CuO + H2 -to-> Cu + H2O
x____x_____x(mol)
Fe2O3 + 3 H2 -to-> 2 Fe + 3 H2O
y_____3y_______2y(mol)
Ta có: \(\left\{{}\begin{matrix}22,4x+22,4.3y=15,68\\64x+2y.56=28,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=> m= mCuO+ mFe2O3= 80x+160y= 80.0,1+160.0,2= 40(g)
=> m=40(g)