9.5 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. Stoichiometry in solution only differs in how the moles
are calculated. We use the amount concentration formula:
n=cxv
Use the following technique when solving solution stoichiometry:
A(aq) +
(Given)
mol
3B(aq) 
(Given)
mol
2C(S)
+
D(aq)
(What quantity are we looking for?)
Step One: Write the quantities given below of the reactants they represent in the reaction
(# 1 on diagram) and use the necessary equation to determine the number of moles.
Step Two: If necessary, decide which reactant is limiting.
Step Three: Use the moles of limiting reactant to decide the moles of product, using a
stoichiometric ratio or mole ratio.
Step Four: Use the amount in moles of the desired product, obtained in step four, to find
an amount in grams, volume or a concentration.
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.
9.5 LIMITING REACTANTS IN SOLUTIONS
Example 1:
If 10.3 L of 1.8 M aqueous copper (II) chloride reacts with 6.3 x 103 mL of 2.30 M sodium
phosphate, calculate the mass of the precipitate produced.
Balanced equation: 3 CuCl2(aq) + 2 Na3PO4(aq) ---> Cu3(PO4)2(s) + 6 NaCl(aq)
(10.3 L) (6.3 x 103 mL)
x
(1.8 mol/L) (2.30 mol/L)
SOLUTION:
1 - Find moles of CuCl2(aq) and Na3PO4(aq)
nCuCl2 = c x v
= 1.8 mol/L x 10.3 L
= 19 mol CuCl2
nNa3PO4 = c x v
= 2.30 mol/L x 6.300 L
= 14.5 mol Na3PO4
2- Determine Limiting Reactant – Compare the mass of the reactant to the
mass of the product
nCu3(PO4)2 = 19 mol CuCl2 (1 mol Cu3(PO4)2) nCu3(PO4)2 = 14.5 mol Na3PO4 (1 mol Cu3(PO4)2)
3 mol CuCl2
2 mol Na3PO4
= 6.3 mol Cu3(PO4)2
= 7.25 mol Cu3(PO4)2
The Limiting Reactant is CuCl2.
3 – Calculate the mass of the precipitate. (You can use the GRASS method)
mCu3(PO4)2 = n x MM
= 6.2 mol x 380.5 g/mol
= 2.4 x 103 g or 2.4 kg
Example 2:
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.
KCl(aq)
19.8 g

Molar mass
Amount (mol)
Amount of ClConcentration of Cl-
K+(aq)
+
Cl-(aq)
cCl = ?
CaCl2(aq)
26.5 g

Ca+(aq)
+
2 Cl-(aq)
cCl = ?
Molar mass
Amount (mol)
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.
Read Page 444-449
Practice Problems: Page 447 PP#1 Page 448 PP#2,3
Review Problems: Page 449 PP#1,3,5,7,9