Question

Total volume occupied by a mixture containing 1 equivalent each of $N_2$, $H_2$ and $N{H}_3$ based on the following equation at STP is
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is:

Answer 1

According to the equation, the oxidation number of $N$ is change from 0 to -3.
So the equivalent mass of $N_2$ = $\frac{Molarmass}{6}$
So volume occupied by 1 equivalent of $N_2$ at STP = $\frac{22.4}{6}L$
Similarly, equivalent mass of $N{H}_3$ = $\frac{Molarmass}{3}$
So volume occupied by 1 equivalent of $N{H}_3$ at STP = $\frac{22.4}{3}L$
Again, equivalent mass of $H_2$ = $\frac{Molarmass}{2}$
So volume occupied by 1 equivalent of $N{H}_3$ at STP = $\frac{22.4}{2}L$
Thus total volume = $(\frac{22.4}{6}+\frac{22.4}{3}+\frac{22.4}{2})L$
$=\frac{(22.4+2\times 22.4+3\times 22.4)}{6}L=22.4L$