1) What is the percent yield of the solid product when 10.68 g of iron(III) nitr
ID: 1073068 • Letter: 1
Question
1) What is the percent yield of the solid product when 10.68 g of iron(III) nitrate reacts in solution with excess sodium phosphate and 4.603 g of the precipitate is experimentally obtained?
2) What is the experimental yield (in g of precipitate) when 19.3 mL of a 0.6 M solution of iron(III) chloride is combined with 16.3 mL of a 0.512 M solution of lead(II) nitrate at a 83.1% yield?
3) What is the theoretical yield (in g of precipitate) when 18.3 mL of a 0.553 M solution of iron(III) chloride is combined with 18.2 mL of a 0.607 M solution of lead(II) nitrate?
Explanation / Answer
1) Fe(NO3)3(aq) + Na3PO4(aq) ----> FePO4(s) + NaNO3(aq)
no of mole of Fe(NO3)3 =w/Mwt = 10.68/241.9 = 0.0441 mole
no of mole of FePO4 formed = 0.0441 mole
mass of FePO4 formed (theoretical yield) = 0.0441*150.8164 = 6.65 g
practical yield = 4.603 g
% yield = practical yield/ theoretical yield *100
= 4.603/6.65*100
= 69.22%
2) 3Pb(NO3)2(aq) + 2FeCl3(aq) ----> 3PbCl2(s) + 2Fe(NO3)3(aq)
no of mole of Pb(NO3)2 = 19.3*0.6/1000 = 0.0116 mole
no of mole of FeCl3 = 16.3*0.512/1000 = 0.00834 mole
limiting reactant = Pb(NO3)2
theoretical yield of precipitate = 0.0116*278.1 = 3.226 g
% yield = practical yield/ theoretical yield *100
83.1 = x/3.226*100
practical yield = x = 2.68 g
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