Question

A mixture of $N{H}_3\left(g\right)$ and $N_2{H}_4\left(g\right)$ is placed in a sealed container at $300K$. The total pressure is $0.5atm$. The container is heated to $1200K$ at which time both substances decompose completely according to the equations $2N{H}_3\left(g\right)\to N_2\left(g\right)+3H_2\left(g\right)$ and $N_2{H}_4\left(g\right)\to N_2\left(g\right)+2H_2\left(g\right)$. After decomposition is complete, the total pressure at $1200K$ is found to be $4.5atm$. Find the $mole\%$ of $N_2{H}_4$ in the original mixture:

• A. $20\%$
• B. $25\%$
• C. $50\%$
• D. $75\%$

Answer 1

Answer: (D)

$75\%$

Let initial mixture contains $n_1$ and $n_2$ moles of $N{H}_3$ and $N_2{H}_4$ respectively.

Total moles of gases originally present $=n_1+n_2$

Total moles of gases after decomposition of gases $=2n_1+3n_2$

$0.5\times V=(n_1+n_2)R\times 300$

$4.5\times V=(2n_1+3n_2)R\times 1200$
$\frac{2n_1+3n_2}{n_1+n_2}=\frac{9}{4}$
$\frac{n_1}{n_2}=\frac{1}{3}$
$\frac{n_2}{n_1+n_2}\times 100=75\%$.