The reaction formed 4 g of copper (II) oxide. Calculate: a) the mass and quantity of the copper substance

The reaction formed 4 g of copper (II) oxide. Calculate: a) the mass and quantity of the copper substance that has entered into the reaction b) the mass and quantity of the thing of the consumed oxygen

Metallic copper reacts with oxygen to form oxide. The reaction is described by the following chemical equation:

Cu + ½ O2 = CuO ↑;

Let’s calculate the chemical amount of copper oxide. To do this, we divide the mass of the available substance by the weight of 1 mole.

M CuO = 64 + 16 = 80 grams / mol;

N CuO = 4/80 = 0.05 mol;

To get this amount of copper oxide, you need to take the same amount of copper.

Find the weight of the copper. For this purpose, we multiply the amount of the substance by the weight of 1 mole of the substance.

M Cu = 64 grams / mol;

m Cu = 0.05 x 64 = 3.2 grams;

To obtain such an amount of copper oxide, it is necessary to take 2 times less oxygen.

Let’s calculate the weight and volume of oxygen.

To find the weight, we multiply the amount of the substance by the weight of 1 mole of the substance.

M O2 = 16 x 2 = 32 grams / mol;

m O2 = 0.05 / 2 x 32 = 0.8 grams;

To find the volume of oxygen, we multiply the amount of the substance by the volume of 1 mole of gas (which is 22.4 liters).

m O2 = 0.05 / 2 x 22.4 = 0.56 liters;