0.64 g of sulfur (4) -oxide was dissolved in 50 g water. Determine the mass fraction of sulfurous acid in the resulting solution.

Given:
m (H2O) = 50 g
m (SO2) = 0.64 g
To find:
W (H2SO3) =?
Decision:
1) Let’s make the equation of the chemical reaction, let the mass of the acid be x g.
H2O + SO2 = H2SO3
2) We find what is in excess and what is in short supply.
n = m / M
M (H2O) = 2 + 16 = 18 g / mol
n (H2O) = 50 g: 18 g / mol = 2.8 mol.
M (SO2) = 32 + 32 = 64 g / mol
n (SO2) = 0.64: 64 = 0.1 mol.
3) The calculation is carried out by deficiency, that is, by H2SO3
M (H2SO3) = 2 + 32 + 48 = 82 g / mol
Find the mass of H2SO3
0.64: 64 = x: 8
x = (0.64 x 8): 64 = 8.2 g.
4) m solution = 50 + 0.64 = 50.64 g
W = (m substance: m solution) x 100%
W (H2SO3) = (8.2: 50.64) x 100% = 16.2%
Answer: W (H2SO3) = 16.2%