$n_{Mg}=\dfrac{2,4}{24}=0,1mol \\PTHH : \\Mg+2HCl\to MgCl_2+H_2↑ \\a.Theo\ pt : \\n_{H_2}=n_{Mg}=0,1mol \\⇒V_{H_2}=0,1.22,4=2,24l \\b.Theo\ pt : \\n_{HCl}=2.n_{Mg}=2.0,1=0,2mol \\⇒V_{dd\ HCl}=\dfrac{0,2}{0,5}=0,4l \\c.Theo\ pt : \\n_{MgCl_2}=n_{Mg}=0,1mol \\⇒C_{M_{MgCl_2}}=\dfrac{0,1}{0,4}=0,25M$