The combustion of propane may be described by the chemical equation C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g) are needed to completely burn 86.6 g C3H8(g)?

Answer 1

This problem is best broken down into three steps:

1) Convert grams of propane (C₃H₈) to moles of propane using conversion factors:

The molar mass of C₃H₈ is 3(12.01 g) + 8(1.008 g) = 44.09 g, so:

86.6 g C₃H₈ × (1 mole C₃H₈/44.09 g) = 1.964 moles C₃H₈

2) Convert moles of propane to moles of oxygen (O₂) based on the stoichiometry of the balanced equation:

There are 5 moles of O₂ for every 1 mole of propane, so:

1.964 moles C₃H₈ × (5 moles O₂/1 mole C₃H₈) = 9.82 moles O₂

3) Convert moles of oxygen to grams of oxygen using conversion factors:

The molar mass of O₂ is 2(16.00 g) = 32.00 g, so:

9.82 moles O₂ × (32.00 g/1 mole O₂) = 314 g O₂

The final answer is 314 g!