pthh CuO + H2 --> Cu+H2O
x x
Fe2O3 + 3H2 --> 2Fe + 3H2O
y 3y
nH2= 19,6/22,4=0,875mol => ta có hệ
\(\left\{{}\begin{matrix}x+y=0,875\\80x+160y=50\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=0,125\\y=0,25\end{matrix}\right.\)
=> mCuO =0,125 * 80 =10g => %mCu = 10*100/50=20%
%mFe= 100- 20 =80%