a) \(\left\{{}\begin{matrix}m_{Fe}=1,4\left(g\right)\\m_{Cu}=6-1,4=4,6\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Fe}=\dfrac{1,4}{56}=0,025\left(mol\right)\\n_{Cu}=\dfrac{4,6}{64}=0,071875\left(mol\right)\end{matrix}\right.\)
PTHH:
\(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\\ CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=0,109375\left(mol\right)\)
`=> V_{H_2} = 0,109375.22,4 = 2,45 (l)
b) Theo PT:
\(n_{Fe_2O_3}=\dfrac{1}{2}.n_{Fe}=0,0125\left(mol\right)\\
n_{CuO}=n_{Cu}=0,071875\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{0,0125.160}{0,0125.160+0,071875.80}.100\%=25,81\%\\\%m_{CuO}=100\%-25,81\%=74,19\%\end{matrix}\right.\)
