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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 COB.0.8 L $C{O_2}$ C.0.6 L $C{O_2}$ and 0.8 L COD.0.8 L $C{O_2}$ and 0.6 L CO

Last updated date: 15th Sep 2024
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Hint: Red – Hot coke can be treated as a form of pure carbon when it is used in reactions at very temperatures. When we carry out an experiment carbon dioxide is made to react with red hot coke, we can consider it to be a reaction between carbon dioxide and carbon. This reaction results in the formation of carbon monoxide.

Before we move forward with the solution of the given question, let us first understand some important basic concepts.
The chemical equation for this reaction can be given as:
$C{O_2} + C \to 2CO$
Thus, we can observe one mole of both carbon dioxide and red – hot coke result in the formation of 2 moles of carbon monoxide. This interpretation can be translated in terms of volume as well. 1 L of $C{O_2}$ would give 2 L of CO in this reaction.
If we consider the case where ‘n’ L of $C{O_2}$ reacts to form ‘2n’ L of CO,
Then the volume of $C{O_2}$ left would be = (1 - n) L
We have been told that total volume of the product is 1.4 L
Hence, we can say that:
$\left( {1 - n} \right)L + \left( {2n} \right)L = 1.4L$
$1 + n = 1.4$
n = 0.4 L
Since the volume of $C{O_2}$ used is 0.4 L, we can say that the volume of CO formed is (2) (0.4) = 0.8 L
Hence, the composition of the product(s) is: 0.6 L $C{O_2}$ and 0.8 L CO
Hence, Option A is the correct option

Note: Another important reaction involving red hot coke is the formation of water gas. When steam is passed over red - hot coke, it results in the formation of water gas. Water gas is a mixture of carbon monoxide and hydrogen gas.