11Some one pls try...
15 ml of a gaseous hydrocarbon was required for combustion in 357 ml of air ( 21% of O2 by volume ) . If the gaseous products occupied 327 ml (at NTP) what is the formula of the hydrocarbon )...
Ans c3h8
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8 Answers
29sry for late reply i was busy studying for boards..so didnt come online...
Find the volume of O2 used = 75ml..
Volume of CxHy = 15ml
that means the ratio of moles used of CxHy : 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 CxHy ...3 moles of CO2 will be produced...
29CxHy + (2x+y/2)/2 O2 → xCO2 + y/2 H2O
Find the volume of O2 used = 75ml..
Volume of CxHy = 15ml
that means 4x + y = 20
so possible values of x and y
x y
3 8
15 ml of CxHy 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
11hey 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 ...
11hey no prob.. dun havta be sry.. thanx a lot... i was trying this for over half an hr :P
1another way
eqn is CmHn + (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