Asked by Mai
Calcium can also be precipitated out of the solution by adding sodium phosphate (Na3PO4) to form solid calcium phosphate.
A.) write a net ionic equation for this process
B.)calculate the mass of this precipitate you would generate from treating 1.000 L of your unknown sample with an excess of sodium phosphate.
A.) write a net ionic equation for this process
B.)calculate the mass of this precipitate you would generate from treating 1.000 L of your unknown sample with an excess of sodium phosphate.
Answers
Answered by
DrBob222
Is this unknown sample Ca^2+. Then
3Ca^2+ + 2PO4^3- --> Ca3(PO4)2(s)
You must know mols Ca^2+ from your unknown.
Then mols Ca^2+ x (2 mols PO4^3-/3 mols Ca^2+) = mols Ca^2+ x 2/3 = ? mols Ca3(PO4)2.
Then g = mols x molar mass.
3Ca^2+ + 2PO4^3- --> Ca3(PO4)2(s)
You must know mols Ca^2+ from your unknown.
Then mols Ca^2+ x (2 mols PO4^3-/3 mols Ca^2+) = mols Ca^2+ x 2/3 = ? mols Ca3(PO4)2.
Then g = mols x molar mass.
Answered by
Mai
@DrBob222: what about the sodium phosphate?
Answered by
DrBob222
Sorry about that. I got carried away with calculating the mass Ca3(PO4)2.
mols
? mols Ca^2+ x (2 mol Na3PO4/3 mol Ca^2+) = ? mols Ca^2+ x 2/3 = ? mols Na3PO4. Does that look familar from above.
g Na3PO4 = mols Na3PO4 x molar mass Na3PO4. And that should look familiar from above, also.
mols
? mols Ca^2+ x (2 mol Na3PO4/3 mol Ca^2+) = ? mols Ca^2+ x 2/3 = ? mols Na3PO4. Does that look familar from above.
g Na3PO4 = mols Na3PO4 x molar mass Na3PO4. And that should look familiar from above, also.
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