Studies of a Precipitation Reaction
Calculations
Predicting the theoretical yield of a precipitate:
Step 1. Write and balance your equation. You did this in step I.B. last week and the balanced equation
you predicted is either
Ca(NO3)2(aq) + 2 KIO3(aq)  Ca(IO3)2(s) + 2 KNO3(aq)
or
Ca(NO3)2(aq) + 2 KIO3(aq)  Ca(IO3)2(s) + 2 KNO3(aq),
Depending on whether you assumed this reaction is reversible or irreversible.
Step 2. Calculate the moles of reactant used.
The units of your stock Ca(NO3)2 and KIO3 solutions were “M”. “M” is the abbreviation for the unit
molarity which means moles/L. If you multiply moles per liter by the number of liters used, you can find
the number of moles. For example, for solution 1
25 mL (1 L/1000 mL) = 0.025 L, and
0.1 moles/L( 0.025 L) = 0.0025 moles or 2.5 x 10-3 moles.
Calculate the moles of Ca(NO3)2(aq) and KIO3(aq) used in solution 2 as illustrated above.
Step 3. Convert moles of reactant(s) to moles of product(s).
You predicted in section I.B. last week that your precipitate is Ca(IO3)2(s). Based on your balanced
equation (given above) you should obtain 1 mole Ca(IO3)2(s) for every 1 mole of Ca(NO3)2 that
reacts and 1 mole of Ca(IO3)2(s) for every 2 moles of KIO3 that reacts. In other words your
stoichiometric factors are 1 Ca(IO3)2(s)/1 Ca(NO3)2 and 1 Ca(IO3)2(s)/ 2 KIO3.
You should estimate the moles of Ca(IO3)2(s) you can theoretically precipitate by converting
both the moles of Ca(NO3)2 and KIO3 reacted to the moles Ca(IO3)2(s) produced using these
stoichiometric factors. For example, for solution 1
2.5 x 10-3 moles Ca(NO3)2 * (1 Ca(IO3)2(s)/1 Ca(NO3)2) = 2.5 x 10-3 moles Ca(IO3)2(s)
and 2.5 x 10-3 moles KIO3 * (1 Ca(IO3)2(s)/2 KIO3) = 1.25 x 10-3 moles Ca(IO3)2(s)
Calculate the moles of Ca(IO3)2(s) produced from solution 2 based on the concentrations of both
Ca(NO3)2 and KIO3 in solution 2 as illustrated above. You should have calculated two different
predictions for solution 2.
Step 4: Determine the limiting reactant
For solution 1 the calculation which produced the lowest number of moles of Ca(IO3)2(s) tells
you which reactant (Ca(NO3)2 or KIO3) is limiting in this case. For example in the previous
calculations fewer moles of Ca(IO3)2(s) were produced based on the number of moles of KIO3 in
solution 1 and KIO3 was therefore the limiting reactant in solution 1.
For solution 2 the calculation which produced the lowest number of moles of Ca(IO3)2(s) tells
you which reactant (Ca(NO3)2 or KIO3) is limiting in this case. You may not obtain the same
answer for solutions 1 and 2.
Step 5: Convert moles of reactant to moles of product based on the limiting reactant and
calculate the theoretical yield.
Calculate the molar mass of Ca(IO3)2(s), and multiply the moles of Ca(IO3)2(s) you predicted
based on the limiting reactants for solutions 1 and 2 by the molar mass of Ca(IO3)2(s). These
values are the theoretical yield for each of these solutions.
Step 6: Calculate the % yield
% yield = (actual yield in grams/theoretical yield in grams) *100.