Mixed a solution weighing 200 g with a mass fraction of zinc sulfate 16.1% with a solution weighing 100 g

Mixed a solution weighing 200 g with a mass fraction of zinc sulfate 16.1% with a solution weighing 100 g with a mass fraction of sodium sulfide 7.8%. Determine the mass fraction of salt in the solution.

Given:
m solution (ZnSO4) = 200 g
ω (ZnSO4) = 16.1%
m solution (Na2S) = 100 g
ω (Na2S) = 7.8%

To find:
ω (salt in solution) -?

Decision:
1) ZnSO4 + Na2S => ZnS ↓ + Na2SO4;
2) m (ZnSO4) = ω * m solution / 100% = 16.1% * 200/100% = 32.2 g;
3) n (ZnSO4) = m / M = 32.2 / 161 = 0.2 mol;
4) m (Na2S) = ω * m solution / 100% = 7.8% * 100/100% = 7.8 g;
5) n (Na2S) = m / M = 7.8 / 78 = 0.1 mol;
6) n react. (ZnSO4) = n (Na2S) = 0.1 mol;
7) n rest. (ZnSO4) = n – n reag. = 0.2 – 0.1 = 0.1 mol;
8) m rest. (ZnSO4) = n rest. * M = 0.1 * 161 = 16.1 g;
9) n (Na2SO4) = n (Na2S) = 0.1 mol;
10) m (Na2SO4) = n * M = 0.1 * 142 = 14.2 g;
11) n (ZnS) = n (Na2S) = 0.1 mol;
12) m (ZnS) = n * M = 0.1 * 97 = 9.7 g;
13) m solution = m solution (ZnSO4) + m solution (Na2S) – m (ZnS) = 200 + 100 – 9.7 = 290.3 g;
14) ω rest. (ZnSO4) = m rest. * 100% / m p-ra = 16.1 * 100% / 290.3 = 5.5%;
15) ω (Na2SO4) = m * 100% / m solution = 14.2 * 100% / 290.3 = 4.9%.

Answer: The mass fraction of ZnSO4 is 5.5%; Na2SO4 – 4.9%.