Question

1 L of \[C{O_2}\] is passed over red hot coke. The volume becomes 1.4 L. The composition of the product(s) is:
A. 0.6 L $CO$
B. 0.8 L $C{O_2}$
C. 0.6 L $C{O_2}$ and 0.8 L $CO$
D. 0.8 L $C{O_2}$ and 0.6 L of $CO$

Answer 1

Hint:The reaction given in the question is a reduction reaction. We can solve this problem by simple stoichiometric calculations .Write the reaction involved in this process and solve by using stoichiometric calculations.

Complete step by step answer:
We know that this reaction is a reduction reaction and the red hot coke will reduce carbon dioxide to carbon monoxide. We can write the balanced chemical reaction for this process is as follows:
$C{O_2} + C \to 2CO$
From this balanced chemical reaction, we know that 1 mole of carbon dioxide will give 2 moles of carbon monoxide.
So, $x$ litres of $C{O_2}$ will give $2x$ litres of $CO$. In the question, it is given that 1 litre of $C{O_2}$ will react to give 1.4 litre of mixture. In the question, the product formed is not 2 litres by reacting 1 litre of $C{O_2}$ which means the total $C{O_2}$ is not reacted completely and some amount of $C{O_2}$ is left unreacted.
Out of 1 litre, $x$ litres of $C{O_2}$ is reacted and the unreacted amount will be $\left( {1 - x} \right)$ L.
The total volume of mixture is 1.4 litres which is the amount of both reacted and unreacted. That is, $\left( {1 - x} \right) + 2x = 1.4L$.
Therefore, the value of $x$ will be 0.4 L.
The volume of $CO$ formed will be$2 \times 0.4L = 0.8L$.
Volume of $C{O_2}$ left = $\left( {1 - 0.4} \right)L = 0.6L$
Therefore, the correct option will be C.


Note:The reaction given above is a reduction reaction. The red hot coke means the carbon coke under high temperature which will act as a reducing agent here and reduces the carbon dioxide to carbon monoxide.