What volume of carbon dioxide will be released during the interaction of sodium

What volume of carbon dioxide will be released during the interaction of sodium carbonate with a mass of 21 g. An excess of hydrochloric acid solution?

The reaction of sodium carbonate with hydrochloric acid is described by the following chemical reaction equation.

NaCO3 + 2HCl = 2NaCl + CO2;

Dissolving 1 mole of sodium carbonate releases 1 mole of carbon dioxide. This requires 2 mol of hydrochloric acid.

Determine the amount of substance contained in 21 grams of sodium carbonate.

M NaCO3 = 23 + 12 +16 x 3 = 83 grams / mol;

N NaCO3 = 21/83 = 0.253 mol;

The reaction of 0.253 mol of sodium carbonate with hydrochloric acid will produce 0.253 mol of carbon dioxide.

One mole of ideal gas under normal conditions takes a volume of 22.4 liters.

Let’s determine the volume of carbon dioxide.

V CO2 = 0.253 x 22.4 = 5.667 liters;