\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=2.n_{Fe}=2.0,1=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ b,n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)