1. Reactions in Aqueous Solution
Double Replacement:
(precipitate)
2 KNO3(aq) + PbI2(s)
2 KI(aq) + Pb(NO3)2(aq) 
Cu(s) + 2 AgNO3(aq) 
2 Ag(s) + Cu(NO3)2(aq)
Single Replacement:

2. Solution Stoichiometry
Rxn: A(aq) + 2 B(aq)  C(aq) + 2 D(aq)
molar mass A
molarity A (M)
g A
mol A
L of A
g A
1 mol A
mol A
1 L
mol-to-mol ratio
g B
1 mol B
mol B
1 L
molar mass B
molarity B (M)
g B
mol B
L of B 2

3. Example #1: L of A  L of B 2 NH3(aq) + H3PO4(aq)  (NH4)2HPO4(aq)
Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer.
What volume of 14.8 M NH3 is needed to react with 120. L of 12.9 M of H3PO4 ?
2 NH3(aq) H3PO4(aq)  (NH4)2HPO4(aq)
V = ? L
V = 120. L
M = 14.8 mol/L
M = 12.9 mol/L
molarity B
12.9
mol H3PO4 2
mol NH3 1
L NH3 x x
120 L H3PO4 x 1
L H3PO4 1
mol H3PO4
14.8
mol NH3
L of A
molarity A
= ____L NH3
209 L NH3

4. Example #2: L of A  M of B
A 159 mL sample of KOH solution reacts completely with 100. mL of 1.50 M H2SO4 . Calculate the molarity of the of the KOH solution. H2SO4(aq) + 2 KOH(aq)  2 H2O(l) + K2SO4(aq)
V = L
V = L
M = 1.50 mol/L
M = ? mol/L
1.50
mol H2SO4 2
mol KOH
mol KOH x =
0.100 L H2SO4 x 1
L H2SO4 1
mol H2SO4
L of A
molarity A
molarity B
0.300 mol KOH
= _____ M KOH
1.89 M KOH
0.159 L KOH

5. Example #3: g of A  L of B
Solid aluminum is added to an aqueous solution of zinc chloride. What volume of 7.50 M ZnCl2(aq) solution is needed to react completely with grams of Al(s) ?
2 Al(s) ZnCl2(aq)  2 AlCl3(aq) Zn(s)
m = 54.0 g
V = ? L
M = 7.50 mol/L
molarity B 1
mol Al 3
mol ZnCl2 1
L ZnCl2 x x
54.0 g Al x
26.98
g Al 2
mol Al
7.50
mol ZnCl2
g of A
molar mass A
= ____ L ZnCl2
0.400 L ZnCl2

6. Example #4: L of A  g of B
Adding 102 mL of M CuCl2(aq) solution to an excess of AgNO3(aq) forms solid AgCl(s) precipitate. What is the maximum mass of AgCl(s) precipitate? CuCl2(aq) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 AgCl(s)
V = L
m = ? g
M = mol/L
molar mass B
0.508
mol CuCl2 2
mol AgCl x
143.32
g AgCl
0.102 L CuCl2 x x 1
L CuCl2 1
mol CuCl2 1
mol AgCl
molarity A
14.9 g AgCl
= ____ g AgCl