What volume of 0.305 M AgNO₃ is required to react exactly with 155.0 mL of 0.274 M Na₂SO₄ solution? Hint: You will want to write a balanced reaction. A) 581 mL B) 173 mL C) 345 mL D) 139 mL E) 278 mL

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The balanced reaction is 2AgNO3 + Na2SO4 → 2NaNO3 + Ag2SO4. Calculate moles of Na2SO4: 0.155 L * 0.274 M = 0.04247 mol. Moles of AgNO3 needed: 2 * 0.04247 mol = 0.08494 mol. Volume of AgNO3: 0.08494 mol / 0.305 M = 0.278 L = 278 mL.

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