Question

Calculate the mole fraction when a sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 2.15 times the amount needed to completely oxidize the propane to CO2 and H2O at constant temperature. Assume all chemical species are in the gaseous phase.

Answer 1

To solve this problem, we need to take a basis of the compound in interest, so we'll take 100 moles of propane, for which, we'll have the following reaction:

C₃H₈ + O₂ -> CO₂ + H₂O

Which we need to balance to:

C₃H₈ + 5O₂ -> 3CO₂ + 4H₂O

For 100 moles of propane, we need 500 moles of oxygen for complete reaction. If we need 500, and what we have is 2.15 times of what needed, then we have:

500 x 2.15 = 1,075 moles of O₂

​If reaction takes place completely, we'll have the following products:

1,075 moles O₂ - 500 needed moles of O₂ = 575 moles O₂

We'll also produce 3 moles of CO₂ and 4 moles of water per mole, which means we'll have 300 moles of carbon dioxide and 400 moles of water. Propane will be consumed totally, therefore mole fractions will be:

Total moles: 575 + 400 + 300 = 1,275 moles

Fraction of O₂ = 575 / 1,275 = 0.45 Fraction of CO₂ = 300/1,275 = 0.24 Fraction of Water = 400/1,275 = 0.31

That is for the resulting product. For the entrance, we have:

Total moles = 100 moles of propane + 1,075 moles of O₂ = 1,175 moles

​Fraction of propane = 100 / 1,175 = 0.085 Fraction of O₂ = 1,075 / 1,175 = 0.915