii: Khi lng mui khan mmui clorua = 3,06+ 11.0 , 6 7 2

2 2 , 4=3,39 . Ta chn A

Cu 13: Hn hp X gm hai mui cacbonat ca 2 kim loi kim thhai chu k lin tip. Cho 7,65 gam Xvo 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

Gii:2

8, 75 7, 65 0,111 11

muoi clorua muoi cacbonatCO muoi cacbonat

m mn n

= = = =

Gi cng thc chung hai mui cabonat l MCO3 c + 60 =7,65

76,5 16,50,1

M= = nn ta chn C

Cu 14: Cho 3,6 gam hn hp A. gm 2 mui cacbonat ca 2 kim loi k tip nhau trong phn nhm chnhnhm II.Cho A. tan ht trong dung dch H

2SO

4long 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 c



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Gii: Do dung dch Ca(OH)2 d nn2 3CO CaCO muoicacbonat

5n n n 0,05 mol n

100= = = = =


72 120,05

M= = nn ta chn C

Cu 15: Ho tan ht 2,25 gam hn hp hai mui cacbonat ca hai kim loi A, B ( k tip nhau trong phn nhchnh nhm II) bng dung dch HCl thu c 0,56 lt kh CO

2(ktc). Hai kim loi A, B l:

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

Gii:2CO muoicacbonat

0,56n 0,025 mol n22,4

= = =


90 300,025

M= = nn ta chn D

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 MgGii:

2CO muoicacbonatn 0, 2 mol n= =


92 32

0,2

M= = nn ta chn C

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

Gii: mmui clorua = mmui cacbonat + 11.nCO2 = 115 +0,448

.1122,4

= 115,22 gam . Ta chn A

Cu 18: Ho tan hon ton 4,68g hn hp mui cacbonat ca hai kim loi A v B k tip trong nhm IIA vdd HCl thu c 1,12 lit CO2ktc. 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.

Gii:2CO muoicacbonat

1,12n 0,05 mol n

22,4= = =


93,6 33,60,05

M= = nn ta chn B

Dng 3 : Bo Ton IN TCH

Cho : dung dch

m+

n+

M :a (mol)X

N :b(mol)

v

x-

z-

X :c(mol)

Z :d (mol)

Bo ton in tch : m.a + n.b = x.c + z.dmmui= khi lng tt c ion = M.a + N.b + X.c + Z.d

Cu 19: Mt dung dch cha 0,2 mol

2+

Ca ; 0,1 mol

2+

Mg ;0,1 mol 3HCO

v x mol Cl

. Tm x ?A. 0,5 B. 0,6 C. 0,7 D. 0,8Gii: 0,2.2 + 0,1.2 = 0,1.1 + x.1 suy ra x = 0,5 chn ACu 20: Mt dung dch cha 0,1 mol 2+M ; 0,05 mol 3+Al ; 0,1 mol Cl v x mol 24SO

. 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 = 0,125mmui = M.0,1 + 27.0,05 + 35,5.0,1 + 96.0,125 =19,3 suy ra M = 24 (Mg), chn A

Cu 21 (H Khi A 2010): Cho dung dch X gm: 0,007 mol Na+; 0,003 mol Ca2+; 0,006 mol Cl-; 0,006

3HCO v 0,001 mol

3NO . loi b ht Ca2+ 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. 0,180Gii: nCa(OH)2 = x. nOH- = 2x v nCa2+ = x. Theo bi:

OH- + HCO3- CO3

2- + H2O2x 0,006 --------- 0,006

Ca2+ + CO32-- CaCO3.

x + 0,003 0,006Ch c x = 0,003 tha mn. Vy a = 0,003.74 = 0,222 (g) , ta chn ACu 22 (H Khi A 2010): Dung dch X c cha: 0,07 mol Na+; 0,02 mol SO4

2- 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 dchZ. Dung dch Z c pH (b qua sin 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 ACu 23 (C 2007): Dung dch A cha cc ion Al3+=0,6 mol, Fe2+=0,3mol, Cl- = a mol, SO4

2- = b mol. C cdung 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 0,3Gii: 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 35,5 96 107,7a b= + + + = + = 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 cdung dch X v 2,688 lt kh H2 (ktc). Dung dch Y gm HCl v H2SO4, t l mol tng ng l 4 : 1. Trungha 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 gamGii: 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

: Gii: Dung dch thu c ch cha mui sunfat duy nht nn ch cha cc ion : Fe3+ ; Cu2+ ; SO42-

3

2+

2-4

Fe Fe2 bao toan nguyen to

Cu Cu2

S SO

n n 0,12.1 0,120,12 mol FeS

n n 0, 2.a 0, 2aa mol Cu S

n n 0,12.2 a.1 0, 24 a

+ = = =

= = =

= = + = +

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

c bao nhiu gam mui khan ?A. 27,9 gam B. 59,7 gam C.30,4 gam D. 22,0 gamGii: Bo ton in tch : 0,1 . 2 + 0,2 . 2 = 0,2 . 1 + x . 1 x = 0,4Nn ch khi b nhit phn th s c phng trnh : 23 3 2 22HCO CO CO H O

+ +

0,4 -------->0,2

2+ 2+ - 2-3

muoi Ca Mg Cl COm m m m m 0,1.40 0, 2.24 35,5.0,2 0, 2.60 27,9= + + + = + + + = , chn A

Dng 4: xit kim loi + Axit mui + H2O

2 n 2 4M O HCl( hay H SO )+ mui + nc2-

+2O ( )+ 2H = H Otrong oxit



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+2O H OH

n =2.n =n v mmui = mkim loi + mgc axit vi mkim loi = m xit mOHoc c th dng cng th tnh nhanh cho trc nghim :

+ i vi H2SO4 (long) : m mui sunfat = m xit + 80. 2 4H SOn

+ i vi HCl : m mui clorua = m xit + 27,5. HCln 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 mA. 13,1 B. 40,2 C. 39,4 D. 41,8

Gii: 0,5.1,6 0,8 0,8 0, 4 16.0,4 6,4( )2H

HCl O OH

nn n n m g

+

+= = = = = = =

19,8 6, 4 13, 4( ), 0,8 0,8.35,5 28, 4kl Cl Cl m g n m = = = = = Vy 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. HCln = 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 dungdch 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. 5,81gGii: nH2SO4 = 0,05 = n SO42 nH+ = 0,12H+ + O2= H2O0,1 0,05 mol

mmui = moxit mO(trong oxit) + mgc axit= 2,81 0,05.16 +0,05.96 = 6,81 gam, chn ACu 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

Gii: TH1 : X + HCl : e-cho5,6

n 2. 0,5 mol22,4

= =

TH2 : X + O2 : 2 22O O 4e + 22 Oe-cho OV

n 4.n 4. mol22,4= =

Do ha tr 2 kim loi khng i nn s mol e cho ca 2 phng trnh bng nhau

2

2

OO

V0,5 4. mol V 2,8 lit

22,4= = , chn A

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 gam

Gii: m mui clorua = m xit + 27,5. HCln = 50 + 27,5. 0,2 . 4 = 72 (g), chn CCu 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 khngi thu c m gam cht rn. Gi tr m l :

A. 6 gam B. 7 gam C. 8 gam D. 9 gam

Gii: + +HCl O FeH H1

n n 0,26 mol , n (trong oxit) n 0,13 mol m 7,68 0,13.16 5,62

g= = = = = =

Fen 0,1 mol= , s hp thc : 2Fe Fe2O30,1 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 png 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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Gii: m mui clorua = m xit + 27,5. HCln 28,2 15 27,5.1,6.V V 0,3 300l ml = + = = , chn A

m mui clorua = m xit + 27,5. HCln 28,2 15 27,5.1,6.V V 0,3 300l 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 khiphn 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. HCln 40,6 27,5.1,6.0,4 23 gamoxitm = = , 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. 2 4H SOn 401 281 80.3.V V 0,5 500l ml = + = = , chn CCu 35 (H Khi A 2008): ho tan hon ton 2,32 gam hn hp gm FeO, Fe3O4 v Fe2O3 (trong smol 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.

2 3FeO Fe On n= coi hn hp ch gm 1 xit duy nht l Fe3O4 v

+3 4Fe O O O HClH

n 0,01 mol n 0,04 mol n 2n 0,08 mol n= = = = = suy ra V = 0,08 lt, chn C

Cu 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 gamFeCl3. Gi tr ca m l

A. 9,75. B. 8,75. C. 6,50. D. 7,80.

Coi hn hp ch gm FeO, Fe2O3 : 2FeO FeCl7,62

n n 0,06 mol127

= = =

2 3 3 2 3Fe O FeCl Fe O

9,12 72.0,06n 0,03 mol n 2n 0,06 mol

160

= = = =

m (FeCl3) = 0,06 . 162,5 = 9,75, chn ADng 5:

320

2 4

HNO2 3 3 4 hayH SO dac,t( , , , )

Odu Fe Fe FeO Fe O Fe O+ mui + sn phm kh+ H2O

Bo ton e : 2 2 2 2Fe oxit Fe

NO NO N O N SO

m m m.3 .2 n 3n 8n 10n 2n

56 16

= + + + + +

3 3

FeFe(NO )

mm = .242

56v

2 4 3

FeFe (SO )

mm = .400

2.56

Cch khc :Quy i hn hp gm Fe : x mol v O : y mol hh Fe Om m m 56 16 (1)x y= + = +

Qu trnh cho nhn e :3

33

o

Fe e Fe

x x

+

v

2

22

o

O e O

y y

+

Suy ra phng trnh sau :2 2 2 2NO NO N O N SO

3x 2y n 3n 8n 10n 2n (2)= + + + + + x, yNu c cho Cu th ta c phng trnh tng qut :

2 2 2 2Fe Cu O NO NO N O N SO3n 2n 2n n 3n 8n 10n 2n (2 ')+ = + + + + + x, y

Vn cn mt cch khc : mFe = 0,7. m (hh xit st) + 5,6 . ncho/ nhnncho/ nhn= mol kim loi .ha tr = gim s xi ha . s mol sp kh


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Cu 37: m gam bt st ngoi khng kh, sau mt thi gian bin thnh hn hp (A) c khi lng 12 ggm Fe , FeO , Fe3O4 , Fe2O3 . Cho (A) td hon ton vi dd HNO3 thy sinh ra 2,24 lkh 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 ) v5+

N ca HNO3 thu (5+

N thu 3e ) :

Qu trnh oxi ha :3

Fe Fe+

+ 3e

56

mmol 3

56

mmol

Qu trnh kh :0

O 2 + 4e 22

O ;5+

N + 3e 2+

N

32

12 m 4

32

12 mmol 0,3mol 0,1mol

Ta c: 356

m= 4

32

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

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

FeFe

56 16 120,18 n

0,18.56 10,08 gam2,243 2 3. 0,3 0,12

22,4

x yx

mx y y

+ == =

= = = = =

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 HNO3long d thu c 1,344 lt kh NO duy nht (ktc) v dung dch X. C cn dung dch X thu c m gam mukhan. Gi tr m l : A. 38,72 B. 35,5 C. 49,09 D,34,36

Gii: Cch 1 : Fe oxit Fe NOm m -m

.3= +3n56 16

vi

m 11,36

1,3440,06

22,4

hh Oxit

NO

m

n

= =

= =

suy ra Fe Fem 11,36-m

.3= +3.0,0656 8

Vy mFe = 8,96 suy ra 3 3Fe

Fe(NO )

m 8,96m = .242= .242=38,7256 56 , chn A (cch 2 hc sinh t gii )

Cu 39: Ha tan han ton 46,4 gam mt oxit kim loi bng dung dch H2SO4c nng (va ) thu c V lkh SO2 (ktc) v 120 gam mui. Xc nh cng thc oxit kim loi v VA. FeO; 1,12 B. Fe2O3;2,24 C.Fe3O4;1,12 D. Fe3O4; 2,24Gii: quy i xit thnh Fe (x mol) v O (y mol) suy ra m (xit) = 56x + 16y = 46,4 (1)

2 4 3( )

1200,3 0,3.2 0,6 ;(1) 0,8

400 Fe SO Fe On n x y n= = = = = = =

0,6 3(C,D)

0,8 4Fe

O

n

n= = ,

2

V3 .2 2 3.0,6 0,8.2 2. V 2, 24( )

22,4SO y n l = + = + = , chn D

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 bngHNO3 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 sdng (1) v (2)

FeFe

56 16 360,54 n

0,54.56 30,24 gam6,723 2 3. 0,9 0,36

22,4

x yx

mx y y

+ == =

= = = = =

, chn A

+ 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 gamCu 41 (H Khi B 2007): Nung m gam bt Fe trong oxi thu c 3 gam hn hp cht rn X. Ha tan ht 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



8/44


8


FeFe

56 16 30,045 n

0,045.56 2,52 gam0,563 2 3. 0,075 0,03

22,4

x yx

mx y y

+ == =

= = = = =

, chn A

Cu 42: Nung nng 16,8 gam bt st ngoi khng kh, sau mt thi gian thu c m gam hn hp X gm oxitv st d. Ha tan ht hn hp X bng H2SO4c nng thu c 5,6 lt SO2 (ktc). Ga tr ca m l?

A. 24g B. 26g C. 20g D. 22g


16,80,3

56Fen x= = = ; 2

2

SOSO

5,63.03 23 2.n 22,43x 2y 2n (2) 0,2

2 2

xy

= + = = =

hh Fe Om m m 56 16 56.0,3 0,2.16 20gx y= + = + = + = , chn C+ Cch khc : n nhn = 2.

2SOn =0,5 mol

mFe = 0,7.m xit + 5,6. n nhn suy ra Feoxitm 5,6.n 16,8 5,6.0,5

m 20 gam0,7 0,7

nhan = = =

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

Gii:3 4

FeFe O

o

n 0,06.313,92n 0,06 mol

n 0,06.4232

== =

=,

0,4480,02 mol

22,4n = =

Gi kl gim s xi ha ca kh+2

Fe O3n 2n . 3.0,18 2.0, 24 .0,02 3 N Ok n k k = + = + = , Chn ACu 44 (H Khi A 2009): Ha tan hon ton 20,88 gam mt xit st bng dung dch H2SO4c nng thuc dung dch X v 3,248 lt kh SO2 (sn phm kh duy nht). C cn dung dch X thu c m gam muisunfat khan. Gi tr m l :

A. 52,2 B. 54,0 C. 58,0 D. 48,4Gii: quy i xit thnh Fe (x mol) v O (y mol) v sdng (1) v (2)

2 4 3

Fe FeFe (SO )

56 16 20,880,29 n n

m .400 58 gam3,2483 2 2. 0,29 0,29 2

22,4

x yx

x y y

+ == =

= = = = =

, chn C

Cu 45 (HQGHN 2000): m (g) phoi bo Fe ngoi khng kh, sau mt thi gian c 12 g cht rn Xgm Fe, FeO, Fe3O4, Fe2O3. Ha tan ht X trong dung dch H2SO4, nng c 2,24 lt SO2 (ktc). Gi trca m l: A. 9,52 B. 9,62 C. 9,42 D. 9,72Gii: quy i xit thnh Fe (x mol) v O (y mol) v sdng (1) v (2)

Fe

Fe

56 16 120,17 n

m 0,17.56 9,52 gam2,243 2 2. 0,2 0,15522,4

x yx

x y y

+ == =

= = = = =

, chn A

Cu 46: Nung m gam bt Cu trong oxi thu c 37,6 gam hn hp rn X gm Cu, CuO v Cu2O. Ha tanhon ton X trong dung dch H2SO4c, 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 gamGii: quy i xit thnh Cu (x mol) v O (y mol) v sdng (1) v (2)

CuCu

64 16 37,60,5 n

m 0,5.64 32 gam3,362 2 2. 0,3 0,35

22,4

x yx

x y y

+ == =

= = = = =

, chn B

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 r



9/44


9

A gm Fe,FeO,Fe2O3 v Fe3O4. Ha tan hon ton A trong HNO3 d thu c dung dch B v 12,096 lt hnhp 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 khcGii: Gi a l s mol NO, b l s mol NO2

S mol hh kh l : a b 0,54 moln = + = ,30a 46b

M 10,167.4 30a 46b 21,96a b

+= = + =

+

Ta c : a 0,18 , b 0,36= = , n nhn =2NO NO

3.n 1.n+ =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 BCu 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. NO2Gii: Cn nhr gim s xi ha tng sn phm kh

FeO Fe On n n 0,12 mol= = = , gi X l gim s xi ha ca sn phm kh

Fe NOFe O 2

3n 2n 3.012 2.0,123n 2n X.n X 8 N O

0,336n22,4

spk

spk

= + = = = , Chn C

Cu 49: m gam bt st ngoi khng kh, sau mt thi gian s chuyn thnh hn hp B gm 4 cht rn c k

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 vkhi lng HNO3 phn ng ?A. 10,08 g v 34,02 g A. 10,8 g v 34,02 gC. 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 mol22,4 = mFe = 0,7.m xit + 5,6. n nhn = 0,7.12 + 5,6 . 0,3 = 10,08 gam

Fe Fe(NO3)30,18 0,18 mol ,

bo ton nguyn t N :3 3 3 3N/HNO N/Fe(NO ) N/NO HNO

n n n 3.0,18 0,1 0,64 mol m 0,64.63 40,32 g= + = + = = =

Ta chn DCu 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

Gii: t3 4

Fe

FeO CuO Fe O O

Cu

n 4

x n n n n 6

n

x

x

=

= = = = =

2Fe Cu O NO NO3n 2n 2n n 3n 3.4 2. 2.6 0,09 3.0,05 0,12 x x x x+ = + + + = + + = , chn A

Cu 51: Cho 22,72 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dung dch HNO3 long dthu 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.

Gii: mui chnh l Fe(NO3)3 :3)3Fe Fe(NO Fe77,44n n 0,32 mol m 0,32.56 17,92 gam242

= = = = =

O hh O Om m m 22,72 17,92 4,8 gam n 0,3 mol= = = = ,NO

Fe O NO

V3n 2n 3 V 2,668

22,4l= + = , chn B

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

Sn phm kh+4 o +2 +1 -3

2 22 4 3N O , N , NO, N O, N H NO i vi HNO3+4 -2 o

2 2S O , H S,S i vi H2SO4c



10/44


10

Mui (kim lai phi ha tr cao nht) v Al, Fe, Cr khng tc dng vi H2SO4 v HNO3c ngui.

.chon mol kimloai hoa tri= v .nhan san phamkhun do giam soOXH n= + i vi H2SO4c :

Sp kh+4

2S O -2

2H S o

S

gim s xi ha 6 4 = 2 6 (-2) = 8 6 0 = 6

22 4 4

( / )( ) ( )2

cho nhan

H SO S S S SO

nn n n n trong sp khu n trong sp khu= = + = +

2 24 4

( / )96. 96.2

cho nhan

muoi kimloai kimloai kimloaiSO SO

nm m m m n m = + = + = +

+ i vi HNO3 :

Sp kh+4

2N O o

2N +2

N O +1

2N O -3

4 3N H NO (mui) gim s

xi ha5 4 = 1 (5-0).2 = 10 5 2 = 3 (5 1).2 = 8 5 (-3) = 8

3 3( / )( ) ( ) HNO N N cho nhan S NOn n n n trong sp khu n n trong sp khu= = + = +

3 3

( / )62. 62.1

cho nhan

muoi kimloai kimloai kimloai NO NO

nm m m m n m = + = + = +

Ch : Nu sp khcNH4NO3th khi lng mui sau phn ng phi cng thm khi lng caNH4NO3Cu 52 (C 2011): Dy gm cc kim loi u tc dng c vi dung dch HCl nhng khng tc dng vdung dch HNO3c , ngui l:

A. Fe, Al, Cr B. Cu, Fe, Al C. Fe, Mg, Al D. Cu, Pb, AgGii: Chn A , HNO3 c ngui khng tc dng Al, Fe, CrCu 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 lA. 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

5+

Nnhn bng nhau .

2 H+

+ 2 e H25+

N + 3e 2+

N0,3 0,15 mol 3x x mol

3x = 0,3 x = 0,1 V = 2,24 lt . chn ACu 54 : Cho 1,35 gam hn hp Cu , Mg , Al tc dng vi HNO3 dc 896 ml (ktc) hn hp gm NO vNO2 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



11/44


11

a = 0,01 = nNO ; b = 0,03 = nNO22nhan NO NO

n 3.n 1.n 3.0,01 0,03.1 0,06 mol = + = + =

m hh mui = mhh kim loi +3NO

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

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 dungdch HCl thu c 1,064 lt kh H2 . Khi ho tan 1,805 gam hn hp trn bng dd HNO3 long d thu c0,896 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 An+ + ne Fe Fe2+ + 2e 2 H

+

+ 2 e H2x nx y 2y 0.095 0.0475

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

A An+ + ne Fe Fe3+ + 3e5+

N + 3e 2+

Nx nx y 3y 0,12 0.04

nx + 3y = 0,12 (3) . T (1) , (2) y = 0.025 .T (1) , (2) nx = 0,045 v Ax = 0,405 A = 9n

n 1 2 3 Chn A = 27 ( Al ) , chn CA 9 18 27Cu 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 N2Okhng 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. 12,04Gii: nnhn=3.0,01 + 8.0,01 =0,11 mmui = mkim loi + mgc axit = 3,6 +0,11.62= 10,42 gam, chn A

Cu 57: Ha tan hon ton hn hp Al, Fe, Mg vo 800ml dung dch HNO3(va ) thu c 0,08 mol NO; 0mol N2O v 0,01 mol N2. Vy nng mol ca dung dch HNO3 lA. 2M B. 1,5M C.1,3M D.1,8MGii: 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=0,22suy ra nHNO3 = nnhn + nN (trong sp kh) = 0,82 + 0,22 = 1,04 suy ra CM(HNO3)=

1,041,3

0,8= , chn C

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

Gii: Kh khng mu khng mi khng chy l N2 , 20,672

0,0322,4N

n = = , gim s OXH l 10

Bo ton e : mol kim loi . ha tr(tm t l n) = gim s OXH . s mol sp khn=1 M=9

2,7 .n=10.0,03 M=9.n, n=2 M=18M

n=3 M=27 (Al)

, chn A

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 H2Phn 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 = 2yGii:s mol Al hai phn bng nhau

Al + NaOH d :2Al H

n .3=2.n =2.x

v Al + dd HNO3 d :2Al N O

n .3 8.n 8y= =



12/44


12

Ta c c 2.x 8.y x 4y= = , chn CCu 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. N2Gii:gi X l gim s xi ha ca kh cn tm

Bo ton e :3,6 2,24

.2 X. X 3 NO24 22,4

= = . Chn A

Cu 61: Cho m gam Al tan hon ton trong dd HNO3, thu c 44,8 lt (ktc) hn hp kh NO, N2O, N2 theot l mol 1 : 2 : 2. Gi tr m lA. 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 0,4 molx =

2 2nhan NO N O Nn 3n 8n 10n 3. 8.2 10.2 39 39.0, 4 15,6 mol x x x x= + + = + + = = =

Bo ton e : .3 15,6 140,4 gam27

mm= = , Chn C

Cc em hy c gng thuc gim s xi ha mi kh lm bi tt hnCu 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 phnng.A. 8,074gam v 0,018mol B. D. 8,4gam v 0,8molC. 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 = 0,01t s molNO : 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 = 0,06125nHNO3 = 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 DCu 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 Bgm NO v mt kh X, vi t l th tch l 1:1. Xc nh kh X?

A. NO B. N2O C. NO2 D. N2Gii: 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

Bo ton electron : 211,2

.3 3.0,15 X.0,15 X 1 NO56

= + = , Chn C

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 gNO2 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 : 1xnhh kh = 4x = 8,96 /22,4 0,1 molx =

Bo ton electron :19,2

.n 3.0,1 1.0,3 M 32n M 64, n 2 (Cu)M

= + = = = , chn B

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 kgm (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 40Gii: 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 molTa c 13,5 gam hn hp gm Al : x mol v Ag : y mol 27x + 108y = 13,5 gamBo ton electron : 3x + 1y = 3.a + 1.b = 0,4 x = y = 0,1 mol

Nn Al Ag27.0,1

%m .100 20% %m 80%13,5

= = = , chn C

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



13/44


13

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

Gii: NO2 : a mol v SO2 : b mol

46 64 2,940,05

1,3440,06 0,01

22,4

a ba

a b b

+ ==

+ = = =

Cho 3 gam hn hp gm Cu : x mol , Ag : y mol 64x + 108y = 3 gamBo ton electron : 2x + 1y = 1.a + 2.b = 0,07 x = 0,03 v y = 0,01 mol

Nn Cu Ag64.0,03%m .100 64% %m 36%3= = = , chn D

Cu 67: Trn 60g bt Fe vi 30g lu hunh ri un nng (khng c kkh ) thu c cht rn A. Ho tan Abng dd axit HCl dc dd B v kh C. t chy C cn V lt O2 (ktc). Tnh V, bit cc phn ng xy rahon ton.

A. 32,928 lt B. 33 lt C. 34 lt D. 35 lt

Gii: nFe > nS =32

30. nn Fe d v S ht

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 .

Fe 2+

Fe + 2e S 4+

S + 4e O2 + 4e 22

O

56

60mol 2

56

60mol

32

30mol 4

32

30mol xmol 4x mol

Theo nh lut bo ton electron : 256

60+ 4

32

30= 4x x = 1,47

2O

V = 32,928 lt, chn A

Cu 68: Th tch dd FeSO4 0,5M cn thit phn ng va vi 100ml dd cha KMnO4 0,2M v K2Cr2O70,1M mi trng axit l :

A. 160 ml B. 320 ml C. 80 ml D. 640 ml

Gii : Ta c : 4KMnOn = 0,02 722 OCrKn = 0,01

+2

Fe Fe3+ + 1e7+

Mn + 5e 2+

Mn 26+

Cr + 6e 23+

Crx mol x mol 0,02 0,1 0,02 0,06

x = 0,1 + 0,06 = 0,16 4FeSO

V = 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 vN2O (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.

Gii: NO : a mol v N2O : b mol0,04 0,03

30 44 16,75.2.0,04 0,01

a b a

a b b

+ = =

+ = =

Bo ton e :3 3( )

17.3 3.0, 03 8.0, 01 1,53 gam n n27 300 Al Al NOm

m= + = = = mol

m mui = 17:300 . 213 = 12,07, chn ACu 70: Ha tan hon ton 14,8g hh (Fe, Cu) vo lng d dung dch hn hp HNO3 v H2SO4c, 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

Gii: Fe : x mol , Cu : y mol

56 64 14,80,15

10,08 2,243 2 1. 2 0,1

22,4 22,4

x yx

x y y

+ ==

+ = + =

56 8,4 gamFem x= = , chn B



14/44


14

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. 80mlGii: nFe = 0,1 mol

Fe Fe2+ + 2e Fe2+ Fe2+ + 1e7+

Mn + 5e 2+

Mn 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

Cu 72: Ho tan hon ton 1,2 gam kim loi M vo dung dch HNO3 d thu c 0,224 lt kh N2ktc (snphm 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

Bo ton elctron :1,2 0, 224

n 10. M 12n M 24, 2 MgM 22,4

n= = = = , ta chn D

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

Bo ton elctron :9,72 0,672

n 3. M 108n M 108, 1 AgM 22, 4

n= = = = , ta chn D

Cu 74: Ha tan hon ton 16,2g mt kim loiha tr III bng dung dch HNO3,thu c 5,6 lt(kc) hnhp X gm NO v N2. Bit t khi hi ca X so vi kh oxi bng 0,9. Xc nh tn kim loiem dng?

A. Al B. Fe C.Cu D.NaGii: NO : x mol , N2 : y mol

0,560,25 0,1

22,40,15

30 28 0,9.32.0,25

hh

hh

n x y x

ym x y

= + = = =

= = + =

Bo ton electron :16,2

3 3.0,1 10.0,15 M 27 AlM

= + = , chn A

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)thora. M l kim loi: A. Al B. Cu C. Fe D. MgGii: gi n l ha tr kim loi

Bo ton elctron :8,1 6,72

n 3. M 9n M 27, 3 AlM 22,4

n= = = = , ta chn A

Dng : To mui NH4NO3 (du hiu nhn bit : tao bo ton electron 2 v khng bng nhau nn phc 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 xyra hon tonthu c 0,896 lt kh NO (ktc) v dung dch X. Khi lng mui khan thu c khi lmbay hidung 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 molBo ton e : 0,09. 2 0,04.3 nn c to mui NH4NO3 : x molBo ton e khi c mui NH4NO3 : 0,09.2 = 0,04.3 + 8.x x = 0,0075 mol

Mg Mg(NO3)20,09 0,09 molm mui =

3 2 4 3( )0,09.148 0,0075.80 13,92 gamMg NO NH NOm m+ = + = , chn D

Cu 77 (H Khi A 2009): Ho tan hon ton 12,42 gam Al bng dung dch HNO3 long (d), thu cdung dch X v1,344 lt (ktc) hn hp kh Y gm hai kh l N2O v N2. T khi ca hn hp kh Y sovi kh H2l 18. C cn dung dch X, thu c m gam cht rn khan. Gi tr ca m l

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15

A. 38,34. B. 34,08. C. 106,38. D. 97,98.Gii: nAl = 0,46 mol, N2O : a mol , N2 : b mol

1,3440,06

22,4 0,03 mol

44 28 2.18.0,06

hh

hh

n a ba b

m a b

= + = =

= = = + =

Bo ton e : 0,46. 3 8.0,03 + 10.0,03 nn c to mui NH4NO3 : x molBo ton e khi c mui NH4NO3 : 0,46. 3 = 8.0,03 + 10.0,03 + 8.x x = 0,105 mol

Al Al(NO3)30,46 0,46 mol

m mui =3 3 4 3( )

0,46.213 0,105.80 106,38 gam Al NO NH NOm m+ = + = , chn C

Cu 78: Cho 1,35 g hn hp gm Cu , Mg , Al tc dng ht vi dung dch HNO3 thu c hn hp kh gm0,01 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 :5+

N + 3e 2+

N5+

N + 1e 4+

N0,03 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 C

Cu 79 : Cho hn hp gm 0,15 mol CuFeS2 v 0,09 mol Cu2FeS2 tc dng vi dung dch HNO3 d thu cdung 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 ithu c a gam cht rn. Gi tr ca m v a l:

A. 111,84g v 157,44g B. 111,84g v 167,44g

C. 112,84g v 157,44g D. 112,84g v 167,44g

Gii: Ta c bn phn ng:

CuFeS2 Cu2+ + Fe3+ + 2SO4

2

0,15 0,15 0,15 0,3

Cu2FeS2 2Cu2+ + Fe3+ + 2SO4

2

0,09 0,18 0,09 0,18

24SO

n 0,48 = mol; Ba2+ + SO4

2 BaSO4

0,48 0,48 m = 0,48 233 = 111,84 gam.

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

Cu CuO 2Fe Fe2O30,33 0,33 0,24 0,12

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

xit kim loi +

2

Al

CO

C

H

ot cao kim loi +

2 3

2

2

2

Al O

CO

COhayCO

H O

, iu kin : kim loi Al>



16/44


16

Quy i xit = kim loi + O Cng thc2 2 2 2O CO CO H O H (H ,CO

n (trong Oxit) n n n n n= = = = =

Cu 80 (H Khi A 2009): Cho lung kh CO (d) i qua 9,1 gam hn hp gm CuO v Al2O3 nung nngn 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 DCu 81 : Dy cc xit b CO khnhit cao l :A. CuO, FeO, ZnO, MgO B. CuO, Fe3O4, Fe2O3, Al2O3C. 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, MgChn DCu 82 : Kh hon ton 6,64 gam hn hp gm Fe, FeO, Fe2O3, Fe3O4 cn dng va 2,24 lt CO (ktc). Khlng 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 , O CO O2,24

n (trong oxit) n 0,1 m 0,1.16 1,6(gam)22,4

= = = = =

Suy ra mFe = mxit - mO = 6,64 1,6 = 5,04 gam. Ta chn BCu 83 : Kh hon ton 6,4 gam hn hp gm CuO v Fe2O3 bng kh H2 thy to ra 1,8 gam H2O. Khi lnhn 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

2O O

1,8n =n 0,1 m =0,1.16=1,6(gam) m 6,4 1,6 4,8g

18 H O kim loai= = = = . Ta chn B

Cu 84 : Cho V lt hn hp kh (ktc) gm CO v H2 phn ng vi mt lng d hn hp rn gm CuO vFe3O4 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

2 2O (H ,CO) (H ,CO)

0,32n = =0,02 n V =V=0,02.22, 4 0, 448

16= = . Ta chn A

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

Gii:2

FeFe O CO

O

n14,56 8,736 0, 26 2n = =0,26, n (oxit)=n = =0,39,

56 22,4 n 0,39 3= = . Ta chn C

Cu 86 (H Khi A 2010) : Cho m gam hn hp bt X gm ba kim loi Zn, Cr, Sn c s mol bng nhautc 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 3oxit th th tch kh O2 (ktc) phn ng lA. 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 O+2, Cr to +3, Sn to +4.

- Gi s mol mi kim loi l a (mol) th: 02,098,8)713

1195265.(3

2==>=+

++= aam

ClM.

- Bo ton (e) cho qu trnh tc dng O2:2

2.0,02 3.0,02 4.0,020,045( ) 1,008( )

4On mol lit

+ += =

Chn DCu 87 (C 2009) : Kh hon ton mt xit st X nhit cao cn va V lt CO (ktc), sau phn



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17

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 0,448C. Fe3O4 v 0,448 D. Fe3O4 v 0,224

Gii: nO (trong xit) = nCO = nCO2 = 0,02 mol ; nFe = 0,015 mol

Fe3 4

O

n 0,015 3Fe O

n 0,02 4= = , V = 0,02.22,4 = 0,448 lt, chn C

Cu 88 (C 2011): t chy hon ton 17,4 gam hn hp Mg v Al trong kh oxi (d) thu c 30,2 gamhn 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

Gii:2 2O O O

1 30,2 17,4n n 0,8 mol V 0,8.22,4 17,92 lit

2 16 16oxit kl m m = = = = = = , chn A

Cu 89 : Thi lng kh CO di qua ng ng hn hp hai xit Fe3O4 v CuO nung nng n khi phn ngxy ra hon ton thu c 2,32 gam hn h p kim loi. Kh thot ra c a vo bnh ng dung dchCa(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 gamGii: Do Ca(OH)2 d nn ta lun c :

2 30,05 mol

CO CaCO O COn n n n= = = =

m xit = m kim loi + mO = 2,32 + 16.0,05 = 3,12 gam , chn A

Cu 90 (sdng 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 tchdung dch HCl 2M va phn ng ht vi Y l

A. 50 ml. B. 57 ml. C. 75 ml. D. 90 ml.

Gii: oxit kim loaiO3,33 2,13

n 0,8 mol16 16

m m = = = , + HCl OH 2.n 0,15 moln n= = =

HCl

0,15V 0,075 lit 75 ml

2 = = = , chn C

Cu 91 (sdng quy i xit = kim loi + O):t chy hon ton 26,8 g hn hp 3 kim loi Fe, Al, Cu thuc 41,4 g hn hp 3 oxit. Th tch dung dch H2SO4 1M cn dng ha tan va hn hp oxit trn lA. 1,8250 lt. B. 0,9125 lt. C. 3,6500 lt. D. 2,7375 lt

Gii:2 4 2 4

oxit kim loaiO H SO H SO

41,4 26,8 0,9125n 0,9125 mol n V 0,912516 16 1

m m = = = = = = lt

Ta chn BCu 92 : Cho 31,9 gam hn hp Al2O3, ZnO, FeO, CaO tc dng ht vi CO d nung nng thu c 28,7gam 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 ltGii: Khi lng nguyn t xi = gim khi lng cht rn

mO = 31,9 28,7 = 3,2 gam2O H CO

n n n 0,2 mol V 0,2.22,4 4,48 lit= = = = = , chn C

Cu 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 FeOun nng th lng Fe thu c l :

A. 24 gam B. 26 gam C. 28 gam D. 30 gamGii:

2H Cu Fe Fe

32n n n 0,5 mol 56.0,5 28 g

64m am= = = = = = , chn C

Cu 94 : Cho 34,8 gam hn hp gm xit v mui cacbonat ca kim loi km R. Ha tan ht hn hp trnbng mt lng va dung dch cha 0,6 mol HCl. Tn kim loi R l :

A. Na B. K C. Li D. CsGii: 2 2 2 3 2 2R O 2HCl 2RCl H O ; R CO 2HCl 2RCl H O CO+ + + + +

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

mol l 0,334,8

M 116 2 16 M 2 60 28 500,3

R R R = = + < < + < < l K (39), chn B



18/44


18

Dng 8: Ha tan hon ton (K, Na, Ca, Ba) + H2O dd kim ( cha ion-OH ) + 2H

Ta c pt ion sau :2e

2 2

2.1

2OH H

ran kimloai OH

n n H O OH H

m m m

+ =

+ = +

Dung dch sau phn ng trung ha bi axit th + -H OHn =n

Cu 95(H khi B 2007): Cho mt mu hp kim Na-Ba tc dng vi nc (d) thu c dung dch X v3,36 lt H2ktc. Th tch dung dch H2SO4 2M cn dung trung ha dd X l

A. 60ml B. 30ml C. 75ml D. 150ml

Gii:2

2 2

1 3,362. 2. 0,3

2 22, 4e

OH H H O OH H n n

+ + = = =

Gi th tch dung dch H2SO4 2M l V +2 4H SOH

n =2.n 2.2.V=4V =

Dung dch sau phn ng trung ha bi axit th

+ -H OHn n 4V 0,3 V 0,075 lit 75ml= = = = , chn C

Cu 96 : Cho a (g) hh Na, K, Ca tc dng vi nc (d) thu c dung dch X v 0,224 lt H2ktc. Th tch

dung dch H2SO4 0,1M cn dung trung ha dd X lA. 0,15 lt B. 0,1 lt C. 0,12 lt D. 0,20 ltGii: gi V l th tch cn tm

+2HH OH

3,36n n 2.n V.2.2 2. V 0,1

22,4= = = = lt, chn B

Cu 97 : Ha tan ht mu hp kim K-Ba vo nc thu c dung dch X v 0,224 lt H2ktc. trung hahon 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

Gii: gi V l th tch cn tm2

2HOH

0,224n 2.n 2. 0,02 mol, [ ] 10 M

22,4H

+ = = = =

1/10 dung dch X =2

OH H OHn 0,002 mol n n 10 .V 0,002 0,2V+

= = = = lt, chn ACu 98 : Cho hh Na, K, Ca vo nc thu c dung dch A v V (lt) kh H2ktc. Trung ha 1/3 dung dch 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

Gii: dung dch X :2

2 222, 4 11, 2HOHV V

n n = = =

1 / 3 dung dch X : 0,1.1.0, 2 0,5.0, 2.2 7,39233,6OH HV

n n V += = = + = , chn B

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 ltGii:

2H

1.1,792n 0,3 mol

0,082.(273 27,3)

PV

RT= = =

+,

+2HCl H HClH OH

n n n 2.n 0,6 mol V 0,6 : 0,2 3= = = = = = lt Ta 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 lH2 (ktc). Tnh pH ca dung dch A.A. 12 B. 11,2 C. 13,1 D. 13,7

Gii: dung dch A :2

14140,1 102 0,1 mol; [OH ] 0,5 [ ] 2.10

0,2 0,5HOHn n M H

+ = = = = = =

Nn pH log[ ] 13,7H+= = , chn D



19/44


19

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 lA. 0,15 lt B. 0,3 lt C. 0,075 lt D. 0,1 lt

Gii:2HOH

3,36n 2.n 2. 0,3 mol

22,4 = = =

dung dch YOH

0,3n =0,15 mol

2 = +

3HNOH OHn n n .1.2 0,15 0,075 75 mlV V l= = = = = ,

chn CDng 9: Cho Cu vo dung dch hn hp cha cc ion 3( , ) H NO

+ NO

hoc v d nh phn ng ca Cu kim loi vi hn hp dung dch NaNO3 v dung dch H2SO4 l

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

Dung dch Cu phn ng c th cha nhiu axit nhng nu c ion3+Fe th cn ch ti phn ng gia Cu v

3+Fe Cn tnh

3

Cu

H

NO

n

n

n

+

sau lp t s3, ,

3 8 2NOCu H

nnn +v sau ta s mol ca cht hay ion c t s nh

nht 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 Na2SO4C. 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 caNaNO3 trong phn ng l?

A. Cht xc tc B. Cht oxi ha C. Mi trng D. Cht khGii: Vai tr ca NaNO3 trong phn ng l cht xi ha , chn BCu 104 : Xem phn ng: a Cu + b NO3

- + c H+ d Cu2+ + e NO + f H2OTng 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. 24Gii: 3Cu + 8H+ + 2NO3

3Cu2+ + 2NO + 4H2O, tng h s = 22 chn CCu 105 (H Khi A 2008): Cho 3,2 gam bt ng tc dng vi 100ml dung dch hn hp gm HNO3 0,8Mv H2SO4 0,2 M. Sau khi phn ng xy ra hon ton thu c V lt NO duy nht (ktc). Gi tr V lA. 0,448ml B. 1,792 C. 0,672 D. 0,746

Gii:3 2 4

3

0,123,20,05; 0,8.0,1 0,08; 0,2.0,1 0,02

0,0864

H

Cu HNO H SO

NO

nn n n

n

+

== = = = = =

=

+ - 2+3 2 23C u + 8H + 2N O 3Cu + 2N O + 4H O

0,12 0, 03

0, 05 0,12 0, 08

3 8 2tinh theo H+

Vy V(NO) = V = 0,03.22,4 = 0,672 . Chn C

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 nhtTh nghim 2 : Cho 3,84 gam Cu phn ng vi 80ml dung dch cha HNO3 1M v H2SO4 0,5M thy thot raV2 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



20/44


20

Gii: TN1:

3

Cu

HNO

3,84n 0,06 mol

64n 0,08 mol

= =

=

3

H

NO

n 0,08 mol

n 0,08 mol

+

= =

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

Ban u :0,06 0,08 0,08 mol H+ phn ng ht

Phn ng: 0,03 0,08 0,02 0,02 mol

V1 tng ng vi 0,02 mol NO.

TN2: nCu = 0,06 mol ;3HNO

n = 0,08 mol ;2 4H SO

n = 0,04 mol. Tng:H

n + = 0,16 mol ;3NO

n = 0,08 mol.

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

Ban u: 0,06 0,16 0,08 mol Cu v H+ phn ng ht

Phn ng: 0,06 0,16 0,04 0,04 mol

V2 tng ng vi 0,04 mol NO. Nh vy V2 = 2V1. Chn BCu 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). Gitr ca V lA. 6,72 B. 8,96 C. 4,48 D. 10,08

+ - 23 2

2 + - 3+3 2

3Cu 8H 2NO 3Cu 2NO 4H O

0,3 0,8 0,2 0,2 mol

3Fe 4H NO 3Fe NO 2H O

0,6 0,8 0,2 0,2 mol

+

+

+ + + +

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

Cu 108 : Ha tan ht 3,6 gam FeO bng HNO3 long, va . Thm H2SO4 long d vo dung dch sau phnng 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 lnlt l :

A. 16 gam v 3,36 lt B. 14,4 gam v 3,36 ltC. 1,6 gam v 3,36 lt D. 16 gam v 4,48 ltGii : FeO 3 3 3Fe(NO )

HNO

0,05 0,05 molThm H2SO4 long d. Dung dch cha cc tc nhn ca phnng Cu l

3+ +3Fe , H , NO

3+ 2+ 22Fe Cu 2Fe Cu

0,05 0,025 mol

++ +

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

0,225 0,15 0,015 molmCu = 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. Saukhi phn ng xong c kh NO l sn phm kh duy nht ca N+5 v 0,92a gam hn hp kim loi. Gi tr al :

A. 11,0 B. 11,2 C. 8,4 D. 5,6

Gii :

2+

+

-3

Cu

H

NO

n 0,1 mol

n 0,08 mol

n 0,28 mol

=

=

=

Kim loi cn d nn Fe+ b chuyn v Fe2+ :3Fe + 8H+ + 2NO3

3Fe2+ + 2NO + 4H2O



21/44


21

0,03 0,08 0,02Fe + Cu2+ Fe2+ + Cu0,1 0,1 0,1

Nn : a 56.(0,03 + 0,1) + 64.0,1 = 0,92a a = 11, chn ACu 110 : Ha tan 1,28 gam Cu vo 50ml dung dch hn hp H2SO4 0,1M v NaNO3 0,5M thu c a mol khNO dung nht. Tnh a ? A. 0,0025 B.0,0133 C. 0,025 D. 0,032

Gii : +

-3

Cu

H

NO

n 0,02 mol

n 0,01 moln 0,025 mol

=

=

=

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

Ban u : 0,02 0,01 0,025 molPhn ng 0,01 0,0025 mol , chn ACu 111 : Cho 0,09 mol Cu vo 400ml dung dch cha HNO3 0,3M v H2SO4 0,1M. n khi phn ng ktthc, 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

Gii : +

-3

Cu

H

NO

n 0,09 mol

n 0,2 mol

n 0,12 mol

= =

=

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

Ban u : 0,09 0,2 0,12 molPhn ng 0,2 0,05 mol

V = 0,05.22,4 = 1,12 lt , chn D

Cu 112 : Ha tan 12,8 gam Cu vo 200ml dung dch hn hp H2SO4 1M v KNO3 0,5M. Th tch kh NO dunht ktc l : A. 2,24 lt B. 2,99 lt C. 4,48 lt D.11,2 lt

Gii : +

-3

Cu

H

NO

n 0,2 moln 0, 4 mol

n 0,1 mol

==

=

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


V = 0,1.22,4 = 2,24 lt , chn ACu 113 : Cho 1,92 gam Cu vo 100 ml dung dch cha ng thi KNO3 0,16M v H2SO4 0,4M thy sinh ramt cht kh c t khi hi so vi H2 l 15. Th tch kh ktc l :

A. 0,672 lt B. 1,446 lt C. 0,3584 lt D. 0,4568 lt

Gii : +

-3

Cu

H

NO

n 0,03 mol

n 0,08 mol

n 0,016 mol

=

=

=

v Kh c M = 15.2 = 30 l NO

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


V = 0,016.22,4 = 0,3584 lt , chn CCu 114 : Cho 1,12 gam Cu vo 50 ml dung dch cha ng thi KNO

30,16M v H

2SO

40,1M thy c kh


22/44


22

NO ( sn phm kh duy nht ca s kh N+5 ) bay ra. kt ta ton b Cu2+ trong dung dch sau phn ngcn 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

Gii : +

-3

Cu

H

NO

n 0,0175 mol

n 0,01 mol

n 0,008 mol

=

=

=

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O0,00375 0,01 0,0025 mol 0,00375

2+OH Cun 2 n 0,0075 mol = = nn V = 0,0075 : 0,5 = 0,015 lt , Chn B

Dng 10: Cho dd chc ion H+ vo dung dch cha 23 3,CO HCO

Yu cu : tnh s mol ca cc ion sau 23 3, , H CO HCO+ ( nu c thm HCO3

-t gi thuyt )

+ u tin s c phn ng + 2- -3 3H + CO HCO

+ Sau nu H+ cn d mi xy ra tip phn ng sau : + -3 2 2H + HCO CO + H O

- Vy : nu c khthot ra th 2-3CO ht v nu khng c khthot ra th +H ht- Nu bi ton hi ngc li : cho 2-3CO vo dung dch cha +H th ch c 1 phn ng duy nht

+ 2-3 2 22H CO CO H O+ +

Cu 115 (H Khi A 2009):Dung dch X cha hn hp gm Na2CO3 1,5M v KHCO3 1M. Nh t t tnggit 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.Gii: n (Na2CO3) = 0,1.1,5 = 0,15 mol = n (CO3

2-)n (KHCO3) = 0,1.1 = 0,1 mol = n ( HCO3

-)n (HCl) = 0,2.1 = 0,2 mol = n (H+)

+ u tin s c phn ng

+ 2- -3 3H + CO HCO

0,15 0,15 0,15

vy tng s mol -3HCO 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 :

+ -3 2 2H + HCO CO + H O

0,05 0,05 0,05

du

vy

2COV =0,05.22,4=1,12 l chn B

Cu 116 (H Khi A 2007):Cho t t dung dch chc a mol HCl vo dung dch cha b mol Na2CO3ngthi khuy u, thu c V lt kh (ktc) v dung dch X. Khi cho d nc vi trong vo dung dch X thy cxut 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)Gii: Khi cho d nc vi trong vo dung dch X thy c xut hin kt ta chng t c mui NaHCO3 :

2 3 2 3 2Ca(OH) 2NaHCO CaCO3 +Na CO 2H O+ +

Vy bi ton c ngha l cho a mol +H vo b mol2-3CO cho kh CO2 v mui NaHCO3

+ 2- -3 3H + CO HCO

b b

tnh theo s mol2-3CO do

+H phi cn dmi to phn ng th hai c+ -

3 2 2 2H du + HCO CO + H O V(CO )=(a-b).22,4

( ) ( ) ( )a b a b a b

, ta chn D


23/44


23

Cu 117 (H Khi A 2010) : Nh t t tng git n ht 30 ml dung dch HCl 1M vo 100ml dung dchcha 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. 0,030Gii: 2

3 3CO HCO H

n 0,02 mol, n 0,02 mol, n 0,03 mol += = =

+ u tin s c phn ng+ 2- -

3 3H + CO HCO

0,02 0,02 0,02 mol

vy tng s mol 3HCO

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 :+

3 2 2H + HCO CO + H O

0,01 0,01 0,01

du

Vy ta chn ACu 118 : Cho t t dung dch HCl vo dung dch Na2CO3n 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 BCu 119 : Cho t t 100ml dung dch HCl 2M vo 200ml dung dch Y cha KHCO3 1M v K2CO3 1M. Thtch kh CO2 thot ra ktc l :

A. 0,0 lt B. 4,48 lt C. 2,24 lt D. 3,36 ltGii: 2

3 3CO HCO H

n 0, 2 mol, n 0, 2 mol, n 0,2 mol += = =

u tin s c phn ng+ 2

3 3H + CO HCO

0,2 0,2 0,2 mol

Do H+ ht phn ng trn nn s khng c kh CO2 thot ra, chn ACu 120 : Nh rt t tn ht 500ml dung dch HCl nng 1,2M vo 480ml dung dch Na2CO3 1M angc 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. 6,288Gii: 2

3CO H

n 0, 48 mol, n 0,6 mol += =

+ u tin s c phn ng+ 2

3 3H + CO HCO

0,48 0,48 0,2 mol

s mol 3HCO l 0,2 mol

+ Sau do H+ cn d ( 0,6 0,48 = 0,12 mol ) nn xy ra tip phn ng sau :+

3 2 2H HCO CO + H O

0,12 0,12 0,12 mol

du +

V (CO2) = 0,12.22,4 = 2,688 mol , chn A

Dng 11 : PHNG PHP QUY I TRONG HA HCPhm vi p dng : nhng bi ton hn hp cho qua Axit c tnh xi ha mnh ra sn phm khri e

dung dch cho vo kim thu kt ta, sau nung kt taHn hp thng gp :

1/

g

gS

S

quy i v

,

,

,

Mg S

gS S

gS Mg


24/44


24

2/

2

Cu

CuS

S

Cu S

quy i v

2

,

,

,

,

..............

Cu S

CuS S

CuS Cu

CuS Cu S

3/

2

Fe

FeS

S

FeS

quy i v 2

2

,,

,

,

..............

Fe S

FeS S

FeS Fe

FeS FeS

Ch : cha a v 2 cht trong nhiu cht sau ta thng dng bo tan e- v pt hn h gii

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 duynht l 53,76 lt kh NO2 (ktc) v dung dch A. Cho NaOH d vo dung dch A thu kt ta, nung kt ta trongkhng 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 gamGii: NO2

: 2,4 molTa 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 )0

33

3

Fe e Fe

x x x

+

03 6

2 15 2

15

FeS e Fe S

y y y

+ + +

Vy tng s mol e- nhng l ncho = 3x + 15yv nnhn =( gim s oxh) . (s mol kh ) = 1.2,4

3.x + 15.y =2,4 (2) =>2

Fe

FeS

0,05 n

0,15 n

x

y

= =

= =ddA

3

24

: 0,2

2.0,15 0,3S

Fe x y mol

SO n

+

+ =

= = =

2,33 2 3( ) : ( )

nung OOH Fe Fe OH Fe O a mol +

32 3 2 3( )

0,2 2 0,1 0,1 (56 2 16 3) 16 Fe Fe Fe O Fe OFen n n a a m g += = = = = + = , chn A

Cu 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 (spkh duy nht ktc) v dung dch Y. Thm Ba(OH)2 d vo dung dch Y thu c m gam kt ta. Tm gi trm. A. 81,55 B. 104,20 C. 110,95 D. 115,85

Gii: NO

: 0,9 molTa quy i hn hp v Cu (x mol) v S(y mol) mhh = 64x + 32y = 30,4 gam (1)0 0 6

+2Cu 2 Cu S 6 S

2 y 6y

e e

x x x

+

Vy tng s mol e- nhng l ncho = 2.x + 6.yv nnhn = ( gim s oxh) . (s mol kh ) = 3.0,9 = 2,7 mol

Ta c h :64 32 30, 4

0,3; y 0, 352 6 2,7

x yx

x y

+ = = =

+ =


25/44


25

2OH2+2Cu Cu(OH)

0,3 0,3 mol

Nn

2Cu(OH)m 0,3.98 29,4 gam= =

2+ 24 4Ba SO BaSO

0,35 0,35 mol

+

Nn

4BaSOm 0,35.233 81,55 gam= = ,

Khi lng kt ta = 29,4 + 81,55 = 110,95 gam, chn C

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 khN2 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. 12,0Gii: N2

: 0,13 mol , Ta quy i hn hp v Mg (x mol) v S(y mol) mhh = 24x + 32y0 0 6

+2Mg 2 Mg S 6 S

2 y 6y

e e

x x x

+

Bo ton electron : 2x + 6y = 10.0,13 (1)2+ 2+ 2

2 4 4Mg 2OH Mg(OH) Ba SO BaSO

x x y y

+ +

58 233 46,55 gamm x y= + = (2)

Suy ra x = 0,2 mol , y = 0,15 mol , mhh = 24x + 32y = 9,6 gam , chn C

Cu 124 : Ha tan hon ton 25,6 g hn hp X gm Fe, FeS, FeS2 v S vo dung dch HNOd thu c V lt kNO 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. 27,58Gii: Ta quy i hn hp v Fe (x mol) v S(y mol) mhh = 56x + 32y = 25,6 gam (1)

3 4Fe(OH) BaSOm m 107 233 126,25m x y= + = + = (2)

(1) v (2) ta c : x = 0,2 mol ; y = 0,45 mol

Bo ton electron : 3x + 6y = 3.V

22,4 V = 24,64 lt, chn C

Cu 125 : Ha tan hon ton 3,76 gam hn hp: S, FeS, FeS2 trong HNO3c dc 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 nkhi 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 gamGii: NO2 : 0, 48 molTa 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

4BaSOm 233 15,145 gamy= =

3 2 32Fe(OH) Fe O0,03 0,015 mol

2 3Fe O0,015.160 2,4 gamm = =

Khi lng cht rn = 15,145 + 2,4 = 17,545 gam , chn ADng 12 :

3HNO 2- 3+ 2+2 2 4FeS, FeS , CuS,Cu S SO ,Fe ,Cu 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 ?


26/44


26

A. 0,04 B. 0,075 C. 0,12 D.0,06Gii: Dung dch thu c ch cha mui sunfat duy nht nn ch cha cc ion : Fe3+ ; Cu2+ ; SO4

2-

3

2+

2-4


Cu Cu2

S SO

n n 0,12.1 0,120,12 mol FeS

n n 0, 2.a 0,2aa mol Cu S

n n 0,12.2 a.1 0,24 a

+ = = =

= = =

= = + = +

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

Cu 127 : Ha tan hon ton hh gm a mol FeS2 ; 0,5a mol FeS v 0,06 mol Cu2S vo axit HNO3 va thuc dung dch X ch cha 2 mui sunfatv kh NO duy nht . Gi tr ca a l :

A.0,24 B.0,2 C.0,12 D.0,06Gii: Dung dch thu c ch cha mui sunfat duy nht nn ch cha cc ion : Fe3+ ; Cu2+ ; SO4

2-

3

2+

2-4

Fe Fe2

bao toan nguyen toCu Cu

2 S SO

n n 0,5 1,5a mol FeS

0,5a mol FeS n n 0,06.2 0,12

0,06 mol Cu S n n 2,5 0,06

a a a

a

+ = = + =

= = =

= = +

Bo ton in tch : 1,5a.3 + 0,12.2 = (0,06 + 2,5.a ).2 a = 0,24 , chn ACu 128: Khi cho hn hp FeS v Cu2S phn ng vi dung dch HNO3ds thu c dung dch cha cc ionA. Cu2+, S2-, Fe2+ , H+, NO3 -. B. Cu2+ , Fe3+ , H+ , NO3- .C. Cu2+, SO4

2 - , Fe3+, H+ , NO3 -. D. Cu2+ , SO42- , Fe2+ , H+ , NO3- .

Gii: Chn C v cn c HNO3 d

Cu 129 : Ha tan hon ton hn hp gm x mol FeS2 v 0,05 mol Cu2S vo acid HNO3 (va ), thu cdung 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

Gii:

3

2+

2-4


Cu Cu2

S SO

n nx mol FeS

n n 0,05.2 0,10,05 mol Cu S n n 2 0,05

x

x

+ = =

= = = = = +

Bo ton in tch : x.3 + 0,1.2 = (0,05 + 2.x).2 x = 0,1 , chn CCu 130 : Cho hn hp A gm x mol FeS2 v y mol Cu2S tc dng vi HNO3 long, dun nng ch thu cmui 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

Gii:

3

2+

2-

4


Cu Cu2

S SO

n nx mol FeS

n n 2y mol Cu S

n n 2

x

y

y

+ = =

= =

= = +

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 dungdch 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. 25Gii: t x v y ln lt l s mol FeS2 v Cu2S


27/44


27

3

2+

2-4


Cu Cu2

S SO

n nx mol FeS

n n 2y mol Cu S

n n 2

x

y

x y

+ = =

= =

= = +

Bo ton in tch : x.3 + 2y.2 = (y + 2.x).2 x : y = 2 hay x = 2y5 2

N 3e N

2,4 0,8 mol

+ +

+

0 +3 6

2

0 +2 6

2

FeS Fe 2 S 15e

x 15x

Cu S 2Cu 1S 10e

y 10y

+

+

+ +

+ +

Bo ton electron : 15x + 10y = 2,4 suy ra x = 0,06 mol , y = 0,12 molVy 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)2Cho 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 boton 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. NaHCO3C. Na2CO3 v NaOH d D. NaHCO3 v Na2CO3

Gii: Nhn xt : ch c th c 2 mui lNa2CO3 ( Na : C = 2 : 1= 2 ) hoc NaHCO3 ( Na : C = 1 : 1 = 1)

nC = nCO2 = 0,1 mol , nNa = n NaOH = 0,2 mollp t l Na

C

n 0,22

n 0,1= = to mui Na2CO3 ( Na : C = 2 : 1 ), chn A

Cu 133 : 0,1 mol CO2 hp th vo dung dch cha 0,18 mol KOH. Tm cc sn phmA. K2CO3 B. KHCO3C. K2CO3 v KOH d D. KHCO3 v K2CO3Gii: Nhn xt : ch c th c 2 mui lK2CO3 ( K : C = 2 : 1 =2 ) hoc KHCO3 ( K : C = 1 : 1 = 1)

ng dng nC = nCO2 = 0,1 , nK= n KOH = 0,18

lp t l 0,181 1,8 20,1

K

C

n

n

< = = < c c 2 mui 2 3

3

K CO

KHCO

Ta chn DCu 134 : 0,5 mol P2O5 cho vo dd cha 0,8 mol Ca(OH)2. Tm cc sn phm ?

A. Ca3(PO4)2 B. Ca(H2PO4)2 v CaHPO4B. Ca(H2PO4)2 v Ca3(PO4)2 D. Ca(H2PO4)2

Gii: Nhn xt : ch c th c 3 mui l Ca3(PO4)2 ( Ca : P = 3 : 2 = 1,5 )Ca(H2PO4)2 ( Ca : P = 1 : 2 = 0,5 )

v CaHPO4 ( Ca : P = 1 : 1 = 1 )n P = 2.0,5 = 1 mol , n Ca = 0,8.1 = 0,8 mol




28/44


28

lp t s( )2 4 2

4

Ca H PO0,80,5 0,8 1

1 CaHPOCa

P

n

n

< = = <

, Chn B

Cu 135 : Trn 450 ml dung dch NaOH 1 vi 150 ml dung dch H3PO4 2M. Dung dch thu c sau phnng cha nhng mui no ?

A. NaH2PO4 v Na2HPO4 B. Na3PO4 v Na2HPO4C. NaH2PO4 v NaOH d D. Na2HPO4

Gii: Nhn xt : : ch c th c 3 mui l Na3PO4 ( Na : P = 3 : 1 = 3 )

NaH2PO4 ( Na : P = 1 : 1 = 1 )v Na2HPO4 ( Na : P = 2 : 1 = 2 )

n(NaOH) = 0,45.1 = 0,45 mol = n Na n(H3PO4) = 0,15.2 =0,3 mol = n P

lp t l0,45

1 1,5 20,3

Na

P

n

n< = = < NaH2PO4 v Na2HPO4 , chn A

Cu 136 : Trn dung dch cha a mol Ca(OH)2 vi dung dch cha b mol H3PO4. Chn t l f = a/b nh thno sau phn ng thu c 2 mui gm Ca(H2PO4)2 v CaHPO4.

A. 1 < f < 2 B. 1 < f < 1,5C. 0,5 < f < 1 D. 0,5 < f < 1,5

Gii: Nu bi ny chng ta ghi phng trnh th bi ton s rt phc tp, khng ph hp cho k thi quc gia

theo phng php trc nghim. Tc gi xin trnh bi cch gii n gin sau :Nhn xt : c 2 mui l Ca(H2PO4)2 ( Ca : P = 1 : 2 = 0,5 )v CaHPO4 ( Ca : P = 1 : 1 = 1 ) ( cho 2 mui ny )nCa = a v n P = b

m Ca

C

naf

b n= = nn ta thy 0,5 < f < 1, chn C

Cu 137 : Cho 2,24 lt H2S vo 100ml dung dch NaOH 1,6M thu c dung dch X. tnh s gam mui thuc khi c cn dung dch X ?

A. 6,92 B. 2,96 C. 9,26 D. 92,6Gii: Nhn xt : c 2 mui l NaHS ( Na : S = 1 : 1 = 1 ) v Na2S ( Na : S = 2 : 1 = 2 )

n(H2S) = 0,1 mol = nS v n (NaOH) = 0,1.1,6 = 0,16 mol = n Na

lp t l2

2 0,16 0,040,161 1,6 2

0,1 0,060,1Na

S

NaHS x mol nNa x y xn

Na S y mol nS x y yn

= + = = < = = <

= + = =

m(mui) = m(NaHS) + m(Na2S) = 0,0,4.56 + 0,06.78 = 6,92 gam, chn ACu 138 (H Khi B 2008) Cho 0,1 mol P2O5 vo 0,35 mol KOH. Dung dch thu c gm nhng cht :A. K3PO4, K2HPO4 B. K2HPO4, KH2PO4C. K3PO4 v KOH D. H3PO4 v KH2PO4Gii: n P = 0,1.2 = 0,2 v n K= 0,35.1 = 0,35

0,351 1,75 2

0,2K

P

n

n< = = < nn ta chn p n B

Cu 139 (TN THPT 2007) Hp th hon ton 4,48 lt SO2 (ktc) vo dung dch cha 16 gam NaOH thuc dung dch X. Khi lng mui tan c trong dung dch X l :A. 20,8 gam B. 18,9 gam C. 23,0 gam D. 25,2 gam

Gii: Nhn xt : ch c th c 2 mui l Na2SO3 ( Na : S = 2 : 1= 2 ) hoc NaHSO3 ( Na : S = 1 : 1 = 1)Ta c nS = nSO2 = 0,2 mol , nNa = n NaOH = 16/40 =0,4 mol

2 3

0,42 2 0,2

0,2Na

S

S

n Na SO amol n

n= = = =

vy khi lng mui l m (mui) = 0,2.126 = 25,2 gamChn cu D ( c th tnh theo s mol Na cng c )



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Cu 140 : Cho0,25a

17mol P2O5 vo 125 gam dung dch NaOH 16% thu c dung dch Y ch cha 2 mui

NaH2PO4 v Na2HPO4. Gi tr ca a nm trong khong no sau y ?A. 8,5 < a < 17 B. 17 < a < 34C. 10,2 < a < 20,4 D. 1207 < a < 2414

Gii: Nhn xt : NaH2PO4 ( Na : P = 1 : 1 = 1 ) v Na2HPO4 ( Na : P = 2 : 1 = 2 )

n(P) =0,25a

17

.2 =0,5a

17

v n (Na) = n( NaOH) =16 125

. 0,5

100 40

=

Ta c PNa

0,5an a17= =n 0,5 17

v t bi ta c P

Na

n a1 = 2 17 34

n 17a< < < < chn p n B

Cu 141 (H Khi B 2009): Cho 100 ml dung dch KOH 1,5M vo 200 ml dung dch H3PO4 0,5M, thuc dung dch X. C cn dung dch X, thu c hn hp gm cc cht sau l :

A. K3PO4, K2HPO4 B. K2HPO4, KH2PO4C. K3PO4 v KOH D. H3PO4 v KH2PO4

Gii: nK

= 0,15 mol , nP

= 0,1 mol

KOH + H3PO4

K

P

n 0,151 1,5 2n 0,1< = = < nn sn phm l KH2PO4 v K2HPO4 , chn B

Cu 142 (C 2010): Hp th hon ton 3,36 lt CO2 (ktc) vo 125 ml dung dch Ba(OH)2 1M thu cdung dch X. Coi th tch dung dch khng thay i, nng mol ca cht tan trong dung dch X l :

A. 0,4M B. 0,2M C. 0,6M D. 0,1M

Gii: CO2 : 0,15 mol + Ba(OH)2 : 0,125 mol

2 2C CO Ba Ba(OH)n n 0,15 mol; n n 0,125 mol= = = =

Ba

C

n 0,125 50,5 1n 0,15 6

< = = < nn c c hai mui BaCO3 : x mol , Ba(HCO3)2 : y mol

nC = x + 2y = 0,15 mol , nBa = x + y = 0,125 mol x = 0,1 mol , y = 0,025 mol[Ba(HCO3)2] = 0,025 : 0,125 = 0,2 M, Chn B

Cu 143 (C 2008): Dn t t V lt kh CO (ktc) i qua ng sng lng d hn hp rn gm CuO,Fe2O3 (nhit cao). Sau khi cc phn ng xy ra hon ton, thu c kh X. Dn ton b kh X trn volng d Ca(OH)2 th to 4 gam kt ta. Gi tr V l :

A. 1,120 B. 0,896 C. 0,448 D. 0,224

Gii: Ca(OH)2 d nn2 3CO CaCO

4n n 0,04 mol

100= = =

K2HPO4 , K : P = 2

KH2PO4 , K : P = 1

K3PO4 , K : P = 3

Ba(HCO3)2 , Ba : C = 1 : 2 = 0,5

BaCO3 , Ba : C = 1



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2CO CO CO

n n 0,04 mol V 0,04.22,4 0,896= = = = lt, Chn B

Cu 144 : Cho a mol CO2 hp th hon ton vo dung dch cha 2a mol NaOH. Dung dch thu c sau phnng c gi tr pH:

A. > 7 B. < 7 C. Khng xc nh. D. = 7Gii: ch c th c 2 mui l

Na2CO3 ( Na : C = 2 : 1= 2 ) hoc NaHCO3 ( Na : C = 1 : 1 = 1)nC = nCO2 = a mol , nNa = n NaOH = 2a mol

lp t l NaC

n 22n

a

a= = to mui Na2CO3 ( Na : C = 2 : 1 ), dung dch trung tnh, pH = 7, chn D

Cu 145 : Hn hp X gm Fe2O3, FeO, Al2O3 c khi lng l 42,4 gam. Khi cho X tc dng vi CO d, unnng thu c 41,6 gam cht rn Y v hn hp gm CO, CO2. Khi cho hn hp kh ny qua dung dchBa(OH)2 d th thu c m gam kt ta. Gi tr m l :A. 39,4 gam B. 9,85 gam C. 9,7 gam D. 4,925 gam

Gii: m O (trong xit) = 41,6 42,4 = 0,8 gam

Ba(OH)2 d :2 3O CO BaCO

0,8n n n 0,05 mol

16= = = =

Khi lng kt ta : m = 0,05.197 = 9,85 gam, chn B

Cu 146 : Qu trnh thi CO2 vo dung dch NaOH mui to thnh theo th t lA. NaHCO3 , Na2CO3 B.Na2CO3 v NaHCO3C. Na2CO3 D. Khng d kin xc nh.

: Gii: Chon ACu 147 : Thi kh CO2 d vo dung dch Ca(OH)2 mui thu c l

A.Ca(HCO3)2 B. CaCO3 C. C A v B D.Khng xc nh c.Gii: Chn A

Cu 148 : Kh m gam hn hp MgO, FeO, CuO bng kh CO nhit cao, sau phn ng thu c 21,6 gamhn hp rn A v hn hp kh B, cho B i qua dung dch Ba(OH)2 thu c 19,7 gam kt ta v dung dch C.Lc b kt ta, un nng dung dch C li xut hin thm 19,7 gam kt ta na. Gi tr m l :

A. 28,4 B. 25,5 C. 26,4 D. 32,4

Gii:3BaCO

19,7n 0,1 mol197

= =

un nng dung dch C li c kt ta chng t :0t

3 2 3 2 2Ba(HCO ) BaCO CO H O

0,1 0,1 0,1 mol

+ +

Khi cho CO2 vo c 2 phng trnh :

2 2 3 2

2 2 3 2

Ba(OH) CO BaCO H O

0,1 0,1

Ba(OH) 2CO Ba(HCO )

0,2 0,1

+ +

+

2CO O (trong oxit)n 0,2 0,1 0,3 n= + = =

p dng nh lut bo ton khi lng : m = mA + mO = 21,6 +0,3.16 = 26,4 gam, chn CCu 149 (H Khi B 2008) : Nhit phn hon ton 40 gam mt loi qung lmit c ln tp cht trsinh ra8,96 lt kh CO2(ktc). Thnh phn phn trm v khi lng ca CaCO3.MgCO3 trong loi qung nu trn l

A. 84%. B. 50%. C. 92%. D. 40%.

Gii:2CO

8,96n 0,4 mol

22,4= = ;

0

3 3 2CaCO ,MgCO CaO.MgO 2CO

0,2 0,4

t +

3 3MgCO .CaCOm 0,2.(100 84) 36,8 gam= + =



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36,8

% m= 100 92%40

= , chn C

Cu 150 (H Khi B 2007) : Nung 13,4 gam hn hp 2 mui cacbonat ca 2 kim loi ha tr 2, thu c 6,8gam cht rn v kh X. Lng kh X sinh ra cho hp th vo 75 ml dung dch NaOH 1M, khi lng mui khanthu c sau phn ng l

A. 6,5 gam. B. 5,8 gam. C. 6,3 gam. D. 4,2 gam.Gii: NaOH Nan 0, 075.1 0, 075 mol n= = =

3

2

MCO MOCO Cm m 13, 4 6,8n 0,15 mol n44 44 = = = =

Na3

C

n 11 NaHCO

n 2= < v CO2 d,

CO2 + NaOH NaHCO30,075 0,075 mol, nn m (NaHCO3) = 0,075.84 = 6,3 gam, chn C

Cu 151 : Hp th ht 0,35 mol CO2 vo duung dch cha 0,225 mol Ca(OH)2. Tnh lng kt ta thu cA. 10 gam B. 11 gam C. 12 gam D. 13 gam

Gii: V OHn

1 1,28 2 100(0,45 0,35) 10 gamn

m

= = =

, chn A

Dng 14 : TON MUI (TON NGC)Cho CO2 vo dung dch Ca(OH)2 hoc Ba(OH)2 thu c kt ta l CaCO3 hoc BaCO3

Vn 1 : Bit2 3Ca(OH) CaCO

n , n , tm2CO

n

TH1 : ch c kt ta 2COn n=

TH2 : c kt ta, sau kt ta tan 1 phn : 2 2CO Ca(OH)n 2.n n=

Vn 2 : Bit3 2CaCO CO

n , n , tm2Ca(OH)

n

Nu3 2CaCO CO

n n th2 3

2

CO CaCOCa(OH)

n nn2

+=

Cu 152 : Hp th ton b x mol CO2 vo dung dch cha 0,03 mol Ca(OH)2 thu c 2 gam kt ta. Tm x ?A. 0,02 mol v 0,04 mol B. 0,02 mol v 0,05 mol

C. 0,01 mol v 0,03 mol D. 0,03 mol v 0,04 mol

Gii:3CaCO

2n n 0, 02 mol

100= = =

TH1 : ch c kt ta 2COn n 0,02 mol= = TH2 : c kt ta, sau kt ta tan 1 phn :

2 2CO Ca(OH)n 2.n n 2.0,03 0,02 0,04 mol= = = , Chn A

Cu 153 : Dn V lt CO2 (ktc) vo 300 ml dung dch Ca(OH)2 0,5M. Sau phn ng thu c 10 gam kt taGi tr V bng ?

A. 2,24 lt B. 3,36 lt C. 4,48 lt D. C A v C

Gii:3 2CaCO Ca(OH)

10n n 0,1 mol; n 0,5.0,3 0,15 mol

100= = = = =



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32

TH1 : ch c kt ta2CO

n n 0,1 mol V 0,1.22,4 2,24= = = = lt

TH2 : c kt ta, sau kt ta tan 1 phn :

2 2CO Ca(OH)n 2.n n 2.0,15 0,1 0,2 mol V 0, 2.22,4 4,48= = = = = lt, Chn D

Cu 154 : Khi cho 0,35 mol CO2 hp th hon ton vo dung dch Ca(OH)2u thu c 0,05 mol kt ta. Smol Ca(OH)2 trong dung dch l

A. 0,15. B. 0,2. C. 0,3. D. 0,05.

Gii: Do3 2CaCO COn n nn

2 3

2

CO CaCOCa(OH) n n 0, 35 0, 05n 0,22 2

+ += = = , chn B

Cu 155 : Hp th ht k mol SO2 bng 150 ml dung dch Ca(OH)2 1M thy c 12 gam kt ta trng to thnh .Tnh k ?A. 0,1 v 0,2 B. 0,15 v 02 C. 0,15 v 0,25 D. 0,1

Gii:3CaSO

12n n 0,1 mol

120= = = ,

2Ca(OH)n 1.0,15 0,15 mol= =

TH1 : ch c kt ta2SO

n n 0,1 mol= =


2 2CO Ca(OH)n 2.n n 2.0,15 0,1 0, 2 mol= = = , Chn A

Cu 156 : Hp th ht a mol SO2 bng 150 ml dung dch Ba(OH)2 1M thy c 21,7 gam kt ta trng to thnh.Tnh gi tr a ?A. 0,1 v 0,2 B. 0,15 v 02 C. 0,15 v 0,25 D. 0,1

Gii:3BaSO

21,7n n 0,1 mol

217= = = ,

2Ba(OH)n 1.0,15 0,15 mol= =

TH1 : ch c kt ta2SO

n n 0,1 mol= =


2 2SO Ba(OH)n 2.n n 2.0,15 0,1 0,2 mol= = = , Chn A

Cu 157 : Cho V lt kh CO2 (ktc) vo 2 lt dung dch Ca(OH)2 0,05M, thu c 7,5 gam kt ta. V c gi tr

l :A. 1,68 lt B. 2,88 ltC. 2,24 lt v 2,8 lt D. 1,68 lt v 2,8 lt

Gii:3 2CaCO Ca(OH)

7,5n n 0,075 mol; n 0,05.2 0,1 mol

100= = = = =

TH1 : ch c kt ta2CO

n n 0,075 mol V 0,075.22,4 1,68= = = = lt


2 2CO Ca(OH)n 2.n n 2.0,1 0,075 0,125 mol V 0,125.22, 4 2,8= = = = = lt, Chn D

Cu 158 (H Khi A 2007): Hp th hon ton 2,688 lt CO2 (ktc) vo 2,5 lt dung dch Ba(OH)2 nng amol/lt, thu c 15,76 gam kt ta. Gi tr a l :

A. 0,032 B. 0,048 C. 0,06 D. 0,04

Gii:2 3CO BaCO

2,688 15,76n 0,12 mol; n 0,08 mol

22,4 197= = = =

Do2 3CO BaCO

n n nn 2 32

CO BaCOBa(OH)

n n 0,12 0, 08n 0,1 mol

2 2

+ += = = ,

0,1a 0,04M

2,5= = , Chn D

Cu 159 : Cho 2,24 lt kh CO2(ktc) vo 20 lt dung dch Ca(OH)2, ta thu c 6 gam kt ta. Vy nng mol/l ca dung dch Ca(OH)2l:


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33

A. 0,004M. B. 0,002M. C. 0,006M. D. 0,008M.

Gii:2 3CO CaCO

2,24 6n 0,1 mol; n 0,06 mol

22,4 100= = = =

Do2 3CO CaCO

n n nn 2 32

CO CaCOCa(OH)

n n 0,1 0, 06n 0,08 mol

2 2

+ += = = ,

2

0,08[Ca(OH) ] 0,004M

20= = , chn A

Cu 160 : Dn V lt kh CO2 (ktc) vo 200 ml dung dch Ca(OH)2 0,1M thu c 1 gam kt ta. Vy V c gitr l :

A. 0,224 hoc 0,896 B. 0,366 hoc 0,672C. 0,224 hoc 0,672 D. 0,896

Gii:3 2CaCO Ca(OH)

1n n 0,01 mol; n 0,02 mol

100= = = =

TH1 : ch c kt ta2CO

n n 0,01 mol V 0,01.22, 4 0, 224= = = = lt


2 2CO Ca(OH)n 2.n n 2.0,02 0,01 0,03 mol V 0,03.22,4 0,672= = = = = lt, Chn C

Cu 161 : Cho V lt CO2 (ktc) vo dung dch A cha 0,2 mol Ca(OH)2 thu c 2,5 gam kt ta. Gi tr V l :A. 0,56 hoc 11,2 B. 0,896 hoc 6,72C. 0,896 hoc 11,2 D. 0,56 hoc 8,40

Gii:3 2CaCO Ca(OH)

2,5n n 0,025 mol; n 0, 2 mol

100= = = =

TH1 : ch c kt ta2CO

n n 0,025 mol V 0,025.22,4 0,56= = = = lt


2 2CO Ca(OH)n 2.n n 2.0,2 0,025 0,375 mol V 0,375.22, 4 8,4= = = = = lt, Chn D

Cu 162 : Hp th hon ton V lt CO2 (ktc) vo 500 lt dung dch Ba(OH)2 1M, thu c 59,1 gam kt ta.

Gi tr ca V l :A. 6,72 hoc 11,2 B. 11,2 hoc 15,68C. 15,68 hoc 8,96 D. 6,72 hoc 15,68

Gii:3BaCO

59,1n 0,3 mol

197= = ,

2Ba(OH)n 0,5 mol=

TH1 : ch c kt ta2CO

n n 0,3 mol V 0,3.22, 4 6,72= = = = lt


2 2CO Ba(OH)n 2.n n 2.0,5 0,3 0,7 mol V 0,7.22,4 15,68= = = = = lt, Chn D

Cu 163 : Cho 4,48 lt CO2 vo 2 lt dung dch Ba(OH)2 thu c 2,955 gam kt ta. Hy tnh nng mol/ltca dung dch Ba(OH)2 :

A. 0,05375M B. 0,07354M C. 0,0875M D. 0,00023M

Gii:2 3CO BaCO

4,48 2,955n 0,2 mol; n 0,015 mol

22,4 197= = = =

Do2 3CO BaCO

n n nn 2 32

CO BaCOBa(OH)

n n 0, 2 0, 015n 0,1075 mol

2 2

+ += = =

[Ba(OH)2] = 0,1075 : 2 = 0,05375 M , chn ADng 15 : HP TH CO2, SO2 VO DD HN HP NaOH v Ca(OH)2

Cn tnh : 2+2CO OH Ca

n ; n ; n

2 3CO OH HCO +



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34

22 3 2CO 2OH CO H O + + , lp t s

2

OH

CO

nf

n

=

2+ 23 3Ca CO CaCO

+ hay 2+ 23 3Ba CO BaCO+

Cu 164 (H Khi A 2008): Hp th hon ton 4,48 lt kh CO2 (ktc) vo 500 ml dung dch hn hp gmNaOH 0,1M v Ba(OH)2 0,2M, sinh ra m gam kt ta. Gi tr ca m l

A. 11,82. B. 19,70. C. 9,85. D. 17,73.2+

2 2NaOH Ba(OH) COBa OHn 0,05 mol; n n 0,1 mol; n 0, 25 mol; n 0, 2 mol= = = = =

2

OH

CO

n 0,251 f 1,25 2

n 0,2

< = = =

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