1. Molarity and StoichiometryV P
mol
M =
mol L
mol = M L
M L
M L
The coefficients in the balanced chemical equation indicate the number of moles of each reactant that is needed and the number of moles of each product that can be produced.
It doesn’t matter if you are dealing with volumes of solutions of reactants or masses of reactants. __ 1
Pb(NO3)2(aq) KI (aq)  PbI2(s) KNO3(aq) 2 1 2
What volume of 4.0 M KI solution is required to yield 89 g PbI2?

2. Stoichiometry for Reactions in Solution
Step 1)
Identify the species present in the combined solution,
and determine what reaction occurs.
Step 2)
Write the balanced net ionic equation for the reaction.
Step 3)
Calculate the moles of reactants.
Step 4)
Determine which reactant is limiting.
Step 5)
Calculate the moles of product or products, as required.
Step 6)
Convert to grams or other units, as required.
Stoichiometry steps for reactions in solution

3. 1 Pb(NO3)2(aq) + 2 KI (aq)  1 PbI2(s) + 2 KNO3(aq)
? L 4.0 M
89 g
What volume of 4.0 M KI solution is required to yield 89 g PbI2?
Strategy:
(1) Find mol KI needed to yield 89 g PbI2.
(2) Based on (1), find volume of 4.0 M KI solution.
1 mol PbI2
2 mol KI
X mol KI = 89 g PbI2
= mol KI
461 g PbI2
1 mol PbI2
M =
mol L
L =
mol M =
0.39 mol KI
4.0 M KI =
0.098 L of 4.0 M KI

4. How many mL of a 0.500 M CuSO4 solution will react
w/excess Al to produce 11.0 g Cu?
Al3+ SO42–
__CuSO4(aq) + __Al (s) 
__Cu(s) +
__Al2(SO4)3(aq) 3
CuSO4(aq) Al (s)  Cu(s) + Al2(SO4)3(aq) 2 3 1
x mol
11 g
1 mol Cu
3 mol CuSO4
X mol CuSO4 = 11 g Cu
= mol CuSO4
63.5 g Cu
3 mol Cu
M =
mol L
L =
mol M
0.173 mol CuSO4
0.500 M CuSO4
= L
1000 mL
0.346 L
= 346 mL
1 L

5. Stoichiometry Problems
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
Cu AgNO3  2Ag Cu(NO3)2
? g
1.5L
0.10M
1.5 L
.10 mol
AgNO3
1 L
1 mol Cu
2 mol
AgNO3
63.55
g Cu
1 mol Cu
= 4.8 g Cu
Courtesy Christy Johannesson

6. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
The concept of limiting reactants applies to reactions that are carried out in solution and reactions that involve pure substances.
If all the reactants but one are present in excess, then the amount of the limiting reactant can be calculated.
When the limiting reactant is not known, one can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation.
Use volumes and concentrations of solutions of reactants to calculate the number of moles of reactants.
Zn HCl  ZnCl H2
79.1 g
0.90 L
2.5M
? L
Courtesy Christy Johannesson

7. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1
g Zn
1 mol Zn
65.39
g Zn
1 mol H2 Zn
22.4 L H2
1 mol
= 27.1 L H2
Courtesy Christy Johannesson

8. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 0.90
2.5 mol
HCl
1 L
1 mol H2
2 mol
HCl
22.4
L H2
1 mol H2
= 25 L H2
Courtesy Christy Johannesson

9. Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
Courtesy Christy Johannesson