a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{1,12}{2,24}=0,05\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,05.56}{5,6}.100\%=50\%\\\%m_{Cu}=50\%\end{matrix}\right.\)
c, Theo PT: \(n_{FeSO_4}=n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,05.152=7,6\left(g\right)\)