Question

In a reaction $N_2+3H_2\to 2N{H}_3$, 40 ml of each $N_2$ and $H_2$ are taken to react together so that $25\%$ yield of $N{H}_3$ was obtained. The volume of $N_2$ present in the container is?

• A. 13.33 ml
• B. 36.67 ml
• C. 26.67 ml
• D. 3.33 ml

Answer 1

The correct option is B 36.67 ml

$N_2+3H_2\to 2N{H}_3$
40 ml 40 ml $\frac{80}{3}ml$
$H_2$ is the limiting reagent
Considering the yield, $V_{N{H}_3}\text{produced}=\frac{80}{3}\times \frac{25}{100}=\frac{20}{3}ml$
$V_{N_2}\text{left}=40-\frac{10}{3}=\frac{110}{3}$