# The Combustion of Propane: Calculating the Mass of Oxygen Required

## The chemical equation below shows the combustion of propane (C3H8)

**C3H8 + 5O2 → 3CO2 + 4H2O**

The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required to completely react with 0.025 g C3H8?

The balanced equation for the reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of the C₃H₈ = 0.025 g

Molar mass of C₃H₈ = 44.1 g/mol

Hence, moles of C₃H₈ = 0.025 g / 44.1 g/mol = 5.67 x 10⁻⁴ mol

The stoichiometric ratio between C₃H₈ and O₂ is 1 : 5.

Hence, moles of O₂ = moles of C₃H₈ x 5

= 5.67 x 10⁻⁴ mol x 5

= 2.835 x 10⁻³ mol

Molar mass of O₂ = 32.00 g/mol

Hence, mass of O₂ = moles x molar mass

= 2.835 x 10⁻³ mol x 32.00 g/mol

= 0.09072 g

Hence, needed O₂ for the reaction is 0.09072 g.

The molar mass of which gases is given in the data?

The molar mass of oxygen gas (O2) is given in the data, which is 32.00 g/mol.