A magnesium ribbon, when burnt in air, left an ash containing $M g O$ and $M g_{3} N_{2}$ . The ash was found to consume 0.6 mole of HCl, when it was taken in solution, according to the following reaction:
$M g_{3} N_{2}$ + $8 H C l$ $\to$ $3 M g C l_{2}$ + $2 N H_{4} C l$
The solution obtained was treated with excess of $N a O H$ , when $0.1 m o l e$ of $N H_{3}$ was evolved. The mass (in $g$ ) of magnesium burnt is

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Solution

According to given reaction
$M g_{3} N_{2}$ + $8 H C l$ $\to$ $3 M g C l_{2}$ + $2 N H_{4} C l$
8 mole of $H C l$ given ⇒ $2 m o l e s N H_{4} C l$
$0.6 m o l e H C l$ give ⇒ $\frac{0.6 \times 2)}{8}$
⇒ $0.15 m o l N H_{4} C l$
$N H_{4} C l + N a O H$ \rightarrow $N H_{3} + N a C l + H_{2} O$
When excess $N a O H$ used, only $0.1 m o l e N H_{3}$ evolved therefore, $0.1 m o l e$ $N H_{4} C l$ react with $N a O H$
For $0.1 m o l N H 4 C l$ , the mole of $H C l = \frac{8 \times 0.1}{2}$
Moles of $M g O$ react with $H C l = 0.4 - 0.1 = 0.25 m o l e s$
$M g + \frac{1}{2} O_{2}$ $\to$ $M g O$
According to above reaction:
Moles of $M g$ = Moles of $M g O$ (molar mass of $M g = 24 g m o l^{- 1}$ )
So, mass of $M g = 0.25 m o l e s \times 24 g m o l^{- 1}$ = $6 g$

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When magnesium ribbon is burnt in air, then identify the type of reaction and colour of ash formed.

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