$120 \mathrm{g} \mathrm{Mg}$ was burnt in air to give a mixture of $\mathrm{MgO}$ and ${\mathrm{Mg}}_3{\mathrm{N}}_2$. The mixture is now dissolved in $\mathrm{HCI}$ to form ${\mathrm{MgCI}}_2$ and ${\mathrm{NH}}_4\mathrm{CI}$, if 107 gram ${\mathrm{NH}}_4\mathrm{CI}$ is produced. Then the moles of ${\mathrm{MgCI}}_2$ formed is :

$\begin{array}{l}\mathrm{Mg}+\frac{1}{2}{\mathrm{O}}_2⟶\mathrm{MgO} ..\left(\mathrm{i}\right)\\ 3\mathrm{Mg}+{\mathrm{N}}_2⟶{\mathrm{Mg}}_3{\mathrm{N}}_2 ...\left(\mathrm{ii}\right)\\ \mathrm{MgO}+2\mathrm{HCl}⟶{\mathrm{MgCl}}_2+{\mathrm{H}}_2\mathrm{O} ...\left(\mathrm{iii}\right)\\ {\mathrm{Mg}}_3{\mathrm{N}}_2+8\mathrm{HCl}⟶2{\mathrm{NH}}_4\mathrm{Cl}+3{\mathrm{MgCl}}_2 ..\left(\mathrm{iv}\right)\end{array}$