\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{HCl}=0,4.36,5=14,6\left(g\right)\\ C\%_{ddHCl}=\dfrac{14,6}{200}.100=7,3\%\)