Given :
Mass of propane used up in reaction = 2.2 g
To find :
Volume of oxygen required at STP
Calculation :
The balanced chemical equation for the combustion of propane is,
C3H8 + 5O2 → 3CO2 + 4H2O
3x12+8x1 5x22.4L
(44g) (112 L)
(∵ 1 mol of ideal gas occupies 22.4 L of volume at STP)
Thus,
44 g of propane require 112 litres of oxygen at STP for complete combustion.
∴ 2.2 g of propane will require
\(\frac{112}{44}\) × 2.2 = 5.6 litres of O2 at STP for complete combustion.
∴ Volume of O2 (at STP) required to bum 2.2 g propane = 5.6 litres
