How many grams of zn (zinc) will react with 11.2L 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 = 11.2 / 22.4 = 0.5 mol;

With this amount of oxygen, 0.5 x 2 = 1 mole of zinc will react.

Let’s calculate its weight.

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

M Zn = 65 grams / mol;

m Zn = 1 x 65 = 65 grams;