Sulfur dioxide with a volume of 4.48 ml was absorbed by 200 g of a 28% potassium hydroxide solution. Determine the mass fraction

Sulfur dioxide with a volume of 4.48 ml was absorbed by 200 g of a 28% potassium hydroxide solution. Determine the mass fraction of the formed salt in the solution.

SO2 + 2KOH → K2SO3 + H2O
n (SO2) = 4.48 / 22400 = 0.0002 mol – deficient
m (KOH) = 200 x 0.28 = 56 g
n (KOH) = 56/56 = 1 mol – in excess
n (K2SO3) = 0.0002 mol
m (K2SO3) = 0.0002 x (39 x 2 + 32 + 48) = 0.0316 g
m (solution) = 200 + (0.0002 x 64) = 200.0128 g
ω (K2SO3) = 0.0316 / 200.0128 x 100% = 0.016%
Answer: the mass fraction of salt (potassium sulfite) in the resulting solution is 0.016%.

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