What volume of carbon monoxide 4 will be released when 200 grams of 20% sodium carbonate

What volume of carbon monoxide 4 will be released when 200 grams of 20% sodium carbonate solution with hydrochloric acid interacts?

The reaction of dissolving sodium carbonate 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 containing in 200 grams of a 20% sodium carbonate solution.

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

N Na2CO3 = 200 x 0.2 / 106 = 0.377 mol;

Thus, when 0.377 mol of soda is dissolved, 0.377 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 = 0.377 x 22.4 = 8.445 liters;