
, it may be a Benzene.
to solve this:
we need to let the unknown hydrocarbon be

from the information given, we conclude that in the resultant 230ml gas, there are 60ml of

, 120ml of

, and 50ml of other gases, which is irrelevant to this question
we then use the formula pv=nRT to find the number of moles of water and carbon dioxide produced , which yields n(

)=0.001537722 mol , n(

)=0.0036754 mol
next, we use the same method to find the original mole number of the unknown hydrocarbon in the reactant gas.
which is n(

) = 0.00061257 mol
to find X, we find the ratio n(

) : n(

)
to find Y, we find

(

) : n(

) ( the multiplication is necessary because there are two hydrogens present in a mole of water as opposed to the presence of only one carbon per carbon dioxide)
following the previous step, we can determine that X is 6, and Y is also 6. so we get
