PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4mol\\n_{H_2}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)