Stoichiometry – Ch. 11
I.
Stoichiometric
Calculations
Background on things you
NEED to know how to do:
1.
2.
3.
4.
5.
Name/write correct chemical
formula
Write chemical equations
Balance chemical equations
Predict Products
Mole/mass conversions
Stoichiometry
o
o
o
o
Stoichiometry uses ratios to determine relative
amounts of reactants or products.
For example If you were to make a bicycle,
you would need one frame and two tires.
1 frame + 2 tires  1 bicycle
If I had 74 tires, what is the most # of bicycles
I could make?
74 tires
1 bicycle
2 tires
= 37 bicycles
Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I
have 5 eggs. How many cookies
can I make?
Ratio of eggs to cookies
5 eggs 5 doz.
2 eggs
= 12.5 dozen cookies
Proportional Relationships
 Stoichiometry
• mass relationships between substances in a
chemical reaction
• based on the mole ratio
 Mole
Ratio
• indicated by coefficients in a balanced
equation
• can be used to determine expected amounts
of products given amounts of reactants.
2 Mg + O2  2 MgO
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
• Mole ratio - moles
molesmoles
moles
• Molar mass - moles  grams
• Molarity moles  liters soln
• Molar volume - moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
6.02 
MOLES
1023
particles/mol
Molarity (mol/L)
LITERS
OF
SOLUTION
NUMBER
OF
PARTICLES
Mole – Mole Conversions

The first type of problems we encounter will
go between moles and moles. For this we
need to use mole ratios.

Ex: Write and balance the reaction between
lead (II) nitrate and potasium iodide.
Pb(NO3)2 + 2KI  2 KNO3 + PbI2
Mole ratio of potasium iodide to lead (II) iodide:
2 moles KI
1 mole PbI2
Mole to Mole Problems
 How
many moles of KClO3 must
decompose in order to produce 9
moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol O2 2 mol KClO3
3 mol O2
9 mol
= 6 mol
KClO3
Mole to Mass
 We
can also convert from moles to mass,
and mass to moles
 For Example:
• 4 Al + 3 O2  2Al2O3
If you know how many grams of Al you start
with, we can write a flow chart to show how to
calculate the # of moles of oxygen need to fully
react with the Al.
 g Al  moles Al  moles of oxygen

Mass to Moles: 4 Al + 3 O2 

2Al2O3
If the reaction starts with .84 moles of aluminum,
how many grams of aluminum oxide are produced?
.84 mol Al
2 mol Al2O3 101.9 grams Al O
2 3
1 mol Al2O3
4 mol Al
= 42.8 grams Al2O3

0.92 g of Aluminum oxide are produced from the
reaction. How much aluminum was used up?
.92 g Al2O3
1 mol Al2O3
4 mol Al
26.9 g Al
101.9 g Al2O3 2 mol Al2O3 1 mol Al
= .49 grams Al
Mass to Mass
 How
many grams of silver will be
formed from 12.0 g copper reacting
with silver nitrate?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
?g
12.0 1 mol 2 mol 107.87
g Cu Cu
Ag
g Ag
= 40.7 g
63.55 1 mol 1 mol
Ag
g Cu
Cu
Ag
Stoichiometry Problems – Mole/Mass
 In
photosynthesis, carbon dioxide and
water react to form glucose, C6H12O6 and
oxygen gas.
___CO
6 2
6 2O  ___C6H12O6 + ___O
6
2 + ___H
If 15.6 grams of carbon dioxide react, how
many moles of glucose will be produced?
15.6 g CO2
1 mol CO2
1 mol C2 H12O6
44.01 g CO2 6 mol CO2
= 0.0591 mol
C2H12O6
How many grams of carbon dioxide must
react to produce 0.25 moles of glucose?
0.25 mol C2 H12O6
6 mol CO2
44.01 g CO2
1 mol C2 H12O6 1 mol CO
2
= 66 g
CO2
Stoichiometry with Gases
If
the pressure and
temperature are constant,
the ratio of moles in the
balanced equation is the
ratio of liters in an all gas
reaction.
Molar Volume at STP
At STP
1 mol of a gas=22.4 L
Standard Temperature
&
0°C and 1 atm
Pressure
Molar Volume

Hydrogen and chlorine gas react to produce
hydrochloric acid. If 7.00 L of hydrogen gas
react, how many liters of HCl gas are formed?
H2 (g) + Cl2 (g)  2 HCl (g)
7.00 L H2 2.0 L HCl
1.0 L H2
= 14.0 L HCl
*only in all gas
reactions!
Molar Volume
 In
the following reaction, if 17 g of Mg
react, how many L of H2 forms?
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
17.0 g
Mg
1 mol
Mg
1 mol
=
15.7
L
H2 22.4 L H2
24.31 g 1 mol
Mg
Mg
1 mol
H2
H2
Molar Volume Problems
many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
 How
2KClO3  2KCl + 3O2
?g
9.00 L
9.00 L
O2
1 mol
O2
2 mol 122.55
KClO3 g KClO3
22.4 L
O2
3 mol
O2
1 mol
KClO3
= 32.8 g
KClO3
Molar Volume Problems
 How
many grams of Cu are required
to react with 1.5 L of 0.10M AgNO3?
Cu + 2AgNO3  2Ag + Cu(NO3)2
?g
1.5L
0.10M
1.5 .10 mol 1 mol 63.55
L AgNO3
Cu
g Cu
1L
= 4.8 g
2 mol 1 mol
Cu
AgNO3 Cu
Molarity
 Molarity
is the number of moles of
solute dissolved in one liter of
solution.
 Units are moles per liter or moles of
solute per liter of solution.
 Molarity abbreviated by a capital M
 Molarity = moles of solute
liter of solution
Molarity Problems

As an example, suppose we dissolve 23 g of
ammonium chloride (NH4Cl) in enough water to
make 145 mL of solution. What is the molarity
of ammonium chloride in this solution?
23 g NH4Cl
1 mole NH4Cl
= .43 mol NH4Cl
53.5 g NH4Cl
145 mL
1 L
= .145 L
1000 mL
.43 mol NH4Cl
.145 L
= 2.97 M NH4Cl
Molarity Problems

Now, suppose we have a beaker with 175 mL
of a 5.5 M HCl solution. How many moles of
HCl is in this beaker?
175 mL 1 L
1000 mL

5.5 mol HCl
= .96 mol HCl
1 L
Suppose you had 70 grams of NaCl and you
dissolved it in exactly 2.00 L of solution. What
would be the molarity of the solution?
1.2 mol NaCl
70 g
1 mol
2.0 L
NaCl
NaCl = 1.2 mol
NaCl
58.44 g
NaCl
= 0.6 M NaCl
Volume of Solutions

A 10.% HCl solution (soln) means:
10 g HCl
(pure)
100 g HCl soln

A solution with a density of 1.5 g/mL means:
1.5 g soln
1 mL soln
Impure Substances
To say a substance is 75% NaCl by mass means:
75 g NaCl (pure)
100 g of NaCl solution
or
100 g of NaCl solution
75 g of NaCL
(pure)
 Or, Iron ore that is 15% iron by mass means:
15 g Fe
100 g ore
or
100 g of ore
15 g of Fe

Energy &
Stoichiometry
Exothermic and Endothermic
process – heat is
released into the surroundings
• Exo = Exit
HEAT
 Exothermic
Process – heat is
absorbed from the surroundings
• Endo = Into
HEAT
 Endothermic
Thermochemical Equations
In a thermochemical equation, the energy of
change for the reaction can be written as either
a reactant or a product
 Enthalpy: the heat content of a system at
constant pressure (ΔH)

Endothermic (positive ΔH)
2NaHCO3 + 129kJ
Na2CO3 + H2O + CO2
 Exothermic (negative ΔH)
CaO + H2O
Ca(OH)2 + 65.2kJ

Write the thermochemical equation for the
oxidation of Iron (III) if its ΔH= -1652 kJ Exo
4Fe(s) + 3 O2(g)→ 2 Fe2O3(s) + 1652 kJ
How much heat is evolved when 10.00g of Iron is
reacted with excess oxygen?
10.00g Fe
1 mol Fe
55.85g Fe
1652 kJ
4 mol Fe
=73.97 kJ of heat
Write the thermochemical equation for the
decomposition of sodium bicarbonate, with
a ΔH = + 129 kJ: Endo
2 NaHCO3 + 129kJ → Na2CO3(s) + H2O + CO2
How much heat is required to break down 50.0g
of sodium bicarbonate?
50.0 g NaHCO3 1 mol NaHCO3 129 kJ
83.9 g NaHCO3 2 mol NaHCO3
=38.4 kJ of heat
Write the thermochemical equation for the
synthesis of calcium oxide and water with a ΔH=
- 65.2 kJ:
Exo
CaO + H2O → Ca(OH)2 + 65.2kJ
How much energy is released when 100 g of
calcium oxide reacts?
100 g CaO
1 mol CaO
65.2 kJ
56.07 g CaO
1 mol CaO
=116 kJ of heat
Write the thermochemical equation for the
decomposition of magnesium oxide with a
ΔH= + 61.5 kJ: Endo
2 MgO + 61.5 → 2 Mg + O2
How many grams of oxygen are produced when
magnesium oxide is decomposed by adding 420
kJ of Energy?
420 kJ
1 mol O2
61.5 kJ
31.98 g O2 =218 g of O
2
1 mol O2
Stoichiometry – Ch. 11
Stoichiometry in the
Real World
Limiting Reactants
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
Limiting Reactants
Available
Ingredients
• 24 graham cracker squares
• 1 bag of marshmallows
• 12 pieces of chocolate
Limiting Reactant
• chocolate
Excess Reactants
• Marshmallows and
graham crackers
Limiting Reactants
Limiting
Reactant
• one that is used up in a reaction
• determines the amount of
product that can be produced
Excess
Reactant
• added to ensure that the other
reactant is completely used up
• cheaper & easier to recycle
Limiting Reactant Steps
1. Write the balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product actually
possible
Limiting Reactants
 79.1
g of zinc react with 68.1 g HCl.
Identify the limiting and excess
reactants. How many grams of
hydrogen can be formed?
Zn + 2HCl 
79.1 g 68.1 g
ZnCl2 + H2
?g
Limiting Reactants
Zn + 2HCl 
79.1 g 68.1 g
68.1 1 mol
g HCl HCl
ZnCl2 + H2
?g
1 mol 2.02 g
H2
H2
= 1.89
36.46 2 mol 1 mol g H2
g HCl HCl
H2
Limiting Reactants
Zn + 2HCl 
79.1 g 68.1 g
ZnCl2 + H2
?g
79.1 1 mol 1 mol 2.02 g
g Zn Zn
H2
H2
= 2.44 g
65.39 1 mol 1 mol
H2
g Zn
Zn
H2
Limiting Reactants
Zn: 2.44 g H2
HCl: 1.89 g H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 1.89 g H2
left over zinc
Limiting Reactants #2
 5.42
g of magnesium ribbon react
with 4.00 g of oxygen gas. Identify
the limiting and excess reactants.
How many grams of magnesium
oxide are formed?
2Mg + O2 
5.42 g 4.00 g
2MgO
?g
Limiting Reactants #2
2Mg + O2
5.42 g
4.00 g

2MgO
?g
5.42 1 mol 2 mol 40.31 g
g Mg Mg MgO MgO
= 8.99 g
24.31 2 mol 1 mol
MgO
g Mg Mg
MgO
Limiting Reactants #2
2Mg + O2
5.42 g
4.00
g O2
1 mol
O2
32.00
g O2
4.00 g

2MgO
?g
2 mol 40.31
MgO g MgO
= 10.1 g
1 mol 1 mol MgO
O2
MgO
A. Limiting Reactants #2
Mg: 8.99 g MgO
O2: 10.1 g MgO
Limiting reactant: Mg
Excess reactant: O2
Product Formed: 8.99 g MgO
Excess oxygen
Limiting Reactants
What
other information could you
find in these problems?
• How much of each reactant is
used – in grams, liters, moles
• How much of excess reactant is
left over – in grams, liters,
moles
Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
Percent Yield

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 +
2 HCl

2 KCl + H2CO3
45.8 g 1 mol 2 mol 74.55 g
=
49.4
K
CO
2
3
KCl
KCl
K2CO3
grams
KCl
1
mol
1
mol
138.2 g
KCl
K2CO3 K2CO3
Percent Yield

When 45.8 g of K2CO3 react with excess HCl,
46.3 g of KCl are formed. Calculate the
theoretical and % yields of KCl.
46.3 grams KCl
x 100
49.4 grams KCl
% yield = 93.7 %