The volume of oxygen required for the oxidation of 25.6 g of copper.

Metallic copper reacts with oxygen to form oxide. The reaction is described by the following chemical equation:

Cu + ½ O2 = CuO ↑;

Let’s calculate the chemical amount of copper. To do this, divide the mass of the available substance by the weight of 1 mole.

M Cu = 64 grams / mol;

N Cu = 25.6 / 64 = 0.4 mol;

To oxidize such an amount of copper, it is necessary to take 2 times less oxygen.

Determine the weight and volume of oxygen.

To find the weight, we multiply the amount of the substance by the weight of 1 mole of the substance.

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

m O2 = 0.4 / 2 x 32 = 6.4 grams;

To find the volume of oxygen, we multiply the amount of the substance by the volume of 1 mole of gas (which is 22.4 liters).

m O2 = 0.4 / 2 x 22.4 = 4.48 liters;