How much oxygen and air will be consumed to burn propane (C3H8) with a mass of 22 g?

The interaction with propane and methane oxygen is described by the following chemical reaction equation:

Propane:

C3H8 + 5O2 = 3CO2 + 4H2O;

Methane:

CH4 + 2O2 = CO2 + 2H2O;

Propane:

Let’s calculate the chemical amount of propane to do this, divide its weight by the weight of 1 mole of the substance.

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

N C3H8 = 22/44 = 0.5 mol;

To burn such an amount of matter, you need 0.5 x 5 = 2.5 mol of oxygen.

Let’s define its volume.

To do this, multiply the amount of the substance by the volume of 1 mole of the substance (which is 22.4 liters).

V O2 = 2.5 x 22.4 = 56 liters;

Methane:

Let’s calculate the chemical amount of propane to do this, divide its weight by the weight of 1 mole of the substance.

M CH4 = 12 + 4 = 16 grams / mol;

N CH4 = 64/16 = 4 mol;

To burn this amount of matter, you need 4 x 2 = 8 moles of oxygen.

Let’s define its volume.

To do this, multiply the amount of the substance by the volume of 1 mole of the substance (which is 22.4 liters).

V O2 = 8 x 22.4 = 179.2 liters;