Question

The mixture of $C{{O}_{2}}$ and CO is passed over red-hot graphite when 1 mole of mixture changes to 33.6 l (converted it to STP). Hence, mole fraction of $C{{O}_{2}}$ , in the mixture is
A. 0.50
B. 0.33
C. 0.66
D. 0.25

Answer 1

Mole fraction of x in the mixture of gases x and y is as follows.
Mole fraction of the gas x = $\dfrac{\text{number of moles of x}}{\text{number of moles x + number of moles of y}}$
One mole of any gas at STP occupies 22.4 lit.

Complete answer:
- In the question it is asked to calculate the mole fraction of the $C{{O}_{2}}$ by using the given data in the question.
- In the question it is given that carbon dioxide gas passes through the red-hot coke.
- The reaction between the red-hot coke and the carbon dioxide is as follows.
$C{{O}_{2}}+C\to 2CO$
- In the above chemical reaction one mole of carbon dioxide is going to react with carbon (graphite) and forms 2 moles of carbon monoxide gas as the product.
- Assume ‘x’ liters of carbon dioxide is going to react with coke to form ‘2x’ liters of the CO.
- Then the volume of the carbon dioxide left is = 22.4 -x
- The total volume given in the question is 33.6 liters
- Therefore,
(22.44-x) + 2x = 33.6
x = 11.2 liters.
- Therefore, the volume of the CO = (2) (11.2) = 22.4 liters.
- The volume of the $C{{O}_{2}}$ = 22.4 – 11.2 = 11.2 liters.
- Number of moles of carbon dioxide in 11.2 liters is = 0.5
- Number of moles of carbon monoxide in 22.4 liters is = 1
- Therefore, the mole fraction of the carbon dioxide = $\dfrac{0.5}{1+0.5}=0.33$

So, the correct option is B.

Note:
We should calculate the number of moles of each gas which are going to be involved in the chemical reaction to find the mole fraction of a particular gas in the mixture of the different gases at STP.