How much oxygen is needed for the complete combustion of propane weighing 8.8 g?

The reaction of propane with oxygen takes place according to the following chemical reaction equation:

C3H8 + 5O2 = 3CO2 + 4H2O;

For the oxidation of 1 mol of propane, 5 mol of oxygen is required.

Let’s find the amount of propane.

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

N C3H8 = 8.8 / 44 = 0.2 mol;

To burn 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;