Sodium phosphate dodecahydate reacts with barium chloride dihydrate. If Na2SO4 is an unkown contaminant of sodium phosphate dodecahydrate, how does it presence affect yeild of sodium phosphate in the exexpiriment?

Question

Sodium phosphate dodecahydate reacts with barium chloride dihydrate. If Na2SO4 is an unkown contaminant of sodium phosphate dodecahydrate, how does it presence affect yeild of sodium phosphate in the exexpiriment?

Answer 1

To determine how the presence of Na2SO4 affects the yield of sodium phosphate in the experiment, you need to understand the reaction that takes place between sodium phosphate dodecahydrate (Na3PO4·12H2O) and barium chloride dihydrate (BaCl2·2H2O).

The balanced chemical equation for this reaction is:
3Na3PO4·12H2O + 2BaCl2·2H2O → Ba3(PO4)2 + 6NaCl + 30H2O

In this reaction, sodium phosphate dodecahydrate (Na3PO4·12H2O) reacts with barium chloride dihydrate (BaCl2·2H2O) to form barium phosphate (Ba3(PO4)2) and sodium chloride (NaCl) as products.

Now, let's consider the presence of Na2SO4 as an unknown contaminant in the sodium phosphate dodecahydrate. The equation for the reaction between Na2SO4 and BaCl2 is:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl

In this reaction, Na2SO4 reacts with BaCl2 to form BaSO4 (barium sulfate) and NaCl. Note that Na2SO4 is not involved in the production of sodium phosphate.

Therefore, the presence of Na2SO4 as a contaminant does not directly affect the yield of sodium phosphate in the experiment. The reaction between sodium phosphate dodecahydrate and barium chloride dihydrate solely depends on the presence of Na3PO4 and BaCl2, and the resulting products are barium phosphate and sodium chloride.

Answer 2

Explain bot has it wrong. First, the equation between Na3PO4 and BaCl2 is not balanced correctly. While it is true that the addition of Na2SO4 may not DIRECTLY affect the mass of Na3PO4 it WILL affect the final mass Ba3(PO4)2 obtain BECAUSE there is less Na3PO4 initially. Follow through this. If the sample of Na3PO4 is contaminated with Na2SO4 that means there is less Na3PO4 in the sample initially; i.e., a 10 gram sample of pure Na3PO4 will yield 18.36 g Ba3(PO4)2 assuming the yield is 100%. If you take away Na3PO4 (say 1 g) and replace that with 1 g Na2SO4, the the 9 g Na3PO4 sample will yield 16.52 g Ba3(PO4)2. The 1 g Na2SO4 will yield 1.64 g BaSO4. The total of BaSO4 + Ba3(PO4)2 is 1.64 + 16.52 is 18.16 and that is less than 18.36 by a small amount. So it DOES matter.