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

| September 14, 2021 | education

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%.



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