Calculate the volumes of oxygen and air required to burn 20 kg. methane (CH4)

Calculate the volumes of oxygen and air required to burn 20 kg. methane (CH4) containing 20% non-combustible impurities. The volume fraction of oxygen in the air is 21%.

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.

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

N CH4 = 20,000 x 0.8 / 16 = 1000 mol;

When burning such an amount of substance, it is necessary to take 1,000 x 3 = 3,000 moles of oxygen.

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 О2 = 3000 x 22.4 = 67 200 liters = 67.2 m3;

The oxygen content in the air is 21%. The required air volume will be:

V air = 67,200 / 0.21 = 320,000 liters = 320 m3;