Đặt :
nCuO = 2x mol
nFe2O3 = x mol
mhh= 160x + 160 x = 160
=> x = 0.5
CuO + H2SO4 --> CuSO4 + H2O
1______1___________1
Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
0.5_______1.5_________0.5
mH2SO4 = 2.5*98=245 g
C%H2SO4 = 30.625%
mCuSO4 = 160 g
mFe2(SO4)3 = 200 g
mdd sau phản ứng = 160 + 800 = 960 g
C%CuSO4 = 16.67%
C%Fe2(SO4)3 = 20.83%