CuO+H2-to>Cu+H2O
x-----------------------x mol
Fe2O3+3H2-to>2Fe+3H2O
y------------------------------3y
ta có :\(\left\{{}\begin{matrix}80x+160y=16\\x+3y=0,25\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
=>%mCu=\(\dfrac{0,1.64}{0,1.64+0,05.2.56}.100\)=53,33%
%m Fe=46,67%
vì dùng dư H=10%
VH2=(0,1+0,15).\(\dfrac{100}{10}\).22,4=56l