# Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g. Experimental tests revealed that Na3PO4 • 12H2O (molar mass = 380.12 g/mol) was the limiting reactant in the formation of the precipitate and the BaCl2 • 2H2O was the excess reactant in the salt mixture. Determine the mass of Na3PO4 • 12H2O in the salt mixture.

###### FREE Expert Solution
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###### FREE Expert Solution
• First step is to write the balanced equation, we need to find the other molecules produced when Na3PO4 • 12H2O is reacted with BaCl2 2H2O.

• In the process NaCl and H2O is produced along with Ba3(PO4)2

82% (499 ratings)
###### Problem Details

The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g. Experimental tests revealed that Na3PO4 • 12H2O (molar mass = 380.12 g/mol) was the limiting reactant in the formation of the precipitate and the BaCl2 • 2H2O was the excess reactant in the salt mixture. Determine the mass of Na3PO4 • 12H2O in the salt mixture.

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.

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Our tutors rated the difficulty ofThe Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that f...as medium difficulty.

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Our expert Chemistry tutor, Dasha took 7 minutes and 14 seconds to solve this problem. You can follow their steps in the video explanation above.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Boyd's class at UT.