\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=\dfrac{73.20\%}{36,5}=0,4\left(mol\right)\\ a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCl.dư\\ n_{H_2}=n_{FeCl_2}=0,1\left(mol\right);n_{HCl\left(p.ứ\right)}=0,1.2=0,2\left(mol\right);n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ b,V_{H_2\left(ĐKTC\right)}=0,1.22,4=2,24\left(l\right)\\ b,m_{ddsau}=5,6+73-0,1.2=78,4\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{78,4}.100\approx9,311\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,1}{78,4}.100\approx16,199\%\)