Calculate the volume of air required to burn 250 meters cubed of methane.

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

CH4 + 3O2 = CO2 + 2H2O;

1 mol of gas is reacted with 2 mol of oxygen. 1 mole of carbon dioxide is synthesized.

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

N CH4 = 250 / 22.4 = 11.161 mol;

To burn such an amount of matter, 11.161 x 3 = 33.483 mol of oxygen is required.

Let’s calculate its volume. 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.

V O2 = 33.483 x 22.4 = 750 liters;

The oxygen content in the air is 20%. The required air volume will be:

V air = 750 / 0.2 = 3750 liters = 3.75 m3;