How much air will it take to burn 84 liters of propane?

The propane combustion reaction is described by the following chemical reaction equation:

C3H8 + 5O2 = 3CO2 + 4H2O;

For the combustion of 1 mole of propane, 5 moles of oxygen are required. This produces 3 moles of carbon dioxide and 4 moles of water.

Under normal conditions, one mole of ideal gas occupies a fixed volume of 22.4 liters.

Therefore, to burn 84 liters of propane, 84 x 5 = 420 liters of oxygen is required.

The oxygen content in the air is 20.95%.

In order to find the required volume of air, we divide the required volume of oxygen by its volume fraction.

The required air volume will be:

V air = 420 / 0.2095 = 2,005 liters;