Question

60ml of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38ml of $N_2$ was formed, calculate the volume of each gas in the mixture.

• A. $NO=44ml;N_{2}O=16ml$
• B. $NO=45ml;N_{2}O=20ml$
• C. $NO=34ml;N_{2}O=22ml$
• D. $NO=20ml;N_{2}O=26ml$

Answer 1

Answer: (A)

$NO=44ml;N_{2}O=16ml$

Let the given mixture contains $2x$ ml of $NO$. The volume of $N_{2}O$ present is $60-2x$ ml.

The balanced chemical equations are as shown below.

$\underset{2x}{2NO}+2H_2\to \underset{x}{N_2}+2H_{2}O$

$\underset{60-2x}{N_{2}O}+H_2\to \underset{60-2x}{N_2}+H_{2}O$

The volume of nitrogen collected is 38 ml.

Hence, $60-2x+x=38$

$x=22ml$

Hence, the volumes of NO and $N_{2}O$ present in the originl mixture are $2x=2\left(22\right)=44ml$ and $60-2\left(22\right)=16ml$ respectively.

So, the correct option is $A$