Gọi x là số mol của CuO, y là số mol của \(Fe_2O_3\)
\(m_{CuO}+m_{Fe_2O_3}=m_{hh}\\ \Rightarrow80x+160y=32\left(1\right)\)
\(PTHH\left(1\right):CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(\left(mol\right)\)_______x__________________x
\(PTHH\left(2\right):Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(\left(mol\right)\)_______y_______3y_____2x____3y
\(m_{H_2O\left(1\right)}+m_{H_2O\left(2\right)}=\sum m_{H_2O}\\ \Rightarrow n_{H_2O\left(1\right)}+n_{H_2O\left(2\right)}=\sum n_{H_2O}\\ \Rightarrow x+3y=\frac{9}{18}=0,5\left(2\right)\)
\(\xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}80x+160y=32\\x+3y=0,5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(V_{H_2}=22,4x+22,4.3y=11,2\left(l\right)\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(\Sigma n_{H_2O}=\frac{9}{18}=0,5\left(mol\right)\)
Theo PT: \(\Sigma n_{H_2}=\Sigma n_{H_2O}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)
Good luck!
(Ko c.chắn lắm nha!)