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Date Posted: 5/16/2008 10:29:06 AM  Status: Live
Help Plz!
Course Textbook Chapter Problem
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Question Details:
In an experiment an unknown salt mixture of N3PO4 • 12H2O(s) and BaCl2 • 2H2O(s) is added to water where they form insoluble Ba3(PO4)2(s). The barium phosphate is collected and weighed and the supernatant liquid is tested to determine the limiting reactant. Then using the stoichiometry for this reation, the mass of the limiting reactant can be determined and hence the salt mixture's composition.

 Equation:
2 N3PO4 • 12H2O + 3 BaCl2 • 2H2O -----> Ba3(PO4)2 + 6 NaCl + 30 H2O                                 (1)

6Na+ + PO3- 4  +  24 H2O + 3 B2+ + 6Cl- + 6H2O ----> Ba3(PO4)2 + 6Na+ + 6Cl- + 30H2O      (2)

2 PO3- 4 + 3 Ba2+ -----> Ba3(PO4)2                                                                           (3)

 
Mass of salt mixture: 1.0137 g
Mass of Ba3(PO4)2 precipitate = 0.3396g
moles of Ba3(PO4)2 = 5.64 x 10-4 moles
Na3PO4 • 12H2O (Mol.Wt. = 380.2)
BaCl2 • 2H2O (Mol.Wt. = 244.2)
Ba3(PO4)2 (Mol.Wt. = 602.0)

Need help finding the salt mixture composition:

 %BaCl2 • 2H2O _____ and %Na3PO4 • 12H2O _____

Answers:

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Date Posted: 6/4/2008 12:08:33 AM  Status: Live
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Response:
(n=moles)

n(Ba2+)=3n(Ba3(PO4)2)=1.69 x 10-3 moles

∴ n(BaCl2)=n(Ba2+)=1.69 x 10-3 moles
    m(BaCl2•2H2O)=n*M=1.69 x 10-3 * 244.2=0.413g
    %(BaCl2•2H2O)=m(BaCl2•2H2O)/m(solution)*100=40.8%

n(PO43-)=2n(Ba3(PO4)2)=1.13 x 10-3moles

∴ n(Na3PO4 • 12H2O)=n(PO43-)=1.13 x 10-3moles
    m(Na3PO4 • 12H2O)=n(Na3PO4 • 12H2O)*M=0.429g
    %(Na3PO4 • 12H2O)=m((Na3PO4 • 12H2O)/m(solution)*100=42.3%




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