\(a,n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\\ b.n_{HCl}=0,2.2=0,4mol\\ a=m_{ddHCl}=\dfrac{0,4.36,5}{14,6}\cdot100=100g\\ c.C_{\%FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}\cdot100=22,92\%\)