\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,4}{1}=\dfrac{0,8}{2}\\ b.\Rightarrow P.ứ.hết,không.dư\\c.n_{FeCl_2}=n_{H_2}=n_{Fe}=0,4\left(mol\right)\\ C\%_{ddFeCl_2}=\dfrac{0,4.127}{22,4+200-0,4.2}.100\approx22,924\%\)