60cm3 of a mixture of carbon(ll) oxide and hydrogen yielded 20cm3 of carbon(lV)oxide after explosion with excess of oxygen.What is the percentage of carbon(ll) oxide in the mixture

Answer ::

2CO(g) + O2(g) ---> 2CO2(g)
2vol 1vol 2vol
From the question, no concrete information is provided for oxygen, so will have to ignore it and focus on CO and CO2.
No information is provided on the actual volume of CO that reacted with the excess oxygen, but we can use the volume of CO2 produced to get that as shown below:
If 2vol of CO2 = 2vol of CO
Then, 20cm^3 of CO2 = 20cm^3 of CO.
The percentage of CO in the mixture is given by:
(Volume of CO that reacted/Volume of mixture) x 100
= (20/60) x 100
= 0.3333 x 100
= 33.33%


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