$120$ g Mg is burnt in air to give a mixture of $M g O$ and $M g_{3} N_{2}$ . The mixture is now dissolved in $H C l$ to form $M g C l_{2}$ and $N H_{4} C l$ . If $107$ g $N H_{4} C l$ is produced, then determine the moles of $M g C l_{2}$ formed.

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Solution

Number of moles of $N H_{4} C l = \frac{107 g m}{53.4 g m o l^{- 1}} = 2 m o l$
$2 M g + O_{2} \to 2 M g O$ $3 M g + N_{2} \to M g_{3} N_{2}$
$M g O + 2 H C l \to M g C l_{2} + H_{2} O$
$M g O + 2 H C l \to M g C l_{2} + H_{2} O$
$M g_{3} N_{2} + 8 H C l \to 3 M g C l_{2} + 2 N H_{4} C l$
$120 g M g = \frac{120}{24} = 5$ moles of $M g$
So $2$ moles of $M g$ gives $2$ moles of $M g O$ and $3$ moles of $M g$ gives one mole of $M g_{3} N_{2}$ . $1$ mole $M g O$ and $1$ mole $M g_{3} N_{2}$ give one mole and $3$ mole $M g C l_{2}$ respectively. Again one mole $M g_{3} N_{2}$ give $2$ moles of $N H_{4} C l$ .
The moles of $M g C l_{2}$ formed= $2 \times 1 + 3 = 5$ moles

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