1. Solution Stoichiometry

2. Solution Stoichiometry
The major difference compared to gravimetric and gas stoichiometry is that we use molarity (mol/L) as a conversion factor rather than molar mass or molar volume
Solution stoichiometry – the procedure for calculating the molar concentration or volume of solution products or reactants

3. Solution Stoichiometry Steps
Write a balanced chemical equation.
Convert the given measurement to moles using molarity as a conversion factor.
Use the mole ratio from the balanced chemical equation to calculate moles of unknown.
Convert the moles of unknown to the final quantity requested using the appropriate conversion factor.

4. Example #1
Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer. What volume of 14.8mol/L NH3(aq) is needed to react with 1000 L of 12.9mol/L of H3PO4(aq)?
2NH3(aq) H3PO4(aq)  (NH4)2HPO4(aq)
V = ? V = 1000 L
M = 14.8mol/L M = 12.9 mol/L
1000 L H3PO4 x mol H3PO4 x
1 L H3PO4
2 mol NH3 x
1 mol H3PO4
1 L NH =
14.8 mol NH3
1.74 x 103 L

5. Example #2
In an experiment, a mL sample of sulfuric acid solution reacts completely with 15.9 mL of mol/L potassium hydroxide. Calculate the molarity of the sulfuric acid.
H2SO4(aq) KOH(aq)  2H2O(l) + K2SO4(aq)
V = mL V = 15.9 mL
M = ? M = mol/L
L KOH x mol KOH x
1 L KOH
1 mol H2SO4 =
2 mol KOH
mol H2SO4
M = mol = L
mol H2SO4 =
L H2SO4
0.119 M H2SO4