Homework Answers
The reaction that occurs is:
3 AgNO3 + Na3PO4 = Ag3PO4 + 3 NaNO3
The moles of the reagents are calculated:
n AgNO3 = g / MM = 33.52 g / 170 g / mol = 0.197 mol
n Na3PO4 = M * V = 0.25 M * 0.7 L = 0.175 mol
The limit reagent is AgNO3.
The moles of used and excess Na3PO4 are calculated:
n Na3PO4 = 0.197 mol AgNO3 * (1 mol Na3PO4 / 3 mol AgNO3) = 0.066 mol
n Na3PO4 excess = 0.175 - 0.066 = 0.109 mol
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