Technical soda, the mass fraction of impurities in which is 15%, was treated with hydrochloric acid. Released 0.3 mol of CO2.

Technical soda, the mass fraction of impurities in which is 15%, was treated with hydrochloric acid. Released 0.3 mol of CO2. Calculate the mass of technical soda that has entered into a chemical reaction.

Given:
ω approx. = 15%
n (CO2) = 0.3 mol

Find:
m tech. (Na2CO3) -?

Solution:
1) Na2CO3 + 2HCl => 2NaCl + CO2 ↑ + H2O;
2) M (Na2CO3) = Mr (Na2CO3) = Ar (Na) * N (Na) + Ar (C) * N (C) + Ar (O) * N (O) = 23 * 2 + 12 * 1 + 16 * 3 = 106 g / mol;
3) n (Na2CO3) = n (CO2) = 0.3 mol;
4) m clean. (Na2CO3) = n (Na2CO3) * M (Na2CO3) = 0.3 * 106 = 31.8 g;
5) ω (Na2CO3) = 100% – ω approx. = 100% – 15% = 85%;
6) m tech. (Na2CO3) = m pure. (Na2CO3) * 100% / ω (Na2CO3) = 31.8 * 100% / 85% = 37.4 g.

Answer: The mass of technical Na2CO3 is 37.4 g.