Question

$120gMg$ was burnt in air to give a mixture of $MgO$ and $M{g}_3{N}_2$. The mixture is now dissolved in $HCl$ to form $MgC{l}_2$ and $N{H}_{4}Cl$. If $107gN{H}_{4}Cl$ is produced, then the moles of $MgC{l}_2$ formed is
(At. wt.: $Mg=24,N=14,Cl=35.5$ )

Answer 1

Answer: 5

$\underset{a}{\mathrm{Mg}}+\frac{1}{2}{O}_2\to \underset{a}{\mathrm{MgO}}$,
$\underset{(5-a)}{3Mg+N_2}\to \underset{\frac{(5-a)}{3}}{M{g}_3{N}_2}$
$\underset{a}{\mathrm{MgO}}+\underset{a}{2HCl}\to MgC{l}_2+H_{2}O$
$\underset{\frac{5-a}{3}}{M{g}_3{N}_2}+8HCl\to \underset{\frac{2(5-a)}{3}}{2N{H}_{4}Cl}+\underset{(5-a)}{3MgC{l}_2}$
$\therefore$ Total moles of $MgC{l}_2=a+(5-a)=5mol$