In a eudiometer, 20 ml of a gaseous hydrocarbon A is mixed with 150 ml of oxygen gas. The combustion of A is initiated by an electric spark. After cooling and condensing the vapour, the volume of the gaseous mixture is 110 ml, of which 80 ml causes turbidity of limewater. The remaining volume of the gas makes a glowing splint to catch fire.

Question

In a eudiometer, 20 ml of a gaseous hydrocarbon A is mixed with 150 ml of oxygen gas. The combustion of A is initiated by an electric spark. After cooling and condensing the vapour, the volume of the gaseous mixture is 110 ml, of which 80 ml causes turbidity of limewater. The remaining volume of the gas makes a glowing splint to catch fire. 
determine the formula of A

DrBob222 · Accepted Answer

There may be an easier (better?) way to do this. If so I would like to see it. Here is what I did. The fact that the remaining gas lighted a glowing splint means the remaining gas is O2 which merans the hydrocarbon was the limiting reagent.
...........CxHy + O2 ==> CO2 + H2O
I.........20 mL....150 mL..0...........0
C.......-20...........-x...................
After combustion total volume is 110 mL.That will be the CO2 + the unused O2 gas.
Volume CO2 from the turbidity test is 80
Volume of O2 left uncombusted = 110-80 = 30
Volume of O2 used = 150-30 = 120 mL
Ratio hydrocarbon to O2 is 1 to 6 (that's 20 mL HC used 120 mL O2). Note: I never used a eudiometer (EVER) in my career so I'm assuming the volume readings are just that and I've made no corrections for the condensation of the water vapor. and how that might affect the volume reading. I then wrote balanced equations for the first four saturated hydrocarbons (alkanes) and found no 1:6 ratios. Then I tried the first few alkenes. You should balance the first 4 to confirm that. C4H8 did it; i.e.,
C4H8 + 6O2 ==> 4CO2 + 4H2O
I think the unknown HC is C4H8.