Calculate the volume of carbon dioxide that will be released during the interaction of hydrochloric acid and 415.52 g

Calculate the volume of carbon dioxide that will be released during the interaction of hydrochloric acid and 415.52 g of sodium carbonate containing 49.82 g of impurities.

Given:
m mixture (Na2CO3) = 415.52 g
w approx. (Na2CO3) = 49.82%
To find:
V (CO2)
Decision:
Na2CO3 + 2HCl = 2NaCl + H2O + CO2
w is clean. islands (Na2CO3) = 100% -49.82% = 50.18% = 0.5018
m in islands (Na2CO3) = m mixture * w pure. in-va = 415.52 g * 0.5018 = 209 g
n (Na2CO3) = m / M = 209 g / 106 g / mol = 2 mol
n (Na2CO3): n (CO2) = 1: 1
n (CO2) = 2 mol
V (CO2) = n * Vm = 2 mol * 22.4 l / mol = 44.8 l
Answer: 44.8 L