Giả sử số mol Fe2O3, CuO là a, b (mol)
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a----------------------->3a
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
b----------------->b
=> 3a : b = 3 : 2
=> 2a = b
\(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{160a}{160a+80b}.100\%=50\%\\\%m_{CuO}=\dfrac{80b}{160a+80b}.100\%=50\%\end{matrix}\right.\)
