SCH3U
SOLUTION STOICHIOMETRY
Solution stoichiometry is often used in quantitative analysis, which involves
determining a quantity (how much) of a substance is present in a sample.
•
Recall that in Stoichiometry the mole ratio provides a necessary conversion factor:
molar mass of x
grams x   moles of x 
molar mass of y
 moles of y   grams of y
We can do something similar for solutions:
mol/L of x
volume x   moles of x 
•
mol/L of y
 moles of y   volume of y
In solution stoichiometry, known volumes and concentrations of reactants or
products are used to determine the volumes, concentrations, or masses of other
reactants or products.
In solution Stoichiometry problems should always include a balanced chemical equation
or the net-ionic equation.
Rules for Solving Solution Stoichiometry Problems
Stoichiometry of Solutions
Example 1
A student dissolved 0.212 mol of iron (III) chloride, FeCl3(s), to make a 175 ml solution.
Find the molar concentration of the solution and the concentration of ions in the solution
FeCl3(aq)Fe3+(aq) + 3Cl-(aq)
Concentration of FeCl3(aq)
c = 0.212 mol = 1.21 mol/L
0.175 L
Molar concentration of Fe3+ = 1.21 mol/L x 1 mol Fe3+ = 1.21 mol/L
1 mol FeCl3
Molar concentration of 3Cl- = 1.21 mol/L x 3 mol Cl- = 3.63 mol/L
1 mol FeCl3
Example 2 (pg. 348 textbook)
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
(a)
(b)
(c)
Total amount of Cl(aq) 0.266 0.478 mol0.744 mol
Total volume of solution 0.100 0.150 L 0.250 L
Total concentration of Cl- (aq) 0.744 mol2.98 mol/L
0.250 L
The concentration of chloride ions when the solutions are mixed is 2.98 mol/L.
Example 3
Given:
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
Volume of AgNO3 solution = 100 mL
Mass of dried Ag(s) precipitate = 1.65 g
What is the molar concentration of silver nitrate solution?
1. Calculate the amount in moles of Ag(s):
n(Ag) = 1.65 g ×
1 mol
= 0.015 296 mol
107.87 g
2. Use moles of Ag to calculate moles of AgNO3:
n =AgNO3 = 2 mol AgNO3 × 0.0015 296 mol Ag = 0.015296 mol AgNO3
2 mol
3. Calculate the concentration of AgNO3:
c = 0.015 296 mol = 0.153 mol/L
0.100 L
Example 4
Calcium hydroxide is sometimes used in water treatment plants to clarify water for
residential use. Calculate the volume of 0.0250 mol/L calcium hydroxide solution that
can be completely reacted with 25.0 mL of 0.125 mol/L aluminum sulfate solution.
Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
Volume of Ca(OH)2 in Liters
0.0250 L Al2(SO4)3 x 0.125 mol Al2(SO4)3 x 3 mol Ca(OH)2 x L Ca(OH)\2 = 0.375 L Ca(OH)2
L Al2(SO4)3
1 mol Al2(SO4)3 0.0250 mol
Example 5
Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What
concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia
solution if 50.0 mL of sulfuric acid is used?
H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Calculate mol H2SO4
# mol H2SO4:
0.0244 L NH3 x 2.20 mol NH3 x 1 mol H2SO4 = 0.02684 mol
1L NH3
2 mol NH3
[H2SO4] in 0.05 L:
0.02684 mol H2SO4 = 0.537 mol/L
0.0500 L
Example 6
A student mixed 100.0 mL of a 0.100 mol/L solution of barium chloride with 100.0 mL of a
0.100 mol/L solution of iron(III) sulphate. The barium sulphate precipitate was filtered, dried, and
was measured to have a mass of 2.0 g. Calculate the % yield of the barium sulfate.
3BaCl2(aq) + Fe2(SO4)3(aq)  3BaSO4(s) + 2FeCl3(aq)
100.0 mL
100.0 mL
m
0.100 mol/L 0.100 mol/L
The theoretical mass of barium sulphate is 2.3 g.
The % yield of the barium sulphate precipitate was 87%.