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A 1-litre mixture of CO and CO2 is passed through red-hot charcoal. The volume becomes 1.6 litres (all volumes at the same conditions). What is the percentage of CO in the original mixture?

1. 30%
2. 60%
3. 40%
4. 50%

Correct answer: 40%

Solution

Let CO = x litres, CO2 = (1-x) litres. CO2 + C -> 2CO: all CO2 converts to 2*(1-x) litres of CO. Total volume after = x + 2(1-x) = 2 - x = 1.6 => x = 0.4 L. Percentage CO = 0.4/1 * 100 = 40%.

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