What volume will be released when 35.7 g of potassium sulfate is dissolved in an excess of hydrochloric acid.

n = m: M.

M (K2SO3) = 158 g / mol.

n = 35.7 g: 158 g / mol = 0.2 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

K2SO3 + 2HCl = 2KCl + SO2 + H2O

According to the reaction equation, there is 1 mol of SO2 for 1 mol of K2SO4. Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (SO2) = n (K2SO3) = 0.2 mol.

Let’s find the volume of SO2.

V = Vn n, where Vn is the molar volume of gas equal to 22.4 l / mol, and n is the amount of substance.

V = 0.2 mol × 22.4 L / mol = 4.48 L.

Answer: 4.48 liters.