An excess of barium nitrate solution was added to a solution of sodium sulfate with a volume of 250

| March 14, 2021 | education

An excess of barium nitrate solution was added to a solution of sodium sulfate with a volume of 250 ml with a mass fraction of Na2SO4 equal to 7.60% and a density of 1.080 g / cm. Determine the mass of the precipitate formed.

Find the mass of the Na2SO4 solution.

m = Vp.

m = 250 cm3 × 1.080 g / cm3 = 270 g.

Find the mass of Na2SO4 in solution.

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (Na2SO4) = (270 g × 7.6%): 100% = 20.52 g.

Let’s find the amount of the substance Na2SO4 by the formula:

n = m: M.

M (Na2SO4) = 142 g / mol.

n = 20.52 g: 142 g / mol = 0.14 mol.

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

Na2SO4 + Ba (NO3) 2 = BaSO4 ↓ + 2NaNO3.

According to the reaction equation, there is 1 mol of BaSO4 for 1 mol of Na2SO4. Substances are in quantitative ratios 1: 1.

n (Na2SO4) = n (BaSO4) = 0.14 mol.

Let’s find the mass of BaSO4 by the formula:

m = n × M,

M (BaSO4) = 233 g / mol.

m = 0.14 mol × 233 g / mol = 32.62 g.

Answer: 32.62 g.



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