Answer:
To determine the grams of silver phosphate produced and excess reagent, we need to use stoichiometry and identify the limiting reagent.
First, we need to find the number of moles of each reactant using their molar masses:
Number of moles of AgNO3 = 200 g / 169.87 g/mol = 1.177 mol
Number of moles of Na3PO4 = 200 g / 380.13 g/mol = 0.526 mol
Next, we need to use the balanced chemical equation to determine the theoretical yield of silver phosphate:
From the balanced equation, 3 moles of AgNO3 react with 1 mole of Na3PO4 to produce 1 mole of Ag3PO4. Therefore, the number of moles of Ag3PO4 produced is:
1.177 mol AgNO3 × (1 mol Ag3PO4 / 3 mol AgNO3) = 0.392 mol Ag3PO4
To convert moles of Ag3PO4 to grams, we use its molar mass:
0.392 mol Ag3PO4 × 418.58 g/mol = 164.0 g Ag3PO4
Therefore, 164.0 grams of Ag3PO4 are produced.
To determine the excess reagent, we need to identify the limiting reagent. The reactant that produces less product is the limiting reagent.
From the stoichiometry calculation above, we can see that 1.177 moles of AgNO3 are required to react with 0.526 moles of Na3PO4, so Na3PO4 is the limiting reagent. Therefore, we need to determine how much of the excess AgNO3 remains after the reaction is complete.
The amount of AgNO3 that reacted is:
0.526 mol Na3PO4 × (3 mol AgNO3 / 1 mol Na3PO4) × (169.87 g/mol AgNO3) = 85.3 g AgNO3
The amount of excess AgNO3 is:
200 g AgNO3 - 85.3 g AgNO3 = 114.7 g AgNO3
Therefore, 114.7 grams of AgNO3 is left over as excess reagent.