Chem Eudiometry question

anpanman SG

15cm3 of a gaseous hydrocarbon, W, was burnt with 95 cm3 of oxygen. After cooling to r.t.p, the residual gaseuous products occupied a volume of 80 cm3. On adding aq. NaOH, the volume decreased to 50cm3. Find the molecular formula of the hydrocarbon, W

CxHy (g) + (x+y/4) O2 (g) -> x CO2 (g) + (y/2) H2O (l)

Right, firstly I find the vol. of reacted O2 which I derived at an answer of 45cm3 and also vol. of CO2 produced ( 80 - 50 = 30cm3)

Then I have top compare the mole ratios of CxHy and CO2 (1:2) and also CxHy and O2 (1:3). My final answer is C2H... which I find really strange. Please help if you can. :)

UltimaOnline SG

Vol of O2 in excess = 50cm3

Vol of O2 used = 95 - 50 = 45cm3

Vol of CO2 produced = 80 - 50 = 30cm3

Since 15cm3 of CxHy produced 30cm3 of CO2, x = 2

Since 15cm3 of CxHy used 45cm3 of O2, 15 (x + y/4) = 45

2 + y/4 = 45 / 15

y/4 = 1

y = 4

Hence hydrocarbon is ethane, C2H4.