Eudiometry

sry for late reply i was busy studying for boards..so didnt come online...
Find the volume of O₂ used = 75ml..
Volume of C_xH_y = 15ml
that means the ratio of moles used of C_xH_y : O₂ = 1 : 5..from here i got the equation...
then 15 ml of hydrocarbon producing 45 ml (sry for the typo...edited..) of CO₂ ...by assuming that x = 3..for 1 mole of C_xH_y ...3 moles of CO₂ will be produced...

Some one pls try...

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

hey but how did ya get 4x+2y =20 int eh first step... and then 15 ml of hydrocarbon producing 30 ml of C0₂ ... but ur ans is spot on and thanx a lot ...

another way
eqn is C_mH_n + (2m+n/2)/2O₂ → mCO₂ + nH₂O
vol of oxygen used= 0.21 x 357(mL) = 75 (mL)
vol of remaining reactants left = 357-75 = 282 (mL)=(2m+n/2)/2
so vol occupied by CO₂=327-282 = 45 (mL) = m
as equal no of moles of gases occupy same volume,
(4m+n)/4=75 and m=45, so n=120 so m : n :: 3 : 8
so compound is C₃H₈

thanx fibonacci...