How many liters of oxygen will it take to interact with 2 g of zinc?

Zinc is oxidized by oxygen. The reaction is described by the following 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 = 2/65 = 0.031 mol;

To burn such an amount of zinc, 0.031 / 2 = 0.0155 mol of oxygen will be needed.

Let’s define its volume.

To do this, multiply the amount of substance by the volume of 1 mole of gas, which is 22.4 liters.

V O2 = 0.0155 x 22.4 = 0.3472 liters;