Question

For the reaction given below, what volume of $N{O}_2\left(g\right)$ is produced from the combustion of $50\text{g}$ of $N{H}_3\left(g\right)$, assuming the reaction takes place at standard temperature and pressure?
$4N{H}_3\left(g\right)+7O_2\left(g\right)\to 4N{O}_2\left(g\right)+6H_{2}O\left(l\right)$

• A. 132 L
• B. 67 L
• C. 198 L
• D. 33 L

Answer 1

The correct option is B 67 L

For reaction,
$4N{H}_3\left(g\right)+7O_2\left(g\right)\to 4N{O}_2\left(g\right)+6H_{2}O\left(l\right)$

$4$ mol of $N{H}_3$ produce $4$ mol of $N{O}_2$
So, $4\times 17\text{g}N{H}_3$ produces $4\times 46\text{g}$ of $N{O}_2$
$50\text{g}N{H}_3$ produces $135.3\text{g}$ of $N{O}_2$
$\therefore$ Volume of $N{O}_2$ produced =$\frac{135.3\times 22.7}{46}\simeq 67L$