In 20 g of 8% sodium hydroxide solution, sulfur oxide (IV), released as a result of firing pyrite

In 20 g of 8% sodium hydroxide solution, sulfur oxide (IV), released as a result of firing pyrite weighing 3.2 g, was dissolved. Determine the mass fraction of salt in the resulting solution.

Given:
m solution (NaOH) = 20 g
ω (NaOH) = 8%
m (FeS2) = 3.2 g

Find:
ω (salt) -?

Solution:
1) 4FeS2 + 11O2 = (tOC) => 2Fe2O3 + 8SO2 ↑;
NaOH + SO2 => NaHSO3;
2) M (NaOH) = Mr (NaOH) = Ar (Na) + Ar (O) + Ar (H) = 23 + 16 + 1 = 40 g / mol;
M (FeS2) = Mr (FeS2) = Ar (Fe) + Ar (S) * 2 = 56 + 32 * 2 = 120 g / mol;
M (SO2) = Mr (SO2) = Ar (S) + Ar (O) * 2 = 32 + 16 * 2 = 64 g / mol;
M (NaHSO3) = Mr (NaHSO3) = Ar (Na) + Ar (H) + Ar (S) + Ar (O) * 3 = 23 + 1 + 32 + 16 * 3 = 104 g / mol;
3) n (FeS2) = m (FeS2) / M (FeS2) = 3.2 / 120 = 0.03 mol;
4) n (SO2) = n (FeS2) * 2 = 0.027 * 2 = 0.06 mol;
5) m (SO2) = n (SO2) * M (SO2) = 0.06 * 64 = 3.84 g;
6) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 8% * 20/100% = 1.6 g;
7) n (NaOH) = m (NaOH) / M (NaOH) = 1.6 / 40 = 0.04 mol;
8) n (NaHSO3) = n (NaOH) = 0.04 mol;
9) m (NaHSO3) = n (NaHSO3) * M (NaHSO3) = 0.04 * 104 = 4.16 g;
10) m solution (NaHSO3) = m solution (NaOH) + m (SO2) = 20 + 3.84 = 23.84 g;
11) ω (NaHSO3) = m (NaHSO3) * 100% / m solution (NaHSO3) = 4.16 * 100% / 23.84 = 17.45%.

Answer: The mass fraction of NaHSO3 is 17.45%.