nFe= 22.4/56=0.4 mol
Fe + H2SO4 --> FeSO4 + H2
0.4__0.4______________0.4
VH2= 0.4*22.4=8.96l
VddH2SO4= 0.4/0.5=0.8l
nCuO= 8/80=0.1 mol
CuO + H2 -to-> Cu + H2O
1_____1
0.1____0.4
Lập tỉ lệ: 0.1/1 < 0.4/1 => CuO hết , H2 dư
nCu=nCuO=0.1 mol
mCu= 0.1*64=6.4g