Question

In the reaction;
$CO+\frac{1}{2}{O}_2\to C{O}_2;N_2+O_2\to 2NO$
10 mL of mixture containing carbon monoxide and nitrogen required 7 mL oxygen to form $C{O}_2$ and NO, on combustion.
The volume of $N_2$ in the mixture will be:

• A. 7/2 mL
• B. 17/2 mL
• C. 4 mL
• D. 7 mL

Answer 1

The correct option is A 7/2 mL

No. of moles is directly proportional to volume

Let $x$ be volume of $N_2$ , this will be equal to volume of $O_2$ required to react with it

Let $y$ be volume of $CO$ then $\frac{y}{2}$ is the volume of $O_2$ reacting with it.

$x+y=10$ .....(1)

$x+\dfrac{y}{2} =7$

Therefore, $y=6,x=4$

volume of $N_2$ =4mL

Also , $\frac{1}{2}{N}_2+\frac{1}{2}{O}_2$ $\to$ NO

If $\frac{1}{2}$ mol $O_2$ is used by CO , $\frac{1}{2}mol{O}_2$ will be used by $N_2$

Therefore, volume of $N_2$ can be $\frac{7}{2}$ mL