How many grams of propane can be burned by 24 grams of O2?

Answer 1

For complete proper combustion of Propane:

C3H8 + 5O2 = 3CO2 + 4H2O

The relative atomic weights of a molecule of Propane and Oxygen are:

Propane: 3 × C + 8 × H = 3 × 12 + 8 × 1 = 44

Oxygen: 2 × O = 2 × 16 = 32

Thus a molecule of propane is 44/32 = 1⅜ times heavier than a molecule of oxygen; and the same amount (number of molecules) of propane as 24 g of oxygen is 24g × 1⅜ = 33g

Each propane molecule takes 5 oxygen molecules, thus:

33 g ÷ 5 = 6 3/5 g = 6.6 g

If the combustion produces the poisonous carbon monoxide instead of carbon dioxide:

2C3H8 + 7O2 = 6CO + 8H2O

→ propane = 33g × 2/7 = 9 3/7 g ≈ 9.4 g

A complete answer is thus between 6 3/5 g (6.6g) and 9 3/7 g (9.4 g) depending upon how much carbon monoxide relative to carbon dioxide is produced by the burning - the safe amount is 6.6 g.