What volume of gas will be released during the reaction of 200 g of a 30% sodium carbonate solution

What volume of gas will be released during the reaction of 200 g of a 30% sodium carbonate solution with an excess of hydrochloric acid, if the practical yield of the reaction is 80% of the theoretical?

The reaction of sodium carbonate dissolution in hydrochloric acid is described by the following chemical reaction equation:

Na2CO3 + 2HCl = 2NaCl + CO2 + H2O;

When 1 mol of sodium carbonate is dissolved in acid, 1 mol of gaseous carbon dioxide is synthesized. This consumes 2 mol of hydrochloric acid.

Let’s calculate the chemical amount of a substance contained in 10.6 grams of sodium carbonate.

M Na2CO3 = 23 x 2 + 12 + 16 x 3 = 106 grams / mol;

N Na2CO3 = 200 x 0.3 / 106 = 0.566 mol;

Thus, taking into account the yield of 80% when dissolving 0.566 mol of soda, 0.566 x 0.8 = 0.453 mol of carbon dioxide is synthesized.

Let’s calculate its volume.

To do this, multiply the amount of the substance and the standard volume of 1 mole of the gaseous substance.

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

V CO2 = 0.453 x 22.4 = 10.147 liters;