$450 c m^{3}$ of nitrogen monoxide and $200 c m^{3}$ of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.
$2 N O + O_{2} \to 2 N O_{2}$

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Solution

$2 N O (g) 2 V + O_{2} (g) 1 V ⟶ 2 N O_{2} (g) 2 V$

From equation, 1 V of $O_{2}$ reacts with = 2 V of $N O$

$200 c m^{3}$ oxygen will react with $= 200 \times 2 = 400 c m^{3}$ of $N O$

hence, remaining $N O$ is $450 - 400 = 50 c m^{3}$

$N O_{2}$ produced $= 400 c m^{3}$ because 1 V oxygen gives 2 V of $N O_{2}$ .

Total mixture $= 400 + 50 = 450 c m^{3}$

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