Answer: 526 grams of Ag3PO4.
Step-by-step explanation:
First, we need to calculate the number of moles of Silver Nitrate (AgNO3) present in 245 grams:
245 g AgNO3 / 169.88 g/mol AgNO3 = 1.444 mol AgNO3
The balanced chemical equation for the reaction between AgNO3 and Ag3PO4 is:
3 AgNO3 + Ag3PO4 → 3 Ag3PO4 + NO3
We can see that for every 3 moles of AgNO3 reacted, we get 1 mole of Ag3PO4 produced. Therefore, the number of moles of Ag3PO4 produced is:
1.444 mol AgNO3 / 3 mol AgNO3 per 1 mol Ag3PO4 = 0.481 mol Ag3PO4
Finally, we can use the molar mass of Ag3PO4 to convert from moles to grams:
0.481 mol Ag3PO4 × 418.58 g/mol Ag3PO4 = 201.18 g Ag3PO4
Therefore, 245 grams of AgNO3 will produce 201.18 grams of Ag3PO4.