Stoichiometry
TOPICS

Everyday Stoichiometry

Simple Stoichiometry

Calculating Amount of Product or Reactant

Limiting Reagent

Percent Yield
Standards Addressed:
Conservation of Matter and Stoichiometry

3a. Students know how to describe chemical reactions by writing balanced
chemical equations

3d.
Students know how to determine the molar mass of a molecule from
its chemical formula and a table of atomic masses an how to convert the
mass of a molecular substance to moles, number of particles, or volume of
gas at STP.

3e. Students know how to calculate the masses of reactants and products
in a chemical reaction from the mass of one of the reactants or products
and the relevant atomic masses.
Definitions

Stoichiometry: the study of mass relationships in
chemical equations.

Excess: more than enough available; won’t run out of
that reactant

Mole-mole ratio: how two substances are numerically
related to each other in units of moles; you get these
numbers from the balanced chemical equation.
Everyday Stoichiometry:

2 wheels + 1 frame → 1 bike

How many wheels and frames do you need to make 5
bikes?

If you have 6 wheels and 4 frames, how many bikes can
you make?

If you have 11 wheels and 3 frames, how many bikes
can you make?
TOPICS

Everyday Stoichiometry

Simple Stoichiometry

Calculating Amount of Product or Reactant

Limiting Reagent

Percent Yield
Simple Stoichiometry:

Here is the “recipe” for water: 2 H2 + O2 → 2 H2O

*This recipe says:

2 molecules of H2 react with 1 molecule of O2 to produce
2 molecules of water

2 moles of H2 react with 1 mole of O2 to produce 2 moles
of water

*The balanced chemical equation gives us the “recipe”
for how reactants combine to form products
TOPICS

Everyday Stoichiometry

Simple Stoichiometry

Calculating Amount of Product or Reactant

Limiting Reagent

Percent Yield
Mole-Mole Ratios




Come from the balanced chemical equation
Ratio made between the “given” substance and
the “get” substance
Use the coefficients of the “given” and “get”
from the balanced chemical equation
Ratios can be between any 2 substances
 Reactant-reactant
 Reactant-product
 Product-product

How do you know between which
substances to make a mole-mole ratio?

Mole-Mole Ratio Examples

2 H 2 + O 2 → 2 H 2O

2 mole H2
2 mole H2O
1 mole O2
2 mole H2
Mole-Mole Ratio Examples
Fe2O3 (s) + 3 CO (g)→ 2 Fe (s) + 3 CO2 (g)
 Write a mol-mol ratio between the reactants…
 Write
a mol-mol ratio between iron and
carbon monoxide
 Write
a mol-mol ratio between the products
Calculating Amount of Reactant
Needed

2 P + 3Cl2 → 2 PCl3

How many moles of phosphorus are consumed if 12
moles of phosphorus trichloride are produced?
How Much Product is Produced?

2 P + 3Cl2 → 2 PCl3

How many moles of PCl3 can be formed from 5 moles
chlorine and excess phosphorus?

How many moles of chlorine are needed
to react with 3.25 moles of phosphorus?
Question

2 H2 + O2 → 2 H2O

How many moles of water are produced by burning 2.72
mol H2 in an excess of O2?
Question

2 H2 + O2 → 2 H2O

How many moles of O2 are consumed in the complete
combustion of 6.86g H2 ?
Question

2 Ag2CO3 (s) → 4 Ag(s) + 2CO2(g) + O2(g)

How many grams of Ag2CO3 must have decomposed if
75.1 grams of Ag were obtained in the reaction?
Question

2 Ag2CO3 (s) → 4 Ag(s) + 2CO2(g) + O2(g)

How many grams of carbon dioxide were produced if
25.0 g of oxygen were produced?
Let’s Summarize the Steps
1.
Is there a balanced chemical equation?
2.
Write down the “given” and the “get”
3.
Mini road map…g-mol-mol-g
4.
Set-up a dimensional analysis equation
5.
Cancel units and calculate
6.
Report answer to correct # of sig. figs.
THE LONGEST
“MINI ROAD MAP”
YOU WILL HAVE TO FOLLOW

GRAMS----MOLES----MOLES----GRAMS
Question

KClO3 + 5 KCl + 6 HNO3 → 6KNO3 + 3Cl2 + 3 H2O

How many grams of KClO3 are required to prepare
10.0g of Cl2 ?
Iron (II) oxide decomposes into iron
and oxygen gas.

2 FeO → 2 Fe + O2

How many grams of FeO are needed to produce 140.0 g
of Fe?
Hydrochloric acid reacts with zinc metal
to produce zinc chloride and hydrogen
gas

2 HCl + 2 Zn → ZnCl2 + H2

How many grams of hydrochloric acid are required to
react completely with 1.00g of zinc?
Topics

Everyday Stoichiometry

Simple Stoichiometry

Calculating Amount of Product or Reactant

Limiting Reagent

Percent Yield
Anticipatory Set

3 cups flour + 2 cups sugar + 1 cup butter + 3 eggs = 1 batch

Flour
13 cups

How many batches of cookies can we make? How did you
Sugar
10 cups
Butter
3 cups
Eggs
20
figure it out?

Flour-4
Sugar-5 Butter-3 Eggs-7, so we can only make 3
batches. The butter yields the smallest amount possible that
can be produced. The butter tells how many bathes we can
make; therefore the butter is the limiting ingredient.
Summarize

Here, we did multiple calculations side by side and
then chose the correct result.

We can use this same type of reasoning/procedure
to calculate the limiting reagent in a given chemical
reaction.
Definition

Limiting reagent: the reactant that is
completely used up in the chemical reaction.
The limiting reagent determines how much
product can be formed.
Calculating Limiting Reactant
 Which reactant will run out first? You can’t tell just by
looking at it, so you need to do some calculations.
 NOTE: In limiting reagent problems, BOTH
REACTANTS are converted to the same PRODUCT
Calculating the Limiting Reagent

P4 + 6 Cl2 → 4 PCl3
What mass of PCl3 forms in the rxn of 125g P4 with 323 g Cl2?
Limiting Reagent

CH4 + 2 O2 → CO2 + 2 H2O

If 12.0g CH4 and 30.0g O2 react to form CO2 and H2O,
how many grams of CO2 will be formed?
Summarize the Steps we followed
1. Is there a balanced chemical equation?
2. Write down the 2 “givens” (reactants) and the “get” (the product)
3. Set-up Dimensional Analysis equations; 1 for each given

g-mol-mol-g
4. Whichever reactant produced the smaller amount of product, that
reactant is the limiting reagent
5. Make your concluding statement
Topics

Everyday Stoichiometry

Simple Stoichiometry

Calculating Amount of Product or Reactant

Limiting Reagent

Percent Yield
Definitions

Theoretical Yield: How much product you expect to get based on
your calculation

Actual Yield: Amount of product actually produced when the reaction
is conducted… this amount is always stated in the problem

Percent Yield: the percentage of theoretical yield obtained from the
reaction
Calculating Percent Yield

% Yield =
actual yield
X 100
theoretical yield
Calculating Percent Yield
Determine the percent yield for the reaction between
98.7g of Sb2S3 and excess oxygen if 72.4 g of
Sb4O6 are recovered along with an unknown
amount of sulfur dioxide gas.
Percent Yield
Determine the percent yield for the reaction
between 46.5 g of ZnS and excess oxygen if
18.4 g of ZnO are recovered along with an
unknown amount of sulfur dioxide gas.
8.7 Enthalpy of a Reaction: A Measure of
Heat Evolved or Absorbed in a Reaction
Enthalpy: The amount of thermal energy
emitted or absorbed by a chemical
reaction, pressure is constant.
 Enthalpy of a Reaction (∆Hrxn): the
amount of thermal energy (or heat) that
flows when a reaction occurs at constant
pressure.

The sign of ∆H indicates the
direction of heat flow.
Negative Sign…heat flows out of the
system into the surroundings…exothermic
reaction
 Positive Sign… heat flows from the
surroundings into the system
…endothermic reaction

Reaction Energy Diagrams
Stoichiometry of ∆Hrxn

Ratios can be made between the moles of
a substance and the enthalpy of the
reaction, ∆Hrxn

g ↔ mol ↔ kJ

Ex 8.7 An LP gas tank in a home BBQ contains 11.8 x 103 g
of propane (C3H8). Calculate the heat in kJ associated with
the complete combustion of all the propane in the tank.

(-5.47 x 105 kJ)
C3H8 (g) + 5 O2(g) → 3 CO2 (g) + 4 H2O (g) ∆Hrxn= -2044 kJ
SB 8.7 Ammonia reacts with oxygen:
4 NH3 (g) + 5 O2(g) → 4 NO (g) + 6 H2O (g)
∆Hrxn= -906 kJ

Calculate the heat in kJ associated with the complete
reaction of 155 g NH3. (-2.06 x 103kJ)


SB+ What mass of butane, in grams, is necessary to
produce 1.5 x 103 kJ of heat? What mass of CO2 is
produced? ( 33 g 99 g CO2)
C4H10 + 13/2 O2 → 4 CO2 (g) + 5 H2O (g) ∆Hrxn= 2658 kJ
C4H10 + 13/2 O2 → 4 CO2 (g) + 5 H2O (g)
∆Hrxn= -2658 kJ

What mass of CO2 is produced?
(99 g CO2)
END