Technical soda, the mass fraction of impurities in which is 5%, was treated with hydrochloric acid.

Technical soda, the mass fraction of impurities in which is 5%, was treated with hydrochloric acid. Allocated 0.5 mol of CO2. Calculate the mass of technical soda that has entered into a chemical reaction.

Given:
w p (Na2CO3) = 5% = 0.05
n (CO2) = 0.5 mol
To find:
m s (Na2CO3)
Decision:
Na2CO3 + 2HCl = H2O + CO2 + 2NaCl
w h.h. (Na2CO3) = 100% -5% = 95% = 0.95
n (CO2): n (Na2CO3) = 1: 1
n (Na2CO3) = 0.5 mol
m in (Na2CO3) = n * M = 0.5 mol * 106 g / mol = 53 g
m s (Na2CO3) = m w / w h.w = 53 g / 0.95 = 55.8 g
Answer: 55.8 g