Question

$50\text{ml}$ of a mixture of $N{H}_3$ and $H_2$ was completely decomposed by sparking into nitrogen and hydrogen. $40\text{ml}$ of oxygen was added and the mixture was sparked again. After cooling to the room temperature, the mixture was shaken with alkaline pyrogallol solution and a contraction of $6\text{ml}$ was observed. Calculate the % by volume of $N{H}_3$ in the original mixture.

• A. 36%
• B. 54%
• C. 27%
• D. 72%

Answer 1

The correct option is D 72%

Let the volume of $N{H}_3$ be $x\text{ml}$, then the volume of $H_2$ would be $(50-x)\text{ml}$.
On sparking,
$N{H}_3\left(g\right)\to \left(1/2\right)N_2\left(g\right)+\left(3/2\right)H_2\left(g\right)$
Since $40\text{ml}$ of $O_2$ is added and sparked, it must have reacted with $H_2$ to form liquid water. Since $6\text{ml}$ contraction in volume is there with alkaline pyrogallol as it is known to absorb $O_2$ , so $34\text{ml}$ is the volume of $O_2$ used up.
$\therefore$ Total volume of $H_2$ would be $68\text{ml}.(\because 2H_2+O_2\to 2H_{2}O$)
$\therefore (50-x)+\frac{3}{2}x=68,50+\frac{x}{2}=68,x=36\text{ml}$
$\therefore$ % volume of $N{H}_3$ in the original mixture $=\frac{36}{50}\times 100=72\%$