Question

$50\text{kg}$ of $N_2\text{(g)}$ and $10\text{kg}$ of $H_2$ are mixed to produce $N{H}_3\text{(g)}$. Calculate the amount of $N{H}_3\text{(g)}$ formed.
(assume the reaction goes to completion)
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

• A. $56.67\text{kg}$
• B. $51.87\text{kg}$
• C. $50.71\text{kg}$
• D. $47.67\text{kg}$

Answer 1

The correct option is A $56.67\text{kg}$

$N_2+3H_2\to 2N{H}_3$
As per stoichiometry,
$28\text{g}$ of nitrogen reacts with $6\text{g}$ of hydrogen.
So, $50\text{kg}$ will react with $\frac{6}{28}\times 50\text{kg}{H}_2=10.7\text{kg}$
But, we have only $10\text{kg}$of $H_2$
So, $H_2$ is the limiting reagent.
$6\text{g}{H}_2$ gives $34\text{g}N{H}_3$
$10\text{kg}{H}_2$ will give $\frac{34}{6}\times 10=56.67\text{kg}$ of $N{H}_3$