Calculate the volume of oxygen and air required for combustion of 5 liters of butane.

Butane has the chemical formula C4H10

The combustion reaction of butane is described by the following chemical reaction equation:

2С4Н10 + 13О2 = 8СО2 + 10Н2О

It takes 13 moles of oxygen to burn 2 moles of butane.

Let’s find the amount of the substance contained in 5 liters of butane.

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters. Accordingly, 5 liters contain 5 / 22.4 = 0.223 mol of butane.

Let’s find the number of moles of oxygen needed to burn that amount of butane.

It will be 0.223 * 13/2 = 1.45 mol

This amount of oxygen has a volume of 1.45 * 22.4 = 32.48 liters

The volume fraction of oxygen in the air is 23.15%. Therefore, more air will be required.

Its volume will be 32.48 / 0.2315 = 140.302 liters