How many liters of hydrogen will it take to reduce 7.2 g of copper oxide (2)?

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;

Divalent copper oxide reacts with hydrogen in equal (equivalent) molar amounts. In the course of the reaction, the same equal chemical 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 = 7.2 / 80 = 0.09 mol;

Thus, a reduction of 0.09 mol of copper oxide can be carried out and 0.09 mol of copper can be obtained.

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

N H2 = 0.09 mol;

V H2 = 0.09 x 22.4 = 2.016 liters;