bài 11
DESCRIPTION
hjnhgnhTRANSCRIPT
Bi 1:
Bi 1: S mol Fe: nFe = = 0,035 (mol) Khi lng dung dch CuSO4: mdd = V. D = 100 . 1,12 = 112 (g)
Khi lng cht tan CuSO4: mCuSO4 = = = 11,2 (g)
S mol CuSO4: n = = 0,07 (mol) PTHH:
Fe
+CuSO4
FeSO4 +
Cu
Theo pt: 1 mol
1 mol
1 mol
1 mol Theo b: 0,035 mol
0,07 mol
T l:
= 0,035 < = 0,07
Vy: Sau phn ng, Fe phn ng ht, CuSO4 d.
PTHH:
Fe
+CuSO4
FeSO4 +
Cu
Theo pt: 1 mol
1 mol
1 mol
1 mol
Theo b: 0,035
0,035 mol
0,035 mol
0,035 mol
Dung dch sau phn ng gm CuSO4 d v FeSO4.
S mol CuSO4 d: n = 0,07 0,035 = 0,035 (mol)
Nng mol cc cht trong dung dch sau phn ng:
CM CuSO4 d = = 0,35 M
CM FeSO4 = = 0,35 M
Bi 2:
S mol Al: nAl = = 0,2 (mol)
S mol H2SO4: n = V.CM = 0,1 . 0,5 = 0,05 (mol)
PTHH:
2Al
+3H2SO4
Al2(SO4)3 +
3H2
Theo pt: 2 mol
3 mol
1 mol
3 mol
Theo b: 0,2 mol
0,05 mol
T l:
= 0,1 > = 0,0166
Vy: Sau phn ng, H2SO4 phn ng ht, Al d.
PTHH:
2Al
+3H2SO4
Al2(SO4)3 +
3H2
Theo pt: 2 mol
3 mol
1 mol
3 mol
Theo b: 0,03mol
0,05 mol 0,02 mol0,05 mol
Th tch H2 (ktc): V = 0,05 . 22,4 = 1,12 (l)
Nng mol cc cht trong dung dch sau phn ng:
CM Al2(SO4)3 = = 0,2 M
Bi 3:
Khi lng cht tan H2SO4: mH2SO4 = = = 11,76 (g)
S mol H2SO4: nH2SO4 = = 0,12 (mol)
Khi lng cht tan BaCl2: mBaCl2 = = = 10,4 (g)
S mol BaCl2: nBaCl2 = = 0,05 (mol)
PTHH:
BaCl2
+H2SO4
BaSO4 +
2HCl
Theo pt: 1 mol
1 mol
1 mol
2 mol
Theo b: 0,05 mol
0,12 mol
T l:
= 0,05 < = 0,12
Vy: Sau phn ng, H2SO4 d, BaCl2 phn ng ht.
PTHH:
BaCl2
+H2SO4
BaSO4 +
2HCl
Theo pt: 1 mol
1 mol
1 mol
2 mol
Theo b: 0,05 mol
0,05 mol 0,05 mol
0,1 mol
Khi lng cht rn sinh ra: mBaSO4 = 0,05 . 233 = 11,65 (g)
Khi lng dung dch sau phn ng: mdd = 58,8 + 200 11,65 = 247,15 (g) Dung dch sau phn ng gm H2SO4 d, HCl.
S mol H2SO4 d: n d = 0,12 0,05 = 0,07 (mol)
Nng phn trm cc cht:
C% H2SO4 d = . 100% 2,78%
C% HCl = . 100% 1,48%
Bi 4:
Gi R2O3 l oxit ca kim loi ha tr III.
S mol HNO3: nHNO3 = CM. V = 3 . 0,8 = 2,4 (g)
Khi lng HCl : mHNO3 = 2,4 . 63 = 151,1 (g)
PTHH:
R2O3
+6HNO3
2R(NO3)3 +3H2O
Theo pt: (2R + 48) g
6 . 63 g Theo b: 64 g
151,2 g
Ta c phng trnh : 151,2 . (2R + 48) = 6 . 63 . 64 R = 56 (Fe) Cng thc ca oxit kim loi: Fe2O3. PTHH:
Fe2O3
+6HNO3
2Fe(NO3)3 +3H2O
Theo pt: 1 mol
6 mol
2 mol
3 mol
Theo b: 0,4 mol
2,4 mol 0,8 mol
1,2 mol
Nng mol ca Fe(NO3)3 :
CM Fe(NO3)3 = = 1M Bi 5: S mol H2 (ktc): nH2 = = 0,2 (mol)
Gi x v y ln lt l s mol Al v Fe c trong 5,5 gam hn hp.
PTHH:
2Al
+6HCl
2AlCl3 +
3H2
Theo pt: 2 mol
6 mol
2 mol
3 mol
Theo b: x mol
3x mol x mol
1,5x mol
PTHH:
Fe
+2HCl
FeCl2 +
H2
Theo pt: 1 mol
2 mol
1 mol
1 mol
Theo b: y mol
2y mol y mol
y mol
Theo bi, c h phng trnh
27x + 56y = 5,5
1,5x + y = 0,2
Gii h, c x = 0,1 (mol) ; y = 0,05 (mol) Khi lng tng kim loi c trong hn hp:
mAl = 0,1 . 27 = 2,7 (g) ; mFe = 5,5 2,7 = 2,8 (g)
Thnh phn phn trm mi kim loi c trong hn hp:
%mAl = . 100% = 49,09 % ; %mFe = 100% - 49,09% = 50,91 %
S mol HCl: nHCl = 3x + 2y = 3 . 0,1 + 2 . 0,05 = 0,4 (mol) Nng mol HCl: CM HCl = = 0,8 M
Bi 6:
S mol H2 (ktc): nH2 = = 0,3 (mol)
8,7 gam cht rn khng tan chnh l khi lng ca Ag.
Hn hp Mg v Zn c khi lng l : 20 8,7 = 11,3 (g)
Gi x v y ln lt l s mol Mg v Zn c trong 11,3 gam hn hp. PTHH:
Mg
+H2SO4
MgSO4 +
H2
Theo pt: 1 mol
1 mol
1 mol
1 mol
Theo b: x mol
x mol
x mol
x mol PTHH:
Zn
+H2SO4
ZnSO4 +
H2
Theo pt: 1 mol
1 mol
1 mol
1 mol
Theo b: y mol
y mol
y mol
y mol
Theo , c h phng trnh: 24x + 65y = 11,3
x + y = 0,3
Gii h, c x = 0,2 (mol) ; y = 0,1 (mol) Khi lng tng kim loi c trong hn hp:
mMg = 0,2 . 24 = 4,8 (g) ; mZn = 0,1 . 65 = 6,5 (g) ; mAg = 8,7
Thnh phn phn trm mi kim loi c trong hn hp:
%mMg = . 100% = 24 % ; %mZn = . 100% = 32,5% ; %mAg = . 100% = 43,5%
S mol H2SO4: nH2SO4 = x + y = 0,2 + 0,1 = 0,3 (mol)
Th tch dung dch H2SO4 dng: V = = = 0,6 (l) = 600 (ml)
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