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Question

A sample of gaseous hydrocarbon occupying 1.12 litres at NTP when completely burnt in air produced 2.2 g CO2 and 1.8 g H2O. The amount of volume (in L) of O2 at NTP required for its combustion is (as the nearest integer) :

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Solution

Let the formula of hydrocarbon be CaHb.
CaHb+[a+(b/4)]O2aCO2+(b/2)H2O
22.4 litres of CaHb gives on combustion = 44a g CO2
1.12 litres of CaHb will give on combustion = 44a×1.1222.4=2.2ag.
2.2a=2.2a=1
Similarly,
22.4 litres of CaHb gives = 18×(b/2)gH2O
1.12 litres of CaHb gives = 18×b2×1.1222.4gH2O=18×b40gH2O
1.8=(1840)×bb=4
Hydrocarbon is CH4
Thus, mass of hydrocarbon taken = (16×1.12)/22.4=0.8g
Also, O2 used for combustion of 16 g CH4 = (a+b)4 =2 mol
= 2×22.4litre
O2 used for combustion of 0.8 g. CH4=2×22.4×0.816=2.24litre
Nearest integer is 2 L.

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