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

Question

Quizplus · Accepted Answer

First, determine the limiting reactant. For AgNO3, moles = 0.306 L * 0.208 mol/L = 0.0637 mol. For Na3PO4, moles = 0.146 L * 0.170 mol/L = 0.0248 mol. The reaction ratio is 3:1 for AgNO3 to Na3PO4, so Na3PO4 is the limiting reactant. Using the molar mass of Ag3PO4 (418.58 g/mol) and the moles of Na3PO4, the mass of Ag3PO4 produced is 0.0248 mol * 418.58 g/mol = 10.38 g, which rounds to 10.4 g, not matching any of the provided options exactly. However, based on the choices given and the closest match, there seems to be a calculation or typographical error in the question or answer choices. Given the context, the correct process was described, but due to a discrepancy in the final matching option, an accurate selection based on the provided answers cannot be made without further clarification. The explanation indicates a calculation aiming towards option D) 10.4 g, but this does not align with the initial selection of A) 8.88 g, suggesting a need for correction either in the calculation or in the interpretation of the question's options.

306 ML of a 0