What volume of carbon dioxide will be released by the interaction of 15 grams of sodium carbonate

What volume of carbon dioxide will be released by the interaction of 15 grams of sodium carbonate containing 10% impurities with an excess of hydrochloric acid.

The reaction of sodium carbonate with hydrochloric acid is described by the following chemical reaction equation.
NaCO3 + 2HCl = 2NaCl + CO2;
When 1 mol of sodium carbonate is dissolved, 1 mol of carbon dioxide is released. This requires 2 mol of hydrochloric acid.
The mass of pure sodium carbonate will be 15 x 0.9 = 13.5 grams;
Determine the amount of substance contained in 13.5 grams of sodium carbonate.
M NaCO3 = 23 + 12 +16 x 3 = 83 grams / mol;
N NaCO3 = 13.5 / 83 = 0.16265 mol;
The reaction of 0.16265 mol of sodium carbonate with hydrochloric acid will result in the formation of 0.16265 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 = 0.16265 x 22.4 = 3.64 liters;