# If 125.0mL of .1100M K3PO4 (aq

If125.0mL of .1100M K3PO4 (aq) and 150.0 mL of .09500M FeCl2 (aq) arecombined, what mass of Fe3(PO4)2 (s) can potentially be recoveredform the reaction mixture after the reaction is complete? Indicatewhich is the limiting reagent and which is the excess. What is theconcentration of the excess reagent that remains in solution whenthe reaction is complete?

The reaction is

2 K3PO4 (aq) + 3 FeCL2 (aq) —————-> Fe3(PO4)2(s) +6KCl(aq)

125x 0.11 =13.75 150×0.095=14.25 0 0 initial mmoles

To know what is the limiting reagent we find the ratio of molesrequired.

13.75/2 = 6.875 14.25/3=4.75

As the ratio of FeCl2 is less, it is the limiting reagent

2 K3PO4 (aq) + 3 FeCL2 (aq) —————->Fe3(PO4)2(s) + 6KCl(aq)

125x 0.11 =13.75 150×0.095=14.25 0 0 initial mmoles

13.75- (2×14.25/3) 0 14.25/3=4.75 – after reaction

= 4.25 0 4.75 – after reaction

1) mmoles of Fe3(PO4)2 formed = 4.75

thus moles formed = 4.75×10-3 mol

mass of  Fe3(PO4)2 formed = moles x molar mass

=4.75×10-3 mol x 357.5g/mol

= 1.698 g

2) The mmoles of excess reagent remained after reaction =4.25

total volume of solution after reaction = 125+150 = 275 mL

Thus concentration of excess reagent after reaction is complete= mmoles /volume mL

= 4.25/275

=0.01545 M

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