306 mL of a 0.208 M solution of silver nitrate will react with 146 mL of a 0.170 M solution of sodium phosphate to produce how many grams of silver phosphate precipitate? The balanced reaction is: 3 AgNO₃ + Na₃PO₄ → Ag₃PO₄ + 3 NaNO₃

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The balanced reaction is: 3 AgNO₃ + Na₃PO₄ → Ag₃PO₄ + 3 NaNO₃Moles of AgNO₃ = 0.306 L * 0.208 M = 0.0635 molMoles of Na₃PO₄ = 0.146 L * 0.170 M = 0.0249 molThe mole ratio is 3:1, so AgNO₃ is the limiting reactant.Moles of Ag₃PO₄ = 0.0635 mol / 3 = 0.0212 molMass of Ag₃PO₄ = 0.0212 mol * 418.55 g/mol = 8.88 g

306 ML of a 0