Find the mass of zinc oxide formed during the combustion of zinc to 40 liters of oxygen.

Zinc enters into an oxidation reaction with oxygen. The reaction is described by the following chemical equation.

Zn + ½ O2 = ZnO;

1 mol of metallic zinc reacts with 0.5 mol of oxygen. This synthesizes 1 mole of zinc oxide.

Let’s calculate the chemical amount of oxygen contained in 11.2 liters.

To do this, we divide the volume of gas by the volume of 1 mole of gas, which is 22.4 liters.

N O2 = 40 / 22.4 = 1.786 mol;

With this amount of oxygen, 1.786 x 2 = 3.572 mol of zinc will react. The same amount of zinc oxide will be synthesized.

Let’s calculate the weight of zinc oxide.

To do this, multiply the amount of the substance by the weight of 1 mole of the substance.

M ZnO = 65 + 16 = 81 grams / mol;

m ZnO = 3.572 x 81 = 289.33 grams;