CuO +H2-->Cu+H2O
x(80x)------x(64x)
Fe2O3+3H2--->2Fe+3H2O
160y--------------112y
Theo bài ra ta có hệ pt
\(\left\{{}\begin{matrix}80x+160y=24\\64x+112y=17,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
%m\(_{CuO}=\frac{80.0,1}{24}.100\%=33,33\%\)
%m\(_{Fe2O3}=100-33,33=66,67\%\)