$n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$R+2HCl\to RCl_2+H_2$
$\Rightarrow n_A=0,4(mol)$
$\overline{M_A}=\dfrac{19,2}{0,4}=48$
$M_{Fe}=56>48\Rightarrow M_R<48$
$n_R=\dfrac{9,2}{M_R}(mol)$
$n_{HCl}=1(mol)$
Dư axit $\Rightarrow \dfrac{9,2.2}{M_R}<1$
$\Leftrightarrow M_R>18,4$
$\to 18,4<M_R<48$
Vậy $M_R=24(Mg)$ hoặc $M_R=40(Ca)$