Calculate the volume of oxygen required for combustion of 2 liters of propane.

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. 3 moles of carbon dioxide are synthesized.

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 = 2 / 22.4 = 0.09 mol;

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

This will require 0.09 x 5 = 0.45 mol of oxygen.

V O2 = 0.45 x 22.4 = 10.08 liters;




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