\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=n_{FeCl_2}=\dfrac{n_{HCl}}{2}=0,2\left(mol\right)\\ a,m_{Fe}=0,2.56=11,2\left(g\right)\\ b,m_{FeCl_2}=0,2.127=25,4\left(g\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)