Let the formula of hydrocarbon be CaHb.
CaHb+[a+(b/4)]O2→aCO2+(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=2.2∴a=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)×b∴b=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.