Ta có:
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:x\left(mol\right)\\n_{Zn}:y\left(mol\right)\end{matrix}\right.\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Giải hệ PT:
\(\left\{{}\begin{matrix}56x+65y=12,1\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=,1.56=5,6\left(g\right)\\m_{Zn}=12,1-5,6=6,5\left(g\right)\end{matrix}\right.\)