CuO+H2-->Cu+H2O(1)
FexOy+yH2-->xFe+yH2O(2)
Fe+2HCl-->FeCl2+H2(3)
Ta có
n H2(3)=4,48/22,4=0,2(mol)
Theo pthh3
n Fe=n H2=0,2(mol)
m Fe=0,2.56=11,2(g)
m Cu=17,6-11,2=6,4(g)
-->n Cu=6,4/64=0,1(mol)
Theo pthh1
n CuO=n Cu=0,1(mol)
m CuO=0,1.80=8(g)
m FexOy=24-8=16(g)
-->m O(fexOy)=16-11,2=4,8(g)
n O=4,8/16=0,3(mol)
Ta có
n Fe:n O=0,2:0,3=2:3
-->CTHH:Fe2O3
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