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Kinetic Theory of Gases

A sample of m...

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A sample of magnesium was burnt in air to give a mixture of $M g O$ and ${M g}_{3} N_{2}$ . The ash was dissolved in $60$ meq. of $H C l$ and the resultant solution was back-titrated with $N a O H$ . $12$ meq. of $N a O H$ was required to reach the end point. An excess of $N a O H$ was then added and the solution was distilled. The ammonia was then trapped in $10$ meq. of a second acid solution. Back-titration of this solution required $6$ meq. of the base. The percentage of magnesium burnt will be (as nearest integer) :

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Solution

Let the total number of moles of $M g$ used for $M g O$ and ${M g}_{3} N_{2}$ be $a$ and $b$ respectively.
$2 M g + O_{2} \to 2 M g O$
Before reaction $a$ $0$
After reaction $0$ $a$

$3 M g + N_{2} \to {M g}_{3} N_{2}$
Before reaction $b$ $0$
After reaction $0$ $\frac{b}{3}$

Now, $a + \frac{b}{3}$ moles of $M g O$ and ${M g}_{3} N_{2}$ are present in the mixture.

The reactions with $H C l$ are as follows:
$M g O + 2 H C l \to M g {C l}_{2} + H_{2} O$

${M g}_{3} N_{2} + 8 H C l \to 3 M g {C l}_{2} + 2 N H_{4} C l$

The solution contains a mole of $M g {C l}_{2}$ from $M g O$ and $b$ mole of $M g {C l}_{2}$ from ${M g}_{3} N_{2}$ and $\frac{2 b}{3}$ mole of $N H_{4} C l$ .

Also, moles of $H C l$ used for this purpose = $2 a + \frac{8 b}{3}$ for $M g O$ for ${M g}_{3} N_{2}$

Now, mole of $H C l$ or meq. of $H C l$ (monobasic acid) $= 60 - 12 = 48$
$∴ 2 a + \frac{8 b}{3} = 48 . . . (i)$

Further,
mole of $N H_{4} C l$ formed $=$ mole of $N H_{3}$ liberated $=$ mole of HCl used for absorbing $N H_{3}$

$∴ \frac{2 b}{3} = 4$
$⟹ b = 6$

From eq. (i), $2 a + \frac{8 \times 6}{3} = 48$
$⟹$ $a = 16$

Thus, $%$ of $M g$ used for ${M g}_{3} N_{2} = \frac{6}{(6 + 16)} \times 100 = 27.27 %$

So, the nearest integer is $27$ .

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