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How many grams (g) are in 0.329 moles of Ba3(PO4)2?

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Answer:

About 203.381 grams

Step-by-step explanation:

We need to first determine the molar mass of Ba3(PO4)2.

We have:

3 Bariums - 137.327*3 = 411.981

2 Potassiums - 39.098*2 = 78.198

8 Oxygens - 16*8 = 128

Add these together and we get 618.179 g/mol

If we only want 0.329 moles, then we multiply those two numbers together to get about 203.381 grams.

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