a)$n_{Mg}$ = $\frac{m_{Mg}}{M_{Mg}}$ = $\frac{2,4}{24}$ = 0,1(Mol)
$Mg_{}$ + 2$HCl_{}$ → $MgCl_{2}$ + $H_{2}$
0,1 → 0,2 0,1 0,1 : mol
b) $V_{H_2}$ = $n_{H_2}$ . 22,4= 0,1. 22,4= 2,24(L)
$m_{MgCl_2}$ = $n_{MgCl_2}$ . $M_{MgCl_2}$ =0,1. 95=9,5(g)=$m_{muối}$
c) 200ml=0,2L
$C_{M}$ = $\frac{n_{HCl}}{V_{dung dịch HCl}}$ = $\frac{0,2}{0,2}$ = 1(M)