Question

Consider the following equation, which represents the combustion of ammonia.
$N{H}_3\left(g\right)+3O_2\left(g\right)\to 2N_2\left(g\right)+6H_{2}Og)$
a) What volume, in liters, of $N_2\left(g\right)$ is formed when $125L$ of $N{H}_3\left(g\right)$ is burned? Assume that both gases are measured under the same conditions.
b) What volume, in liters, of $O_2\left(g\right)$ is required to form $36L$ of $H_{2}O\left(g\right)$? Assume that both gases are measured under the same conditions.

Answer 1

$4N{H}_3\left(g\right)+3O_2\left(g\right)⟶2N_2\left(g\right)+6H_{2}O\left(g\right)$

(a) $4$ moles of $N{H}_3$ will produce $2$ moles of $N_2$

Hence $125L$ of $N{H}_3$ will produce

$\Rightarrow \frac{125\times 2}{4}=62.5L$ of $N_2$

(b) $6$ moles of $H_{2}O$ is produced from $3$ moles of $O_2$

Hence $36L$ of $H_{2}O$ is produced from

$\Rightarrow \frac{36\times 3}{6}=18L$ of $O_2$ .