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.