During the interaction of ammonia with a volume of 10.64 liters with oxygen under the action

During the interaction of ammonia with a volume of 10.64 liters with oxygen under the action of a catholyzer, nitrogen (II) oxide with a mass of 11.4 g was obtained. It is necessary to calculate the mass fraction of the yield of the reaction product from the theoretically possible (in percent).

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

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

Let’s calculate the chemical amount of a substance in 100 liters of ammonia.

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

N NH3 = 10.64 / 22.4 = 0.475 mol;

Let’s calculate the amount of nitric oxide substance.

M NO = 14 + 16 = 30 grams / mol;

N NO = 11.4 / 30 = 0.38 mol;

The reaction yield is:

K = 0.38 / 0.475 = 0.8 = 80%;