How many litres of oxygen at STP are required to burn completely 2.2 g of propane, C3H8?

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