Question

3MgCl2(aq) + 2Na3PO4(aq) ------> 6NaCl(aq) + Mg3(PO4)2(s)

1.) If given 29 grams of each reactant, determine the limiting reagent.  

2.) Based on your answer from number 1, what is the theoretical yield of this reaction for Mg3(PO4)2?

3.) Based on your answers from questions 1 and 2, if you were to perform this reaction and generate 20 grams of material, what is the % yield of the reaction?

4.) Based on your answers to the previous 3 questions, when 4 moles of MgCl2 are allowed to react with an excess of Na3PO4 how many moles of NaCl are produced?

Answer 1

3MgCl2(aq) + 2Na3PO4(aq) ------> 6NaCl(aq) + Mg3(PO4)2(s)

no of moles of MgCl2 = W/G.M.Wt

                                     =29/95   = 0.305 moles

no of moles of Na3PO4 = W/G.M.Wt

                                        = 29/164 = 0.177 moles

3 moles of MgCl2 react with 2 moles of Na3PO4

0.305 moles of MgCl2 react with 2*0.305/3    = 0.203 moles ofNa3PO4

Na3PO4 is limiting reagent

2 moles of Na2PO4 react with MgCl2 to gives 1 moles of Mg3(PO4)2

0.177 moles of Na2PO4 react with MgCl2 to gives = 1*0.177/2 = 0.0885 moles of Mg3(PO4)2

mass of Mg3PO4)2 = no of moles * gram molar mass

                                 = 0.0885*262.87   =23.26g