What volume and what gas will be released during the interaction of potassium carbonate weighing 13.8 g with sulfuric acid.

The reaction of dissolving potassium carbonate in sulfuric acid is described by the following chemical reaction equation:

К2СО3 + Н2SO4 = 2К2SO4 + CO2 + H2O;

When 1 mol of potassium carbonate is dissolved in acid, 1 mol of gaseous carbon dioxide is synthesized. This consumes 1 mol of sulfuric acid.

Let’s calculate the available chemical amount of potassium carbonate.

М К2СО3 = 39 x 2 + 12 + 16 x 3 = 138 grams / mol;

N K2CO3 = 13.8 / 138 = 0.1 mol;

Thus, when 0.1 mol of potassium carbonate is dissolved, 0.1 mol of carbon dioxide is synthesized.

This will consume 0.1 mol of acid.

Let’s calculate its volume. To do this, we multiply the amount of substance and the standard volume of 1 mole of gaseous substance.

1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.

V CO2 = 0.1 x 22.4 = 2.24 liters;