Question

$10mL$ $N_2$ and $25mL$ $H_2$ at same P and T are allowed to react to give ${NH}_3$ quantitatively. Predict (i) the column of ${NH}_3$ formed, (ii) limiting reagent.

Answer 1

At same $P$ and $T$, volume $\alpha$ no. of moles.

$N_2+3H_2⟶2N{H}_3$

By stichiometry, $\frac{n{N}_2}{n{H}_2}=\frac{1}{3}⟶\left(i\right)$

As per given values, $\frac{n{N}_2}{n{H}_2}=\frac{10}{25}=\frac{1}{2.5}⟶\left(ii\right)$

$\frac{1}{3}<\frac{1}{2.5}$

$\therefore$ $\frac{n{N}_2}{n{H}_2}\left(ii\right)>\frac{n{N}_2}{n{H}_2}\left(i\right)$

$\therefore$ $H_2$ is a limiting reagent.

$\therefore$ $H_2$ reacted= $25mL$

$\therefore$ $N{H}_3$ formed= $\frac{2\times 25}{3}mL$

=$16.67mL$

Answer :- (i) The column of $N{H}_3$ formed is $16.67mL$

(ii) $H_2$ is a limiting reagent.