# Determine the amount of hydrogen (in moles) reacted with 300 g of copper (II) oxide, which contains 20% of impurities.

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;

Cupric oxide reacts with hydrogen in equal molar amounts. In this case, equal (equivalent) amounts of metallic 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 = 300 x 0.8 / 80 = 3 mol;

During the reaction, 3 moles of copper oxide can be reduced and 3 moles of copper can be obtained.

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

N H2 = 3 mol;

V H2 = 3 x 22.4 = 67.2 liters;

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