Bài 8:
Trong 7.92 g hh có : x (mol) Fe2O3, y (mol) CuO , z (mol) Fe3O4
Trong 0.275 mol hh có : kx (mol) Fe2O3, ky (mol) CuO , kx (mol) Fe3O4
mhh = 160x + 80y + 232z = 7.92 (1)
nHCl = 0.27 mol
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
x________6x
CuO + 2HCl --> CuCl2 + H2O
y______2y
Fe3O4 + 8HCl --> FeCl2 + 2FeCl3 + 4H2O
z________8z
<=> 6x + 2y + 8z = 0.27 (2)
TC :
k( x + y + z ) = 0.275 (3)
Fe2O3 + 3H2 -to-> 2Fe + 3H2O
kx_______________2kx____3kx
CuO + H2 -to-> Cu + H2O
ky____________ky____ky
Fe3O4 + 4H2 -to-> 3Fe + 4H2O
kz_______________3ky_____4kz
<=> k( 3x + y + 4z ) = 0.675 (4)
(4) : (3)
0.675( x + y + z) = 0.275( 3x + y + 4z)
<=> 0.675x + 0.675y + 0.675z - 0.825x - 0.275y - 1.1z = 0
<=> 0.15x + 0.4y - 0.425z = 0 (5)
Giải (1) , (2) và (5) :
x = 0.025
y = 0.02
z = 0.01
mCr= 0.25*56 + 0.1*64 + 0.15*56= 28.8 (g)
mFe2O3 = 4 g
mCuO = 1.6 g
mFe3O4 = 2.32 g