
Nearly everything we use is
manufactured from chemicals.
›Soaps, shampoos, conditioners,
cd’s, cosmetics, medications, and
clothes.
For a manufacturer to make a
profit the cost of making any of
these items can’t be more than
the money paid for them.
 Chemical processes carried out in
industry must be economical, this
is where balanced equations help.

 Equations
are a chemist’s recipe.
◦ They tell chemists what amounts of
reactants to mix and what amounts
of products to expect.
 When
you know the quantity of
one substance in a reaction, you
can calculate the quantity of any
other substance consumed or
created in the rxn.
◦ Quantity meaning the amount of a
substance in grams, liters,
molecules, or moles.

YouTube - How an Airbag works

The calculation of quantities in
chemical reactions is called
stoichiometry.
 Assume
that the major
components of the bike are the
frame (F), the seat (S), the wheels
(W), the handlebars (H), and the
pedals (P).
 The finished bike has a “formula”
of FSW2HP2.
 The balanced equation for the
production of 1 bike is.
F +S+2W+H+2P  FSW2HP2


Now in a 5 day workweek, A company is
scheduled to make 640 bikes. How many
wheels should be in the plant on Monday
morning to make these bikes?
What do we know?
◦ Number of bikes = 640 bikes
• F +S+2W+H+2P  FSW2HP2
◦ 1 FSW2HP2=2W (balanced eqn)

What is unknown?
◦ # of wheels = ? wheels

The connection between wheels and
bikes is 2 wheels per bike. We can
use this information as a conversion
factor to do the calculation.
640 FSW2HP2
2W
1 FSW2HP2
= 1280
wheels
• We can make the same kinds of
connections from a chemical rxn eqn.
N2(g) + 3H2(g)  2NH3(g)
• The key is the “coefficient ratio”.
◦ The coefficients of the balanced chemical equation
indicate the numbers of moles of reactants and
products in a chemical rxn.
N2(g) + 3H2(g)  2NH3(g)


1 mole of N2 reacts with 3 moles of H2 to
produce 2 moles of NH3.
◦ N2 and H2 will always react to form ammonia
in this 1:3:2 ratio of moles.
So if you started with 10 moles of N2 it would
take 30 moles of H2 and would produce 20
moles of NH3

Using the coefficients, from the balanced
equation to make connections between
reactants and products, is the most
important information that a rxn equation
provides.
› Using this information, you can calculate
the amounts of the reactants involved and
the amount of product you might expect.
› Any calculation done with the next
process is a theoretical number, the real
world isn’t always perfect.
 The
following rxn shows the
synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
• If you only had 1.8 mols of Al how
much product could you make?
Given: 1.8 moles of Al
Uknown: ____ moles of Al2O3
 Solve
for the unknown:
3O2(g) + 4Al(s)  2Al2O3(s)
1.8 mol Al
2 mol Al2O3
4 mol Al
Mole Ratio
= 0.90mol
Al2O3
 The
following rxn shows the
synthesis of aluminum oxide.
3O2(g) + 4Al(s)  2Al2O3(s)
• If you wanted to produce 24 mols of
product how many mols of each
reactant would you need?
Given: 24 moles of Al2O3
Uknown: ____ moles of Al
____ moles of O2
 Solve
for the unknowns:
3O2(g) + 4Al(s)  2Al2O3(s)
24 mol Al2O3
24 mol Al2O3
4 mol Al
2 mol Al2O3
3 mol O2
2 mol Al2O3
= 48 mol Al
= 36 mol O2

1. How many moles of potassium chlorate,
KClO3 are needed to produce 5 mol O2 given
the following BALANCED EQUATION?
2KClO3  2 KCl + 3O2

2. Sodium metal reacts with chlorine gas to
produce sodium chloride. Write a balanced
chemical equation. If 3.75 mol Na react with
enough chlorine gas, how much sodium
chloride is produced?
No lab balance measures moles
directly, generally mass is the unit
of choice.
 From the mass of 1 reactant or
product, the mass of any other
reactant or product in a given
chemical equation can be
calculated, provided you have a
balanced rxn equation.
 As in mole-mole calcs, the
unknown can be either a reactant
or a product.

Acetylene gas (C2H2) is produced by
adding water to calcium carbide (CaC2).
CaC2 + 2H2O  C2H2 + Ca(OH)2
How many grams of C2H2 are produced
by adding water to 5.00 g CaC2?


CaC2 + 2H2O  C2H2 + Ca(OH)2
What do we know?
◦ Given mass = 5.0 g CaC2
◦ Mole ratio: 1 mol CaC2 = 1 mol C2H2 (from
balanced equation)
◦ Molar Mass (MM) of CaC2 = 64.0 g/mol CaC2
◦ Molar Mass of C2H2 = 26.0g/mol C2H2
What are we asked for?
◦ grams of C2H2 produced
mass A  moles A  moles B  mass B

What mass of copper is required to react with
4.00g of silver nitrate ?
Cu + 2AgNO3  2Ag + Cu(NO3)2
How many grams of O2 are produced when a
sample of 29.2 g of H2O is decomposed by
electrolysis according to this balanced equation:
2H2O  2H2 + O2
Aspirin can be made from a chemical rxn between
the reactants salicylic acid and acetic anhydride.
The products of the rxn are acetyl-salicylic acid
(aspirin) and acetic acid (vinegar). Our factory
makes 125,000 100-count bottles of Bayer
Aspirin/day. Each bottle contains 100 tablets, and
each tablet contains 325mg of aspirin. How much
in kgs + 10% for production problems, of each
reactant must we have in order to meet
production?
C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
Salicylic
acid
Acetic
anhydride
aspirin
vinegar
C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2


What are we asked for?
◦ Mass of salicylic acid in kgs + 10%
◦ Mass of acetic anhydride in kgs + 10%
What do we know?
◦ Make 125,000 aspirin bottles/day
◦ 100 aspirin/bottle
◦ 325 mg aspirin/tablet
◦ Mole ratio of aspirin to salicylic acid (1:1) and
acetic anhydride (1:1)
◦ MM aspirin = 180.11g/mol
◦ MM C7H6O3 = 138.10g/mol
◦ MM C4H6O3 = 102.06g / mol
How many grams of aspirin do I need to produce?
C7H6O3 + C4H6O3  C9H8O4 + HC2H3O2
125,000
bottles
100 tablets
1 bottle
325mg asp.
1 tablet
1g
1000 mg





Limiting reactant (reagent): reactant that
determines the amount of product that can be
formed in the reaction.
Excess reactant (reagent): reactant that is not
completely used up in a reaction.
Theoretical yield: The maximum quantity of
product that a reaction could theoretically
make (calculated based upon limiting reactant).
Actual yield : The amount of product that was
obtained experimentally. This is the amount
you really got.
Percent Yield = Actual Yield
x 100
Theoretical Yield
Fe + CuSO4  Cu + FeSO4
 g Cu produced with:
12.5g CuSO4

2.2 g Fe
Limiting reactant:
Theoretical yield:
Actual yield:
Percent yield:
Calculate how much product (Cu2S) each amount
of reactant produces respectively.