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Question
Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H5).
(Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm3 at STP.)
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Solution
\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]
Molar weight of propane = 12 × 3 + 8 = 44
Volume of 5O2 = 5 × 22.4 = 112 litres
44 g of propane requires = 112 litres of oxygen
1 g of propane requires = `112/44` litres
8.8 g of propane requires = `112/44 xx 8.8`
= 112 × 0.2
= 22.4 litres
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