Question

What is the percent yield of the solid product when 18.77 g of iron(III) nitrate reacts...

What is the percent yield of the solid product when 18.77 g of iron(III) nitrate reacts in solution with excess sodium phosphate and 5.446 g of the precipitate is experimentally obtained?

Fe(NO3)3(aq) + Na3PO4(aq) --> FePO4(s) + NaNO3(aq) [unbalanced]

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Answer #1

Fe(NO3)3(aq) + Na3PO4(aq) --> FePO4(s) + 3NaNO3(aq)

1 mole of Fe(NO3)3 react with Na3PO4 to gives 1 mole of FePO4

241.86g of Fe(NO3)3 react with Na3PO4 to gives 150.82g of FePO4

18.77g of Fe(NO3)3 react with Na3PO4 to gives = 150.82*18.77/241.86   = 11.7g of FePO4

percentage yield = Actual Yield*100/theoretical yield

                              = 5.446*100/11.7 = 46.55%

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