Question

$120\, g\, Mg$ was burnt in air to give a mixture of $MgO$ and $Mg _{3} N _{2}$. The mixture is now dissolved in $HCl$ to form $MgCl _{2}$ and $NH _{4} Cl$. If $107\, g\, NH _{4} Cl$ is produced, then the moles of $MgCl _{2}$ formed is
(At. wt.: $Mg =24, N =14, Cl =35.5$ )

Answer 1

$\underset{a}{Mg} + \frac{1}{2}O_2 \rightarrow \underset{a}{MgO}$,
$\underset{(5-a)}{3Mg + N_2} \rightarrow \underset{\frac{(5-a)}{3}}{Mg_3N_2}$
$\underset{a}{MgO} + \underset{a}{2HCl} \rightarrow MgCl_2 + H_2O$
$\underset{\frac{5-a}{3}}{Mg_3N_2} + 8HCl \rightarrow \underset{\frac{2(5-a)}{3}}{2NH_4Cl} + \underset{(5 - a)}{3MgCl_2}$
$\therefore $ Total moles of $MgCl _{2}=a+(5-a)=5\, mol$