Question

One liter of a mixture of carbon monoxide and carbon dioxide is passed through a tube containing red hot charcoal. The volume now becomes $1.6$ liter. Volumes are measured under similar conditions. Find the composition of the gaseous mixture.

Answer 1

We have given the mixture of carbon monoxide $CO$ and carbon dioxide $C{O_2}$ which is passed through a tube containing red hot charcoal which increases the volume of the total mixture. Charcoal is nothing but the lightweight black carbon residue which is obtained by strongly heating wood.

Complete step by step answer:
First, we will understand the question properly and discuss what we have given and what we need to calculate. So, we have given the mixture of carbon monoxide $CO$ and carbon dioxide $C{O_2}$ and the volume of the given mixture is one liter. It is given that the mixture was passed through a tube which results in an increase in volume to $1.6$ liter. So, now using some basic considerations we will calculate the composition of the gaseous mixture.
We know that when carbon dioxide reacts with red hot charcoal gives carbon monoxide. The following reaction is given $C + C{O_2} \to 2CO$ . Now we have given the initial volume $1L$ and the final volume of the mixture $1.6L$ . So, if in the reaction $x\;L$ of carbon dioxide $C{O_2}$ is used from the total initial volume which will form $2x\;L$ of carbon monoxide $CO$ . So, the total final volume of a mixture will be given as the sum of the volumes of carbon monoxide $CO$ and carbon dioxide $C{O_2}$ . Hence, $(1 - x) + 2x = 1.6,x = 0.6\;L$
Now we got the used volume as $x = 0.6L$ which is the volume of $CO$ and $(1 - 0.6) = 0.4L$ is the volume of $C{O_2}$ .

Note:
It is observed that the final volume of a mixture is increased which is due to the formation of extra carbon monoxide. Extra carbon monoxide was formed when the mixture of carbon dioxide and carbon monoxide was passed through red hot charcoal.