Stoichiometry
Chapter 12
Stoichiometry
The study of the
quantitative, or
measurable, relationships
that exist in chemical
formulas and chemical
reactions.
Interpreting Balanced
Chemical Equations
2 H2 + O2  2 H2O
• Based on the mole ratio
Mole Ratio
• indicated by coefficients in a
balanced equation
2 H2 + O2  2 H2O
Molar ratio of H2 to H2O is 2:2
(Simplify 1:1)
Molar ratio of O2 to H2O is 1:2
In terms of moles
2 H2 + O2  2 H2O
2 moles of hydrogen react with
1 mole of oxygen to produce 2
moles of water.
In terms of mass
2 H2 + O2  2 H2O
4 grams of hydrogen react with
32 grams of oxygen to produce
36 grams of water.
In terms of molecules
2 H2 + O2  2 H2O
2 hydrogen molecules react with 1
oxygen molecule to produce 2
water molecules.
• Notice that the number of
molecules is NOT the same on
each side of the arrow.
In terms of atoms
2 H2 + O2  2 H2O
4 atoms of hydrogen react with 2
atoms of oxygen to produce 2
water molecules, which are 4
atoms of hydrogen and 2 atoms
of oxygen.
In terms of volumes
2 H2 + O2  2 H2O
44.8 L of hydrogen gas react with
22.4 L of oxygen gas to produce
44.8 L of water vapor.
• Notice that the number of liters of
gas is NOT the same on each
side of the arrow.
VERIFIED!
Law of Conservation of Matter & Mass
2 H2 + O2  2 H2O
2 moles H2 react with 1 mole of
O2 to form 2 moles of H2O.
O2
H2
2 mol 2 g
1 mol 32 g
1 mol
4g
H2 O
+
2 mol 18 g
1 mol
32 g
1 mol
=
36 g
Practice Problem #1
Lead will react
with hydrochloric
acid to produce
lead (II) chloride
and hydrogen.
How many moles
of hydrochloric
acid are needed
to completely
react with 0.36
moles of lead?
Oh, no …
where do I
start?
Write an equation and balance it.
Lead will react with hydrochloric acid
to produce lead chloride and
hydrogen.
Pb + 2HCl  PbCl2 + H2
Determine Mole Ratio
How many moles of hydrochloric acid
are needed to completely react with
0.36 moles of lead?
Pb + 2 HCl  PbCl2 + H2
lead
“given”
Coefficient = 1
hydrochloric
“wanted”
acid
Coefficient = 2
Set up mole ratio “wanted” to “given”
How many moles of hydrochloric acid
are needed to completely react with
0.36 moles of lead?
Pb + 2HCl  PbCl2 + H2
wanted
given
=
hydrochloric acid
X moles of HCl
0.36 moles of lead
lead
=
2
1
=
2
1
Solve for “wanted”
How many moles of hydrochloric acid
are needed to completely react with
0.36 moles of lead?
Pb + 2HCl  PbCl2 + H2
X moles of HCl
=
0.36 moles of lead
2
1
X = 2 (0.36 moles)
X = 0.72 moles HCl
Umm … It would
require 0.72 moles
of hydrochloric
acid. Is that right?
That’s Correct!
Hem, hem –
you have 9 more
problems …
ready, go.
“Stoich” Problems
How to solve, given
an amount of one
substance to
another
substance.
Stoichiometry Steps
1. Write a balanced equation.
2. Identify “wanted” & “given”.
3. Convert given information to moles.
Determine
Mole
ratio
4. 4.Determine
Mole
Ratio.
(molesof
of “wanted”
“wanted” to: Moles
moles ofof“given”)
(Moles
“given”)
5. Calculate Moles of “wanted”.
6. Convert to required units.
KEY step in all stoichiometry problems!
Mass-Mass Problems
What mass of bromine is produced
when fluorine reacts with 1.72 g of
potassium bromide?
Help … I really
need to know
this right now!
1. Write a balanced equation.
What mass of bromine is produced
when fluorine reacts with 1.72 g of
potassium bromide?
F2 + 2KBr  2KF + Br2
Help … I really
need to know
this right now!
2. Identify “wanted” and “given”
What mass of bromine is produced
when fluorine reacts with 1.72 g of
potassium bromide?
F2 + 2KBr  2KF + Br2
Help … I really
need to know
this right now!
“given”
“wanted”
3. Convert “given” information to moles
1.72 g of potassium bromide
Molar Mass of KBr = 39 + 80 = 119 g/mol
1.72 g
1 mol
119 g
Help … I really
need to know
right now!
= this
0.01445
mol
KBr
4. Determine Mole Ratio
F2 + 2KBr  2KF + Br2
“wanted”
“given”
Help … I really
need to know
thisMole
right now!
tell
Ratio
Coefficients
“wanted” to “given”
1
:
2
5. Calculate Moles of “wanted”
F2 + 2KBr  2KF + Br2
Moles Br2
Moles KBr
=
X
0.01445 mol KBr
1
2
=
2 X = 0.01445 mol
Help … I really
need to know
this right now!
1
2
X = 0.007225 mol
6. Convert to required units
X = 0.007225 mol Br2
Molar Mass Br2 80 + 80 = 160 g/mol
0.007225 mol
160 g
1 mol
=
1.16 g Br2
Stoichiometry
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
• Copper Wire
• 0.5 g AgNO3
Limiting Reactant
• 0.5 grams AgNO3
Excess Reactants
• Copper Wire
Limiting Reactant
The reactant that limits the
amount of product that can be
formed.
When quantities of reactants are
available in the exact ratio
described by the balanced
equation, they are said to be in
Stoichiometric proportions.
Limiting Reactants
Limiting Reactant
• used up in a reaction
• determines the amount of all
products formed
Excess Reactant
• added to ensure that the other
reactant is completely used up
• usually cheaper & easier to recycle
Solving Problems – Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. The reactant that produces the
smaller amount of product is the
limiting reactant.
Very similar to mass-mass problems!
Step 1: Write a balanced
equation.
Identify the limiting reactant when
1.22 g of oxygen reacts with 1.05 g
of hydrogen to produce water.
O2 + 2H2

2 H2O
Step 2:
For each reactant,
calculate the amount of
product formed.
Identify the limiting reactant when
1.22 g of oxygen reacts with 1.05 g
of hydrogen to produce water.
O2 + 2H2

2 H2O
Step 2:
1.22 g oxygen
1 mole
= 0.038 mol O2
32 g
O2 + 2H2

given
wanted
given
2 H2O
wanted
X
0.038 mol
X = 0.076 mol
=
2
1
Step 2:
1.05 g H2
1 mole
= 0.525 mol H2
2g

O2 + 2H2
wanted
given
wanted
given
2 H2O
X
0.525 mol
X = 0.525 mol H2O
=
2
2
Step 3:
The one that produces the
smallest amount is your
limiting reactant.
1.22 g of O2 would produce 0.0763 mol H2O
Oxygen is your limiting reactant!
1.05 g of H2 would produce .525 mol H2O
Limiting Reactants
Identify the limiting reactant when
1.7 g of sodium reacts with 2.6 L of
chlorine gas at STP to produce
sodium chloride.
Step 1: Write a balanced
equation.
Identify the limiting reactant when
1.7 g of sodium reacts with 2.6 L of
chlorine gas at STP to produce
sodium chloride.
2Na + Cl2

2NaCl
Step 2:
For each reactant,
calculate the amount of
product formed.
Identify the limiting reactant when
1.7 g of sodium reacts with 2.6 L of
chlorine gas at STP to produce
sodium chloride.
2Na + Cl2

2NaCl
Step 2:
1.7 g Na
1 mole
23 g
= 0.0739 mol Na
2Na + Cl2

wanted
given
wanted
given
2NaCl
X
0.0739 mol
X = 0.0739 mol NaCl
=
2
2
Step 2:
2.6 L Cl2
1 mole
22.4 L
= 0.116 mol Cl2
2Na + Cl2

wanted
given
wanted
given
2NaCl
X
0.116 mol
X = 0.232 mol NaCl
=
2
1
Step 3:
1.7 g Na
The one that produces the
smallest amount is your
limiting reactant.
would produce 0.0739 mol NaCl
Sodium is your limiting reactant!
2.6 L Cl2 would produce 0.232 mol NaCl
Click on the real player file called
Sodium_Chlorine_2 to see a demo of this reaction
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
yield and % yield of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
45.8 g 1 mol 2 mol 74
K2CO3 K2CO3
KCl g KCl
138 g
K2CO3
= 49.1
1 mol 1 mol g KCl
K2CO3 KCl
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.1 g
actual: 46.3 g
Theoretical Yield = 49.1 g KCl
% Yield =
46.3 g
49.1 g
 100 = 94.3%