$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (l)$
If the total volume of the reactants is 27 L, calculate the volume of nitric oxide produced in the reaction.

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Solution

$4 N H_{3} (g) 4 vol. + 5 O_{2} (g) 5 vol. \to 4 N O (g) 4 vol. + 6 H_{2} O (l)$
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 $4 x$ and that of $O_{2}$ be $5 x$ (Because they react in the ratio of $4 : 5$ )
Volume of reactants given is $27 L$ :
Hence, $9 x = 27 L$
So, $x = 3 L$ .
Volume of nitric oxide produced $= 4 x = 12 L$ .

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