Calculate the mass of copper formed during the reduction of copper (II) oxide with hydrogen

Calculate the mass of copper formed during the reduction of copper (II) oxide with hydrogen with a mass of 60 g with a mass fraction of an impurity of 10%

Divalent copper oxide is reduced with hydrogen gas when heated. 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 molar amounts. When the reaction proceeds, the same equal chemical amounts of copper and water are synthesized.

Let’s find 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 = 60 x 0.9 / 80 = 0.675 mol;

During the reaction, 0.675 mol of copper oxide can be reduced to obtain 0.675 mol of copper.

Its weight will be:

M Cu = 64 grams / mol;

m Cu = 64 x 0.675 = 43.2 grams;