$n_{Fe}=\dfrac{2,8}{56}=0,05mol \\PTHH : \\Fe+2HCl\to FeCl_2+H_2↑ \\a.Theo\ pt : \\n_{HCl}=2.n_{Fe}=2.0,05=0,1mol \\⇒C_{M_{HCl}}=\dfrac{0,1}{0,4}=0,25M \\b.Theo\ pt : \\n_{FeCl_2}=n_{Fe}=0,05mol \\⇒C_{M_{FeCl_2}}=\dfrac{0,05}{0,4}=0,125M \\c.Theo\ pt : \\n_{H_2}=n_{Fe}=0,05mol \\⇒V=V_{H_2}=0,05.22,4=1,12l$