you will produce a precipitate of calcium iodate by reacting aqueous solutions o
ID: 530145 • Letter: Y
Question
you will produce a precipitate of calcium iodate by reacting aqueous solutions of potassium iodate and calcium nitrate.
You will react 10 mL of 1 mol L-1 calcium nitrate with 25 mL of 0.2 mol L-1 potassium iodate. What is the percentage yield of the reaction if 0.68 g of calcium iodate is produced?
Record your answer to 1 decimal place, do not include the percentage symbol in your answer.
Hints:
Consider the stoichiometry of the reaction;
Calculate the number of moles of each of the starting materials used;
Explanation / Answer
The balanced chemical equation is
Ca(NO3)2 (aq) + 2 KIO3 (aq) -------> Ca(IO3)2 (s) + 2 KNO3 (aq)
As per the stoichiometric equation above,
1 mole Ca(NO3)2 = 2 moles KIO3 = 1 mole Ca(IO3)2
Millimoles of Ca(NO3)2 added = (10 mL)*(1 mol/L) = 10 mmole.
Millimoles of KIO3 added = (25 mL)*(0.2 mol/L) = 5 mmole.
Now, 10 mmole Ca(NO3)2 = (10 mmole Ca(NO3)2)*(2 mole KIO3/1 mole Ca(NO3)2) = 20 mmole KIO3.
Offocurse, we do not have 20 mmoles KIO3; we have just 5 mmole KIO3.
Therefore, KIO3 is the limiting reactant and the amount of Ca(IO3)2 produced will be governed by KIO3.
5 mmole KIO3 = (5 mmole KIO3)*(1 mole Ca(IO3)2/2 moles KIO3) = 2.50 mmole Ca(IO3)2.
Molar mass of Ca(IO3)2 = (1*40.078 + 2*126.90447 + 6*15.9994) g/mol = 389.88334 g/mol.
Mass of Ca(IO3)2 produced = (2.50 mmole)*(389.88334 g/mol) = 974.70835 mg = (974.70835 mg)*(1 g/1000 mg) = 0.9747 g 0.97 g.
The actual yield of Ca(IO3)2 = 0.68 g.
Percent yield of Ca(IO3)2 = (0.68 g)/(0.97 g)*100 = 70.10 70.1 (ans).
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