Question

You mix together 25.0 mL of a 0.025 M aqueous solution of copper(II) sulfate and 50.0 mL of 0.01 M sodium phosphate and isolate 68.3 mg of a precipitate.

1.Write the balanced equation for this reaction. Label all states.

2.What is the identity of the precipitate?

3.What is the theoretical yield of the reaction?

4.What is the percent yield of the reaction?

You mix together aqueous solutions of 100.0 mL of 0.05 M sodium nitrate and 10.0 mL of potassium phosphate and the reaction proceeds to completion.

1.What is the balanced chemical equation for this reaction? Label all states.

2.What is the concentration of phosphate ion in solution after the reaction?

You mix together 50.0 mL of 0.005 M lead(II) nitrate and 20 mL of YY M ammonium iodide and the reaction proceeds to completion.

1.What is the balanced chemical equation for this reaction? Label all states.

2.What is the concentration of iodide ions in solution after the reaction? This will take more than 1 calculation.

Answer 1

You mix together 25.0 mL of a 0.025 M aqueous solution of copper(II) sulfate and 50.0 mL of 0.01 M sodium phosphate and isolate 68.3 mg of a precipitate.

1.Write the balanced equation for this reaction. Label all states.

CusO4 + Na3PO4 --> NaSO4 + Cu3(PO4)2(s)

balance

3CuSO4 + 2Na3PO4 --> 3Na2SO4 + Cu3(PO4)2(s)

2.What is the identity of the precipitate?

Cu3(PO4)2(s), since phosphtate are not soluble

3.What is the theoretical yield of the reaction?

mmol of CuSO4 = 0.025*25; mmol of Na3PO4 = MV = 50*0.01

mmol of CuSO4 = 0.625

mmol of Na3PO4 = 0.50

verify limiting reaciton

theoreticall -->

0.50 mmol of Na3PO4 required 1/2*0.50 = 0.25 mmol

theoreticall -->

0.625 mmol of CuSO4 required 1/3*0.625= 0.20833 mmol

CuSO4 limits

yield = mass = mol*MW = 0.20833 * 310.1767 = 64.61 mg

4.What is the percent yield of the reaction?

% yield = obtained / Theoretical * 100% = 68.3/64.61*100 = 105% which is not possible, that is, you are obtaining much more mass than it can be obtained