\(n_Fe=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ a,n_{H_2}=n_{Fe}=0,2(mol)\\ \Rightarrow V_{H_2(đktc)}=0,2.22,4=4,48(l)\\ b,n_{HCl}=2n_{Fe}=0,4(mol)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6(g)\\ c,n_{FeCl_2}=n_{Fe}=0,2(mol)\\ \Rightarrow m_{FeCl_2}=0,2.127=25,4(g)\)