Question

Propane gas (C3H8) is combusted at a constant temperature and pressure. If 34.35L of propane react, what volume of oxygen is required? What volume of carbon dioxide is produced?

Answer 1

First, we must write the reaction equation for the combustion of propane:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Moreover, we note that under standard conditions, propane is a gas and under standard conditions, one mole of gas occupies 22.4 liters of volume. Using this, we can calculate the moles of propane present:
Moles propane = 34.35/22.4 = 1.53 moles

Next, the molar ratio, as visible from the reaction equation, between propane and oxygen is 1:5. We can use this to calculate the moles of oxygen required:
1.53 * 5 = 7.65 moles
Volume of oxygen = 7.65 * 22.4 = 171.36 liters

Next, using the same method, we calculate the moles of carbon dioxide then the volume of carbon dioxide:
Moles = 4.59
Volume = 102.82 liters