More Stoichiometry!
II. Stoichiometry in the
Real World
* Limiting
Reagents
A. Limiting Reactants
X
Available
Ingredients
• 4 slices of bread
• 1 jar of peanut butter
• 1/2 jar of jelly
Limiting
Reactant
• bread
Excess
Reactants
• peanut butter and jelly
Jam Sandwich Equation
X
2 loaves bread + 1 jar of jam  20 sandwiches

Mole Ratio
2 : 1 : 20

What would happen if we had 1 loaf of bread and
1 jar of jam?
• Bread = limiting reactant
• Jam = excess reactant (1/2 jar left)
• Produce 10 sandwiches
Limiting
Reactant
• used up in a reaction
• determines the amount of
product
Excess
Reactant
• added to ensure that the other
reactant is completely used up
• cheaper & easier to recycle
Example 1

Iron burns in air to form a solid black oxide
(FeO). If you had 50.0 g of iron and 60.0 g of
oxygen, what is the limiting and excess
reactant? How much iron oxide could be
produced?
2 Fe (s) + O2 (g)  2 FeO (s)
2 Fe (s) +
mass = 50 g
O2 (g)
mass = 60 g

2 FeO (s)
mass = ?
Example 2
 79.1
g of zinc react with 0.90 L of
2.5M HCl. Identify the limiting and
excess reactants. How many
grams of zinc chloride would be
produced?
Zn + 2HCl  ZnCl2 + H2
mass = ?
79.1 g 0.90 L
2.5M
Zn +
mass = 79.1 g
2 HCl
V = 0.90 L
of 2.5M

ZnCl2 + H2
mass = ?
Example 2 Results
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: x g ZnCl2
left over zinc
Student Example

Sodium carbonate is needed in the manufacturing of glass,
but very little occurs naturally. It can be made from the
double replacement reaction between calcium carbonate and
sodium chloride. If you had 5.00 g of each what is the
limiting and excess reactant? How much sodium carbonate
would be formed?
CaCO3 + 2 NaCl  CaCl2 + Na2CO3
CaCO3
mass = 5.0 g
+ 2 NaCl 
mass = 5.0 g
CaCl2 + Na2CO3
mass = ?
Assignment
Stoichiometry: Limiting Reagent
Worksheet
B. Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
Ivan
Buz
When
45.8 g of K2CO3 react
with excess HCl, 46.3 g of KCl
are formed. Calculate the
theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Example
K2CO3 + 2HCl 
mass = ?
actual: 46.3 g
mass = 45.8 g
# moles =
# moles =
2KCl + H2O + CO2
sm
mm
45.8 g
138.2 g/mol
# moles = 0.331 mol
# moles = 0.662 mol
mass = (# mol) (mm)
mass = (0.662 mol) (74.6 g/mol)
mass KCl = 49.4 g
Theoretical Yield:
Example Continued
Theoretical Yield = 49.4 g KCl
% Yield =
=
Actual Yield
Theoretical Yield
46.3 g
49.4 g
= 93.7 %
 100 %
 100 %
Assignment

Pg 295
# 2, 5, 7, 11, 12, 15,
18, 20, 25 - 30