Calculate the amount of oxygen that will need to be consumed to burn 1 liter of propane

The oxidation reaction of propane with oxygen is described by the following chemical reaction equation:

C3H8 + 5O2 = 3CO2 + 4H2O;

1 mol of gas is reacted with 5 mol of oxygen. 3 moles of carbon dioxide are synthesized.

Let’s calculate the available chemical amount of propane. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

N C3H8 = 1 / 22.4 = 0.045 mol;

Let’s calculate the volume of oxygen. To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

This will require 0.045 x 5 = 0.225 mol of oxygen.

V O2 = 0.225 x 22.4 = 5.04 liters;