How much oxygen is required for the catalytic oxidation of 11.2 liters of ammonia?

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

4NH3 + 5O2 = 4NO + 6H2O;

Let’s find the chemical amount of the substance in 11.2 liters of ammonia.

To do this, we divide its volume by the volume of 1 mole of gas (equal to 22.4 liters).

N NH3 = 11.2 / 22.4 = 0.5 mol;

To oxidize such an amount of a substance, it is necessary to take (in accordance with the coefficients of the chemical equation) 0.5 x 5/4 = 0.625 mol of oxygen.

Let’s calculate the volume of oxygen gas.

To do this, multiply its amount by the volume of 1 mole of gas (assuming a volume of 22.4 liters).

V O2 = 0.625 x 22.4 = 14 liters;