\(Fe+HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=3n_{Fe}=0,3\left(mol\right)\)
\(\dfrac{0,1}{1}< \dfrac{0,3}{2}\rightarrow HCl\) dư
\(n_{HCl}=0,3-0,1.2=0,1\left(mol\right)\)
\(n_{FeCl_2}=n_{HCl}=0,1\left(mol\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)