15ml of a gaseous hydrocarbon was required for complete combustion in 357ml of air (21% of oxygen by volume ) and the gaseous products occupied 327ml (all volumes being measured at NTP) . what is the formula of hydrocarbon?

C_xH_y + (2x+y/2)/2 O₂ → xCO₂ + y/2 H₂O
Find the volume of O₂ used = 75ml..
Volume of C_xH_y = 15ml
that means 4x + y = 20
so possible values of x and y
x y
3 8
15 ml of C_xH_y will produce 45 ml of CO₂
volume of product = volume of CO₂ + volume of left over air
327 = volume of CO₂ + 357 - 75
volume of CO₂ = 45ml
that means x = 3..so the case we took is ryt..hence the answer is C₃H₈

PS : The trick in the problem is that..water will not be in gaseous form at NTP