Limiting Reagent, Solutions 1) Write the balanced equation for the reaction that occurs when iron (II) chloride is mixed with sodium phosphate forming iron (II) phosphate and sodium chloride. 3 FeCl2 + 2 Na3PO4  Fe3(PO4)2 + 6 NaCl 2) If 23 grams of iron (II) chloride reacts with 41 grams of sodium phosphate, what is the limiting reagent? How much sodium chloride can be formed? 23 g FeCl2 x 1 mole FeCl2 x 2 mole Na3PO4 x 163.94 g Na3PO4 = 126.75 g FeCl2 3 mole FeCl2 1 mole Na3PO4 = 20. g Na3PO4 Since we have 41 g Na3PO4, FeCl2 is the limiting reagent. 23 g FeCl2 x 1 mole FeCl2 x 6 mole NaCl x 58.44 g NaCl = 126.75 g FeCl2 3 mole FeCl2 1 mole NaCl = 21 g NaCl 3) How much of the excess reagent remains when this reaction has gone to completion? 41 g Na3PO4 - 20. g Na3PO4 = 21 g Na3PO4 4) If 16.1 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction? 16.1 g NaCl x 100 = 77% 21 g NaCl