Percent Yield following limiting reactant

Question

During an experiment, a student reacted 1.785 g of trisodium phosphate with 20.00 mL of 0.678 M copper (II) chloride to produce solid copper (II) phosphate and aqueous sodium chloride. Determine the percent yield for the above reaction if 1.324 g of copper (II) phosphate were  produced

Answer 1

Hi Duch,

This is a multi-step problem. First you need to determine which compound is the limiting reagent, so you can use the amount of that compound to determine how much product there should be. You also need to know how to write this reaction, complete with correct chemical formulas and balancing the equation:

2 Na3PO4 + 3 CuCl2 --> Cu3(PO4)2 + 6 NaCl

By using Stoichiometry, you can figure out that if you use 1.785 g of Na3PO4, you will need .0163 moles of CuCl2 to run this reaction fully.

By using the Molarity and knowing you have 20ml of CuCl2, you can conclude that you have access to more than enough CuCl2 to run this reaction. So, Na3PO4 is the limiting reagent.

We will use the amount of Na3PO4 we have to determine how much Cu3(PO4)2 we should make.

Again, by using Stoichiometry, you can conclude that 2.08 g should have been produced theoretically, but only 1.324g were produced. Which is a 64% yield.