\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
\(2R+2nHCl\rightarrow2RCl_n+nH_2\)
\(\dfrac{1.2}{n}......1.2...............0.6\)
\(M_R=\dfrac{14.4}{\dfrac{1.2}{n}}=12n\)
\(BL:n=2\Rightarrow R=24\)
\(R:Mg\)
\(m_{MgCl_2}=0.6\cdot95=57\left(g\right)\)
\(m_{dd}=14.4+146-0.6\cdot2=159.2\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{57}{159.2}\cdot100\%=35.8\%\)