How much hydrogen will be consumed to reduce 24 g of copper (II) oxide

Copper oxide is reduced with hydrogen gas. This synthesizes metallic copper and water. This reaction is described by the following equation:

CuO + H2 = Cu + H2O;

Cupric oxide reacts with hydrogen in equal (equivalent) molar amounts. During this reaction, the same equal chemical amounts of metallic copper and water are synthesized.

Let’s calculate the chemical amount of copper oxide.

To do this, divide the weight of the oxide by the weight of 1 mole of oxide.

M CuO = 64 + 16 = 80 grams / mol;

N CuO = 24/80 = 0.3 mol;

In the course of the reaction, 0.3 mol of copper oxide can be reduced and 0.3 mol of copper can be obtained.

The same amount of hydrogen will be required. Let’s determine the volume of hydrogen. To do this, multiply the amount of substance and the volume of 1 mole of gas (22.4 liters).

N H2 = 0.3 mol;

V H2 = 0.3 x 22.4 = 6.72 liters;