What mass of copper is reduced by the interaction of 12 g of copper (II) oxide with hydrogen?

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;

Cupric oxide reacts with hydrogen in equal molar amounts. In this case, equal (equivalent) amounts of metallic copper and water are synthesized.

Determine 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 = 12/80 = 0.15 mol;

Thus, it is possible to carry out the reduction of 0.15 mol of copper oxide and obtain 0.15 mol of copper.

Its weight will be:

M Cu = 64 grams / mol;

m Cu = 64 x 0.15 = 9.6 grams;