Question

$50kg$ of $N_2\left(g\right)$ and $10.0kg$ of $H_2\left(g\right)$ are mixed to produce ${NH}_3\left(g\right)$. Calculate the ${NH}_3\left(g\right)$ formed. Identify the limiting reagent in the production of ${NH}_3$ in this situation.

Answer 1

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

$50kg$ $10kg$

3 mole $H_2$ reacts with 1 mol $N_2$

6 gm $H_2$ reacts with 28 gm $N_2$

1 gm $H_2$ reacts with $\frac{28}{6}g$ of $N_2$

1000 gm $H_2\to$ will react with $\frac{28}{6}\times 1000gm{N}_2$

$=4666.6gm{N}_2$

1 kg $H_2$ will react with $=4.6kg{N}_2$

10 kg $H_2$ will react with $=46.6kg$ of $N_2$

Therefore, $H_2$ is limiting reagent.

Now, 3 mole (6gm) $H_2$ produce 34gm ${NH}_3$

1 gm (6gm) $H_2$ produce $\frac{34}{6}$ gm ${NH}_3$

1 kg (6gm) $H_2$ produce $\frac{34}{6}$ gm ${NH}_3$

10 kg (6gm) $H_2$ produce $\frac{34}{6}\times 10kg$ ${NH}_3$

$=56.6kg$ of ${NH}_3$