Question

1L mixture of $CO$ and $C{{O}_{2}}$ gases were passed over hot coke and the final volume of the mixture became 1200mL. Volume of gas$CO$in the initial mixture was:
(1) 800mL
(2) 200mL
(3) 400mL
(4) 600mL

Answer 1

The answer here to this question is based on the concept of high temperature reactions that uses red hot coke. Basically, the reaction is between carbon dioxide and carbon which gives carbon monoxide.

Complete step – by – step solution:
We have studied in our lower classes about the reactions involving temperature, catalysts etc., and also about the types of reactions like high temperature and low temperature, high pressure and low pressure reactions and so on.
Now, here in this above question given the reaction will be between the carbon dioxide and that of carbon as hot coke can only reduce$C{{O}_{2}}$ which is the indication of high temperature reaction.
We know that the reaction involves high temperature, that is the two gases are passed through red hot coke and data given is that the mixture initially is 1L.
The reaction taking place is,
$C{{O}_{2}}+C\to 2CO$
Here, one mole of both carbon and carbon dioxide gives two moles of carbon monoxide.
Since mixture is initially having 1L=1000mL volume and final volume is 1200mL, now we have nL of mixture gives 2nL of carbon monoxide and thus the volume of carbon dioxide will be (1-n)L…..(1)
Therefore, $\left( 1-n \right)L+\left( 2n \right)L=1200mL=1.2L$
$\Rightarrow 1+n=1.2L$
Thus, $n=0.2L$ which is volume of $C{{O}_{2}}$
Now, amount of $CO$ from equation (1) will be,
1-0.2=0.8L=800mL
Therefore the amount of $CO$initially will be 800 mL

Thus, the correct answer is option (1) 800mL.

Note: The point to be noted here is that the red hot coke does not reduce carbon monoxide and another example for such type of reaction is the formation of water gas when steam is passed over red hot coke.