Question

One litre of mixture of $CO$ and $C{O_2}$ is passed through red hot charcoal in a tube. The new volume becomes $1.4$ litres. Find out the percentage composition of original mixture by volume where all measurements are made at the same Pressure and Temperature.

Answer 1

In the given compounds, carbon monoxide does not react with carbon. So, only carbon dioxide reacts with carbon in charcoal to form carbon monoxide. Find the amount of carbon dioxide reacted to form charcoal and we can find the percentages accordingly.

Complete answer:
In the given one litre mixture of carbon monoxide and carbon dioxide, only carbon dioxide reacts with carbon to produce carbon monoxide. We write this reaction as follows.
$C{O_2} + C \to 2CO$
Now, let us assume the volume of $C{O_2}$ to be $X$ , then the volume of $CO$ would be $1 - X$
After the reaction has occurred, the entire carbon dioxide is converted to carbon monoxide so now the volume of carbon monoxide would be $2X$
Now, we write the total volume as $1 - X + 2X = 1.4$
By solving we get $1 + X = 1.4$ and the value of $X$ calculated would be $0.4$
Which means the volume of carbon dioxide is $0.4$ and the percentage can be calculated as follows
$\dfrac{{0.4}}{1} \times 100\% = 40\%$
The volume of carbon monoxide is $1 - X = 1 - 0.4 = 0.6$ and we can calculate the percentage as
$\dfrac{{0.6}}{1} \times 100\% = 60\%$

**Therefore the percentage composition of carbon dioxide and carbon monoxide are $40\%$ and $60\%$ respectively.

Note:**
In order to calculate the percentage composition, we find out the volume of both carbon dioxide and carbon monoxide in the given mixture. To do that, we calculate the amount of carbon dioxide that reacts with carbon to form carbon monoxide and since carbon monoxide does not react with carbon, the complete volume at the end of the reaction would only be carbon monoxide.