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(C) According to the balanced chemical equation: $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$.From the stoichiometry,$2 \ volumes$ of $NO$ require

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$250$

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(C) According to the balanced chemical equation: $2 NO_{(g)} + O_{2(g)} ightarrow 2 NO_{2(g)}$.From the stoichiometry,$2 \ volumes$ of $NO$ require $1 \ volume$ of $O_2$.Therefore,$100 \ cm^3$ of $NO$ requires $\frac{1}{2} 	imes 100 \ cm^3 = 50 \ cm^3$ of pure $O_2$.Since air contains $20 \%$ $O_2$ by volume,the volume of air required is calculated as: $	ext{Volume of air} = \frac{	ext{Volume of } O_2}{0.20} = \frac{50 \ cm^3}{0.20} = 250 \ cm^3$.

The volume of air needed for complete conversion of $100 \ cm^3$ of $NO$ to $NO_2$ will be (assuming $20 \%$ $O_2$ by volume in air).
$2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ (in $cm^3$)

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The volume of air needed for complete conversion of $100 \ cm^3$ of $NO$ to $NO_2$ will be (assuming $20 \%$ $O_2$ by volume in air).$2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ (in $cm^3$)