On repeated sparking, $10$ mL of a mixture of carbon monoxide and nitrogen required $7$ mL of oxygen for combustion. What was the volume of nitrogen? (All volumes are measured under identical conditions)

\frac{7}{2}

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4

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\frac{17}{2}

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Solution

The correct option is A $4$ mL
Total volume of the mixture containing $C O$ and $N_{2}$ is $10$ mL.
Let volume of $C O =$ $x$ mL, volume of $N_{2}$ $=$ $(10 - x)$ mL.
The reactions are shown below:
$C O_{2} + \frac{1}{2} O_{2} \to C O_{2}$
$N_{2} + O_{2} \to 2 N O$
Volume of $C O_{2}$ required by $O_{2}$ $= \frac{x}{2} m L$ and volume of $N_{2}$ required by $O_{2}$ $= (10 - x)$ mL
$\frac{x}{2} + (10 - x) = 7, x = 6$ mL
Hence, volume of $N_{2} = 10 - 6 = 4$ mL

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On repeated sparking, 10 mL of a mixture of carbon monoxide and nitrogen required 7 mL of oxygen for combustion. What was the volume of nitrogen? (All volumes are measured under identical conditions)

On repeated sparking, $10$ mL of a mixture of carbon monoxide and nitrogen required $7$ mL of oxygen for combustion. What was the volume of nitrogen? (All volumes are measured under identical conditions)