Question

$4N{H}_3\left(\text{g}\right)+5O_2\left(\text{g}\right)\to 4NO\left(\text{g}\right)+6H_{2}O\left(\text{l}\right)$. If the total volume of the reactants is 27 L, calculate the volume of nitric oxide produced in the reaction.

• A. 12 L
• B. 8 L
• C. 14 L
• D. 9 L

Answer 1

The correct option is A 12 L

$\underset{4vol.}{4N{H}_3}+\underset{5vol.}{5O_2\text{(g)}}\to \underset{4\text{vol.}}{4NO\left(\text{g}\right)}+6H_{2}O\left(\text{l}\right)$
Gay-Lussac’s law of combining volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure conditions remain constant. Let the volume of $N{H}_3$ be $4x$ and that of $O_2$ be $5x$ (because they react in the ratio of $4:5$). Volume of reactants given is 27 L; hence $9x=27\text{L}$
So, $x=3\text{L}$
Volume of Nitric oxide produced =$4x=12\text{L}$