To calculate the mass of sodium sulfate required for the precipitation of barium sulfate, we need to use stoichiometry and the given information.
The balanced chemical equation for the precipitation reaction between barium chloride (BaCl₂) and sodium sulfate (Na₂SO₄) is:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl
From the equation, we can see that one mole of barium chloride reacts with one mole of sodium sulfate to produce one mole of barium sulfate.
First, we need to determine the number of moles of barium chloride present in 100 mL of a solution with a mass concentration of 20,820 g/L.
Mass concentration (g/L) = Mass (g) / Volume (L)
Mass = Mass concentration × Volume Mass = 20,820 g/L × 0.100 L Mass = 2,082 g
Next, we need to convert the mass of barium chloride to moles using its molar mass. The molar mass of BaCl₂ is 208.23 g/mol.
Moles of BaCl₂ = Mass / Molar mass Moles of BaCl₂ = 2,082 g / 208.23 g/mol Moles of BaCl₂ ≈ 9.999 mol (approximately 10 mol)
Since the stoichiometry of the reaction indicates that 1 mole of barium chloride reacts with 1 mole of sodium sulfate, we would require an equal number of moles of sodium sulfate for complete precipitation.
Therefore, the mass of sodium sulfate required can be calculated using its molar mass. The molar mass of Na₂SO₄ is 142.04 g/mol.
Mass of Na₂SO₄ = Moles × Molar mass Mass of Na₂SO₄ = 10 mol × 142.04 g/mol Mass of Na₂SO₄ = 1420.4 g
Therefore, approximately 1420.4 grams of sodium sulfate are required for the precipitation of barium sulfate from 100 mL of barium chloride solution with a mass concentration of 20,820 g/L.