\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=n_{FeCl_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ a,m_{Fe}=0,2.56=11,2\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)