How much hydrogen was used to reduce 12.8 g of copper oxide?

Copper oxide is reduced with hydrogen gas. During the reaction, metallic copper and water are synthesized. The reaction is described by the following chemical reaction equation:

CuO + H2 = Cu + H2O;

Bivalent copper oxide reacts with hydrogen in equal molar amounts. In this case, the same identical amounts of metallic copper and water are synthesized.

Let’s calculate the chemical amount of copper oxide.

For this purpose, we divide the weight of the oxide by the weight of 1 mole of oxide.

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

N CuO = 12.8 / 80 = 0.16 mol;

Thus, 0.16 mol of copper oxide can be reduced and 0.16 mol of copper can be synthesized.

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

N H2 = 0.16 mol;

V H2 = 0.16 x 22.4 = 3.584 liters;