$1\ L$ of a mixture of $CO$ and $CO_{2}$ becomes $1.6\ L$ when passes through a red-hot charcoal. The $CO_{2}$ in the mixture is:

When the mixture is passed over red-hot charcoal, any CO2 present reacts to form CO based on the reaction CO2 plus C yielding 2CO. Assuming the initial volume of CO2 is x liters, the original CO volume is (1 - x) liters, and the final total volume of CO becomes (1 - x + 2x) liters. Setting the final volume expression equal to 1.6 liters gives 1 + x = 1.6, which solves to x = 0.6 liters. The percentage of CO2 in the original mixture is therefore 0.6 divided by 1, resulting in 60 percent.

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