3FeCl2 + 2K3PO4 = 1Fe3(PO4)2 + 6KCl you have 8.107g FeCl2 and 9.15g K3PO4 find l

Question

3FeCl2 + 2K3PO4 = 1Fe3(PO4)2 + 6KCl you have 8.107g FeCl2 and 9.15g K3PO4 find limiting reagent, and calculate the mass of Fe3(PO4)2 expected. LR= Fe .063 mol .063x1mol Fe3(PO4)2=.063 moles of Fe3(PO4)2 how much you have

Explanation / Answer

FeCl2 ; 126.751 g/mol (molar mass) 8.107g FeCl2 -> 8.107/126.751 =0.064 moles K3PO4 : 212.27 g/mol (molar mass) 9.15g K3PO4 -> 9.15/212.27 =0.043 moles 3moles FeCl2 reacts with 2 moles of K3PO4 0.064 moles FeCl2 reacts with 2*0.064/3 =0.0426 moles of K3PO4 limiting reactant is FeCl2 : 3moles FeCl2 gives 1 moles of Fe3(PO4)2 0.064 mol ->1*0.064/3 = 0.0213 mol of Fe3(PO4)2 1mol of Fe3(PO4)2 = 357.48 g 0.0213 mol = 0.0213*357.48 = 7.61 g of Fe3(PO4)2

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