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Question

A 1.20 g sample of KOH and 2.50 g of FeCl3, are dissolved in water, then mixed together. What mass of Fe(OH)3 solid is formed? Aqueous KCl is the other product. If the actual yield of Fe(OH)3 is 0.681 g, what is the percent yield?

Answer 1

First thing, write the correctly balanced equation:

3KOH(aq) + FeCl₃(aq) ==> Fe(OH)₃(s) + 3KCl(aq) ... balanced equation

1.20 g............2.50 g.................? g............................

Next, find which reactant, if any, is limiting. We do this by finding the mass of Fe(OH)3 produced from each reactant:

1.20 g KOH x 1 mol KOH/56.1 g x 1 mol Fe(OH)₃ / 3 mol KOH x 107 g Fe(OH)₃/ mol = 0.763 g Fe(OH)₃

2.50 g FeCl₃ x 1 nol FeCl₃/162 g x 1 mol Fe(OH)₃ / 1 mol FeCl₃ x 107 g Fe(OH)₃/mol = 1.65 g Fe(OH)₃

Therefore, KOH is the limiting reactant and it will determine how much Fe(OH)₃ will form.

Mass of Fe(OH)₃ that will form = 0.763 g

Percent yield = actual yield/theoretical yield (x100%) = 0.681 g/0.763 g (x100%) = 89.3% yield