\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(LTL:0.1>\dfrac{0.1}{1}\Rightarrow Fedư\)
\(n_{H_2}=\dfrac{1}{2}\cdot0.1=0.05\left(mol\right)\)
\(V=0.05\cdot22.4=1.12\left(l\right)\)