Technical soda, the mass fraction of impurities in which is 15%, was treated with hydrochloric acid. 0.1 mol CO2 was released.

Technical soda, the mass fraction of impurities in which is 15%, was treated with hydrochloric acid. 0.1 mol CO2 was released. Calculate the mass of technical soda that has entered into a chemical reaction.

Given:
w approx (Na2CO3) = 15%
n (CO2) = 0.1 mol
To find:
m mixture (Na2CO3)
Decision:
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
w is clean. islands (Na2CO3) = 100% -15% = 85% = 0.85
n (CO2): n (Na2CO3) = 1: 1
n (Na2CO3) = 0.1 mol
m in islands (Na2CO3) = n * M = 0.1 mol * 106 g / mol = 10.6 g
m mixture (Na2CO3) = m water / w pure. in-va = 10.6 g / 0.85 = 12.47 g
Answer: 12.47 g