Calculate the volume of hydrogen that is formed during the decomposition of 9 g of water by electric current

Calculate the volume of hydrogen that is formed during the decomposition of 9 g of water by electric current, and also calculate whether the obtained hydrogen is enough to completely reduce copper from 24 g of copper (II) oxide.

Copper oxide is reduced with hydrogen gas. During the reaction, metallic copper and water are synthesized. This reaction is described by the following 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 define the chemical (molar) 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;

During the course of this 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.

When 1 mole of water is decomposed, 1 mole of hydrogen is synthesized.

Let’s calculate the amount of substance in 9 grams of water.

M H2O = 2 + 16 = 18 grams / mol;

N H2O = N H2 = 9/18 = 0.5 mol;

Therefore, there will be enough hydrogen.