Calculate the volume of oxygen required for the complete combustion of 8 liters of propane (C3H8, gas)

Calculate the volume of oxygen required for the complete combustion of 8 liters of propane (C3H8, gas) Calculate the volume of gaseous substances formed as a result of the reaction. (All gas volumes are given under the same conditions).

The oxidation reaction of propane with oxygen is described by the following chemical reaction equation:

C3H8 + 5O2 = 3CO2 + 4H2O;

1 mole of gas is reacted with 5 moles of oxygen. Synthesized 3 moles of carbon dioxide.

Let’s calculate the available chemical amount of propane. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

N C3H8 = 8 / 22.4 = 0.357 mol;

Let’s calculate the volume of oxygen and combustion products. To do this, multiply the amount of substance and the standard volume of 1 mole of gaseous substance. 1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

This will release 0.357 x 3 = 1.071 mol of carbon dioxide. V CO2 = 1.071 x 22.4 = 24 liters;

This will release 0.357 x 4 = 1.428 mol of water vapor. V H2O = 1.428 x 22.4 = 32 liters;

This will require 0.357 x 5 = 1.785 mol of oxygen. V O2 = 1.785 x 22.4 = 40 liters;