A 1.543 gram sample containing sulfate ion was treated with barium chloride reagent, and 0.2243 grams of barium sulfate was isolated. calculate the percentage of sulfate ion in the sample.

Question

Chemistry

Paxton Chapman

9 March, 02:28

A 1.543 gram sample containing sulfate ion was treated with barium chloride reagent, and 0.2243 grams of barium sulfate was isolated. calculate the percentage of sulfate ion in the sample.

+3

Answers (1)

Know the Answer?

Mila · Answer 1

Chemical reaction: SO₄²⁻ + Ba²⁺ → BaSO₄.

m (sample) = 1,543 g.

m (BaSO₄) = 0,2243 g.

n (BaSO₄) = m (BaSO₄) : M (BaSO₄).

n (BaSO₄) = 0,2243 g : 233,4 g/mol.

n (BaSO₄) = 0,00096 mol.

n (BaSO₄) = n (SO₄²⁻).

ω (SO₄²⁻) = m (SO₄²⁻) : m (sample).

ω (SO₄²⁻) = 0,00096 mol · 96 g/mol : 1,543 g.

ω (SO₄²⁻) = 0,059 = 5,9%.