Question

40 mL

of ammonia gas taken in an eudiometer tube was subjected to sparks till the volume did not change any further. It was observed that the volume increased by

40 mL

.

40 mL

of oxygen was then mixed and the mixture was further exploded. The gases remained accounted to

30 mL

. Deduce the formula of ammonia.

Accepted Answer

Let the formula of ammonia be $N_{x} H_{y}$ . The reaction in eudiometer is $40 mL$

The volume of nitrogen and hydrogen formed will be $\frac{x}{2}$ and $\frac{y}{2}$ times of ammonia, respectively.

Gay-Lussac law states that whenever gases react together, the volume of reacting gases as well as the product bear a simple whole number ratio.

According to the statement, ratio of

$N_{x} H_{y} : N_{2} : H_{2} = 2 : x : y$

Thus, $40 mL$ of $N_{x} H_{y}$ gives $(20 mL) x and (20 mL) y$

Given, the total volume doubles.

Thus, final volume $40 mL$

Therefore,

$(20 mL) x + (20 mL) y = 80 mL 20 mL (x + y) = 80 mL \Rightarrow x + y = 4$

The second reaction is:

Hydrogen is more reactive than nitrogen and reacts with oxygen in $2 : 1$ ratio.

$N_{x} H_{y} \to \frac{x}{2} N_{2} + \frac{y}{2} H_{2} initial \to 40 0 0 final 0 (20 mL) x (20 mL) y \frac{x}{2} N_{2} + \frac{y}{2} H_{2} + \frac{y}{4} O_{2 (g)} \to \frac{y}{2} H_{2} O_{(l)} initial (20 mL) x (20 mL) y 40 mL final (20 mL) x 0 (40 - 10 y)$

Thus, $(20 mL) y$ of hydrogen reacts with $(10 mL) y of O_{2}$

Thus, amount of oxygen un-reacted $= (40 mL - 10 ymL)$

Thus, total volume is due to remaining oxygen gas and un-reacted nitrogen gas $= 30 mL$

Thus,

$(20 mL) x + (40 mL - 10 ymL) = 30 mL 20 x + 40 - 10 y = 30 10 = 10 y - 20 x y - 2 x = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Solving equations $(1)$ and $(2)$ simultaneously, we get

$x = 1 and y = 3$

Hence, the molecular formula of ammonia is ${NH}_{3}$ .

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