How many mL of 0.218 M sodium sulfate react with exactly 25.34 mL of 0.113 M BaCl₂ given the reaction:
BaCl₂ (aq) + Na₂SO₄(aq) → BaSO₄ (s) + 2NaCl (aq)

Question

How many mL of 0.218 M sodium sulfate react with exactly 25.34 mL of 0.113 M BaCl₂ given the reaction: BaCl₂ (aq)+ Na₂SO₄(aq)→ BaSO₄ (s)+ 2NaCl (aq)

Quizplus · Accepted Answer

The balanced equation shows a 1:1 molar ratio between BaCl₂ and Na₂SO₄. Calculate moles of BaCl₂: 0.113 M * 0.02534 L = 0.002867 moles. Since the ratio is 1:1, moles of Na₂SO₄ needed = 0.002867. Volume of Na₂SO₄ = moles / concentration = 0.002867 / 0.218 M = 0.0131 L = 13.1 mL.

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