If 125.0mL of .1100M K3PO4 (aq
Answer:
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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