Question

20 ml of mixture of CO and $H_2$ were mixed with excess of $O_2$ and exploded and cooled . There was a volume contraction of 18 ml . All volume measurements corresponds to room temperature $\left({27}^{0}C\right)$ and one atmosphere pressure . Determine the volume ratio $V_1:V_2$ of CO and $H_2$ in the original mixture.

• A. 1 : 2
• B. 3 : 2
• C. 2 : 3
• D. 4 : 1

Answer 1

The correct option is B 3 : 2

Since, the reaction is $2CO+O_2\to 2C{O}_2$ and again another reaction is $2H_2+O_2\to 2H_{2}O$.
Now we let that the volume of $CO$ is x so volume of $H_2$ is (20-x) when excess $O_2$ is given in this reaction then the entire reaction is completed thus we get the initial volume to be 3x/2 mL.
Now, for the second reaction the initial volume will be 30 - 3x/2 and hence the total volume of reaction will be 3x/2 +30 - 3x/2 = 30. And the volume of the product is x mL.
So we know that the volume of contraction is volume of reaction - volume of product.
18 = 30 -x.
So, x = 12.
Hence, volume of $CO$ is 12 and the volume of $H_2$ is 20 - 12 = 8.
So, ratio of volume of $CO$ to $H_2$ is 12/8 = 3:2. {Option B}