# To 150 g of a 19.8% barium bromide solution was added 23.7 g of potassium sulfate.

**To 150 g of a 19.8% barium bromide solution was added 23.7 g of potassium sulfate. Find the mass of the sediment obtained and the mass fractions of substances in the resulting solution.**

Reaction of barium bromide with potassium sulfate.

BaBr2 + K2SO4 = BaSO4 = 2KBr.

Find the mass of barium bromide in the original solution.

m (BaBr2) = W (BaBr2) • m (solution) / 100% = 19.8% • 150 g / 100% = 29.7 g.

Let’s calculate the number of moles of substances that have reacted.

n (BaBr2) = m / Mr = 29.7 g / 297 g / mol = 0.1 mol.

n (K2SO4) = 23.7 g / 174 g / mol = 0.14 mol.

Sulfate is in excess, so the calculation is based on bromide.

m (BaSO4) = n • Mr = 0.1 mol • 233 g / mol = 23.3 g

The number of moles of potassium bromide according to the reaction is twice as large.

n (KBr) = 0.2 mol.

m (KBr) = 0.2 mol * 119 g / mol = 23.8 g.

The mass of the resulting solution.

m (solution) = m (original solution) + m (K2SO4) = 150 g + 23.7 g = 173.7 g.

Mass fractions of salts in the resulting solution are.

W (BaSO4) = m (in-va) / m (solution) • 100% = 23.3 g / 173.7 g • 100% = 13.4%.

W (KBr) = 23.8 g / 173.7 g * 100% = 13.7%.