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0.2 mol of Na3PO4 and 0.5 mol of Ba(NO3)2 are mixed in 1 L of solution. Which of the following statements are correct? (Reaction: 2 Na3PO4 + 3 Ba(NO3)2 -> Ba3(PO4)2 (precipitate) + 6 NaNO3)

1. 0.2 mol of Ba3(PO4)2 is obtained
2. 0.1 mol of Ba3(PO4)2 is obtained
3. Molarity of Ba3(PO4)2 in solution is 0.1 M
4. Molarity of Ba3(PO4)2 in solution is 0.2 M

Correct answer: 0.1 mol of Ba3(PO4)2 is obtained

Solution

Balanced reaction: 2 Na3PO4 + 3 Ba(NO3)2 -> Ba3(PO4)2 + 6 NaNO3. For 0.2 mol Na3PO4, need 0.3 mol Ba(NO3)2. Since 0.5 mol Ba(NO3)2 is available (excess), Na3PO4 is limiting. Moles of Ba3(PO4)2 = 0.2/2 = 0.1 mol. Since Ba3(PO4)2 precipitates, its molarity in solution is essentially 0 (it's not dissolved). However, if the question asks about the amount produced: 0.1 mol.

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