STOICHIOMETRY IN SOLUTION CHEMISTRY
Stoichiometry involves calculating the amounts of reactants and products in chemical reactions. Stoichiometry problems in
solution chemistry involve a few additional steps. If a precipitate forms, the net ionic equation may be easier to use than the
chemical equation. Some problems may require you to calculate the amount of a reactant, given the volume and concentration of
the solution.
If an amount in grams is needed, the moles should be multiplied by molar mass. If a concentration is needed, the amount in
moles should be divided by the final volume.
Example:
Calculate the concentration (in mol/L) of chloride ions in each solution.
a)
19.8 g of potassium chloride dissolved in 100 mL of solution.
b)
26.5 g of calcium chloride dissolved in 150 mL of solution.
c)
A mixture of the two solutions in parts (a) and (b), assuming that the volumes are additive.
Solution (a) and (b)
Solution
Molar mass
Amount (mol)
KCl
39.10 + 35.15 =
g
19.8 g x 1 mol = ______ mol KCl
g
KCl(s)  K+(aq) + Cl-(aq)
______ mol x KCl x 1 mol Cl- = ______ mol Cl1 mol KCl
mol = ______mol/L Cl0.100 L
Dissociation equation
Amount of ClConcentration of Cl-
Solution
Molar mass
Amount (mol)
CaCl2
40.08 + ( 2 x 35.45) = _______ g
26.5 g x 1 mol = ______ mol CaCl2
110.98 g
CaCl2(aq)  Ca2+(aq) + _ Cl-(aq)
_______mol CaCl2 x 2 mol Cl- = ______ mol Cl1 mol CaCl2
mol = ______mol/L Cl0.150 L
Dissociation equation
Amount of ClConcentration of Cl-
The concentration of chloride ions when 19.8 g of potassium chloride is dissolved in 100 mL of solution is 2.66 mol/L. The
concentration of chloride ions when 26.5 g of calcium chloride is dissolved in 150 mL of solution is 3.19 mol/L.
Solution (c)
Total amount of Cl-(aq) = _________ mol + _______ mol
= ______ mol ClTotal volume of solution = 0.100 L + 0.150 L
= ________L
Total concentration of Cl-(aq) =
mol
L
= ________ mol/L Cl-
The concentration of chloride ions when the solutions are mixed is ______ mol/L.