How much oxygen is needed to completely burn 11.2 liters 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 = 11.2 / 22.4 = 0.5 mol;

To burn such an amount of matter, 0.5 x 3 = 1.5 mol of oxygen is required.

Let’s calculate its weight.

To do this, multiply the molar weight by the amount of substance.

M O2 = 16 x 2 = 32 grams / mol;

m O2 = 32 x 1.5 = 48 grams;