Percentage by volume of CO in the original mixture

Luke works as a Melbourne chemistry tutor at high school and University levels for one-to-one and student-organised group tutoring in person or online since 1997.

A 50 mL mixture of CO2 and CO gases was mixed with 20 mL O2 gas and exploded. The resulting gas mixture was bubbled through concentrated LiOH and 14 mL of gas remained. If all gases were measured at the same temperature and pressure, what was the percentage by volume of CO in the original mixture?

The aim of this question:
is to test students’ understanding the concepts of the limiting and excess reactants and their use in calculation.

My Solutions
To solve this question, we have first to reason what has happened during the mixing and explosion.   When CO, CO2 and O2 are mixed and exploded, CO2 does not react with O2 or CO and the only reaction that occurs is 2CO + O2 -> 2CO2. When the products pass through LiOH, all CO2 are absorbed. The only CO or O2 can passes through LiOH. The gas remained must be either CO or O2.

To solve this question, we first assume that the 20 ml of O2 is the limiting reactant. According the chemical equation above, there must be at least 40 ml of CO that reacts with O2, because of the molar ratio. Because O2 is the limiting factor, there must be 40 to 50 ml of CO in the original mixture. The gas remaining would be then smaller than 10 ml, which is inconsistent with the remaining gas being 14ml. Therefore O2 is not the limiting reactant.

We can thus conclude that CO is the limiting reactant. The 14 ml gas remained must be O2 only. The O2 reacted is 6 ml (20 ml of reactant O2 – 14 ml of O2 remained).   There must be 12 ml of CO in the initial mixture. The percentage by volume of CO is the original mixture is 12ml /50 ml *100% = 24%.

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