Question

What is the theoretical yield (in g of precipitate) when 15 mL of a 0.739 M solution of iron(III) chloride is combined with 17.2 mL of a 0.673 M solution of lead(II) nitrate?

Answer 1

2FeCl₃(aq) + 3Pb(NO₃)₂(aq) 2Fe(NO₃)₃(aq) + 3PbCl₂(s)

From the balanced equation, mole ratio of iron(III) chloride : lead(II) nitrate = 2 : 3

15 mL of a 0.739 M solution of iron(III) chloride = 15 mL x 0.739 M = 11.1 mmol of iron(III) chloride

17.2 mL of a 0.673 M solution of lead(II) nitrate = 17.2 mL x 0.673 M = 11.6 mmol of lead(II) nitrate

Thus, the available mole ratio of iron(III) chloride : lead(II) nitrate = 11.1 : 11.6 1 : 1 = 2 : 2

Hence, lead nitrate is the limiting reagent here as its amount is lower than the stoichiometric requirement.

Now, from the balanced equation, 3 mol of lead nitrate gives the precipitate of 3 mol of lead chloride

Therefore, 11.6 mmol of lead nitrate gives = (3 mol x 11.6 mmol)/3 mol = 11.6 mmol of lead chloride

                                                                                                                 = 11.6 x 10^-3 mol of lead chloride

( 1 mmol = 10^-3 mol)

Molar mass of lead chloride = 278.1 g/mol

Hence, the theoretical yield of the product = 11.6 x 10^-3 mol x 278.1 g/mol = 3.23 g