Through a solution of potassium carbonate weighing 138 g with w (K2CO3) = 40%, carefully passed

Through a solution of potassium carbonate weighing 138 g with w (K2CO3) = 40%, carefully passed hydrogen chloride obtained with a yield of 80% from table salt weighing 16.25 g, containing 10% impurities by weight. Find the mass fractions of substances in the final solution, assuming that carbon monoxide (IV) did not evolve from the solution.

Given:
m solution (K2CO3) = 138 g
ω (K2CO3) = 40%
η (HCl) = 80%
m (NaCl) = 16.25 g
ω approx. = 10%

To find:
ω (substances) -?

Decision:
1) NaCl + H2SO4 => HCl + NaHSO4;
K2CO3 + 2HCl => H2CO3 + 2KCl;
2) ω (NaCl) = 100% – ω approx. = 100% – 10% = 90%;
3) m clean. (NaCl) = ω * m / 100% = 90% * 16.25 / 100% = 14.625 g;
4) n (NaCl) = m / M = 14.625 / 58.5 = 0.25 mol;
5) n theory. (HCl) = n (NaCl) = 0.25 mol;
6) n practical (HCl) = η * n theory. / 100% = 80% * 0.25 / 100% = 0.2 mol;
7) m (K2CO3) = ω * m solution / 100% = 40% * 138/100% = 55.2 g;
8) n total (K2CO3) = m / M = 55.2 / 138 = 0.4 mol;
9) n (H2CO3) = n practical. (HCl) / 2 = 0.2 / 2 = 0.1 mol;
10) m (H2CO3) = n * M = 0.1 * 62 = 6.2 g;
11) n (KCl) = n practical. (HCl) = 0.2 mol;
12) m (KCl) = n * M = 0.2 * 74.5 = 14.9 g;
13) n react. (K2CO3) = n practical (HCl) / 2 = 0.2 / 2 = 0.1 mol;
14) n rest. (K2CO3) = n total (K2CO3) – n react. (K2CO3) = 0.4 – 0.1 = 0.3 mol;
15) m rest. (K2CO3) = n rest. * M = 0.3 * 138 = 41.4 g;
16) m (HCl) = n practical. * M = 0.2 * 36.5 = 7.3 g;
17) m solution = m solution (K2CO3) + m (HCl) = 138 + 7.3 = 145.3 g;
18) ω (H2CO3) = m * 100% / m solution = 6.2 * 100% / 145.3 = 4.27%;
19) ω (KCl) = m * 100% / m solution = 14.9 * 100% / 145.3 = 10.25%;
20) ω rest. (K2CO3) = m * 100% / m solution = 41.4 * 100% / 145.3 = 28.49%.

Answer: Mass fraction of H2CO3 is 4.27%; KCl – 10.25%; K2CO3 – 28.49%.