$n_{H_2}=\dfrac{4,48}{22,4}=0,2 mol$
$2A+2nHCl\to 2ACl_n+nH_2$
$\Rightarrow n_A=\dfrac{0,4}{n}$
$M_A=\dfrac{4,8n}{0,4}=12n$
$n=2\Rightarrow M_A=24(Mg)$
$n_{HCl}=2n_{Mg}=0,4 mol$
$\Rightarrow m_{dd HCl}=\dfrac{0,4.36,5.100}{10}=146g$
$m_{dd X}=4,8+146-0,2.2=150,4g$
$n_{MgCl_2}=0,2 mol$
$\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95.100}{150,4}=12,63\%$