$a.PTHH : \\Fe+2HCl\to FeCl_2+H_2 \\n_{HCl}=0,2.0,1=0,02mol \\Theo\ pt : \\n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,02=0,01mol \\⇒V=V_{H_2}=0,01.22,4=0,224l \\b.Theo\ pt : \\n_{FeCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,02=0,01mol \\⇒C_{M_{FeCl_2}}=\dfrac{0,01}{0,2}=0,05M$