Find the mass of oxygen that was required to burn 19.5 g of zinc.

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 a substance contained in 180 grams of zinc.

M Zn = 65 grams / mol;

N Zn = 19.5 / 65 = 0.3 mol;

To oxidize such an amount of zinc, 0.3 / 2 = 0.15 mol of oxygen is needed.

Let’s calculate its weight.

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

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

m O2 = 0.15 x 32 = 4.8 grams;