nH2= 4.48/22.4=0.2 mol
CuO + H2 -to-> Cu + H2O
0.1___________0.1
FexOy + yH2 -to-> xFe + yH2O
0.2/x____________0.2
Fe + 2HCl --> FeCl2 + H2
0.2_________________0.2
mkl= mFe + mCu= 17.6 g
=> 0.2*56 + mCu= 17.6
=> mCu= 6.4 g
=> nCu= 0.1 mol
mhh= 0.1*80 + 0.2/x * (56x + 16y ) = 24
=> 11.2x + 3.2y = 16x
=> 3.2y = 4.8x
=> x/y= 3.2/4.8=2/3
Vậy: CT của oxit : Fe2O3