450 cm³ of nitogen monoxide and 200 cm³ of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.
$\underset{2 vol .}{\underset{Nitrogen monoxide}{2 NO (g)}} + \underset{1 vol .}{\underset{Oxygen}{O_{2} (g)}} \to \underset{2 vol .}{\underset{Nitrogen dioxide}{2 {NO}_{2} (g)}}$

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Solution

Let us consider the following reaction :
$\underset{2 vol .}{\underset{Nitrogen monoxide}{2 NO (g)}} + \underset{1 vol .}{\underset{Oxygen}{O_{2} (g)}} \to \underset{2 vol .}{\underset{Nitrogen dioxide}{2 {NO}_{2} (g)}}$
From the above equation, 1 vol of oxygen reacts with 2 vol. of nitrogen monoxide.
So, 200 cm³ of oxygen will react with = 200 x 2 = 400 cm³ $NO$
Unused $NO$ nitrogen monoxide = (450-400) cm³ = 50 cm³

Also, 1 vol of oxygen $O_{2}$ is needed to produce with 2 vol. of nitrogen dioxide.
So, 200 cm³ of oxygen will produce = 2 x 200 cm³ = 400 cm³ of nitrogen dioxide ${NO}_{2}$
Total composition of the resulting mixture = (400 + 50) cm³
= 450 cm³

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450 cm3 of nitogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.2NO(g)Nitrogenmonoxide2vol.+O2(g)Oxygen1vol.→2NO2(g)Nitrogendioxide2vol.