How much oxygen is needed to burn 8.8 g of propane.

The oxidation reaction of propane with oxygen occurs in accordance with the following chemical reaction equation:

C3H8 + 5O2 = 3CO2 + 4H2O;

Combustion of 1 mol of propane requires 5 mol of oxygen.

Let’s find the required amount of propane.

M C3H8 = 12 x 3 + 8 = 44 grams / mol;

N C3H8 = 8.8 / 44 = 0.2 mol;

To oxidize this amount of propane, 0.2 x 5 = 1 mole of oxygen is required.

Let’s find its volume.

One mole of gas under normal conditions fills a volume of 22.4 liters.

The oxygen volume will be equal to:

V O2 = 1 x 22.4 = 22.4 liters;