Stoichiometry
Chapter
12
 Turn
in:
 Forensic Farming Reading
 Fill Out Goal Sheet
 Our
Plan:
 Begin
Lab
 Daily Challenge – Smore’s Stoichiometry
 Notes – Stoichiometry Conversions
 Worksheet #1 & Worksheet #2
 Wrap Up - Pyramid
 Homework (Write in Planner):
 WS#1 & WS#2 Due Monday
www.doh.state.fl.us
Follow
the procedure
for Day 1 and begin
recording observations
Use the internet to
answer the Pre-Lab
Questions
Stoy-key-ahm-a-tree
Comes
from the
combination of the Greek
words stoikheioin,
meaning “element”, and
metron, meaning “to
measure”
The
study of the
quantitative, or
measurable,
relationships in a
chemical reaction.
Using
stoichiometry,
you can determine the
quantities of reactants
& products in a reaction
from the BALANCED
equation.
Like the Recipe!
Do
show the number of
moles of each substance
involved in the rxn.
Do Not indicate the
actual number of grams
of the substance.
N2 + 3H2 --> 2NH3
2 Atoms N
6 Atoms H
On both the reactant and
product side!
N2 + 3H2 --> 2NH3
1 Molecule N2
3 Molecules H2
2 Molecules NH
3
N2 + 3H2 --> 2NH3
1 Mole N2
3 Moles H2
2 Moles NH
3
N2 + 3H2 --> 2NH3
28 Grams N2
6 Grams H2
34 Grams NH
3
Which
were
conserved?
Atoms and Mass, but
NOT Moles and
Molecules!
A
conversion factor
that relates the
amounts in moles of
any 2 substances
involved in a
chemical reaction.
N2 + 3H2 --> 2NH3
1 mole N2
3 moles H2
3 moles H2
2 moles NH3
1 mole N2
2 moles NH3
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
The
conversion of
moles of one type of
substance to moles
of another type of
substance.
Moles
Given
Molar Ratio
Equation
Moles
Unknown
H2
+ I2 --> 2HI
If 4 moles of H2
react, how many
moles of HI will
form?
H2 + I2 --> 2HI
2 moles HI
4 moles H2 X
1 mole H2
= 8 moles HI
If
0.8 moles of
HI form, how
many moles of I2
were used in the
rxn?
H2 + I2 --> 2HI
0.8 moles HI X
1 mole I2
2 moles HI
=0.4 moles I2
2H2 + O2 --> 2H2O
1.If 1.3 moles of H2O form,
how many moles of O2 were
used in the rxn? (0.65 moles)
2.If 0.21 moles of H2 react,
how many moles of H2O will
form? (0.21 moles)
Given
the moles of
one substance &
asked to determine
the mass of
another substance.
molar
Moles
Given
ratio
molar
mass
Moles
equation Unknown
PT
Mass
Unknown
C6H12O6 +6O2 --> 6CO2 + 6H2O
What
mass of
sucrose was used if 3
moles of water
formed?
C6H12O6 +6O2 --> 6CO2 + 6H2O
3 moles H2O X 1 mole C6H12O6 X
6 moles H2O
90g
C6H12O6
180g C6H12O6
1 mole C6H12O6
=
What
mass of
carbon dioxide
was used to form
5.0 moles of
water?
C6H12O6 +6O2 --> 6CO2 + 6H2O
5.0 moles H2O X 6 moles CO2 X
6 moles H2O
44g CO2
1 mole CO2
220 g CO2
=
N2O5 + H2O --> 2HNO3
1.What mass of water was
used to form 2.5 moles of
HNO3? (23 g H2O)
2.What mass of HNO3 will
form from 12.00 moles of
N2O5? (1512 g HNO3)
63g. HNO3
12.00 moles N2O5 X 2 moles HNO3
X
1 mole N2O5
1 mole HNO3
= 1512 g. HNO3
Given
mass of one
substance and
asked to find moles
of another
substance.
molar
Mass
Given
mass
PT
molar
Moles
Given
ratio
equation
Moles
Unknown
2NO + 3H2 --> 2NH3 + O2
How
many moles
of NO were used to
form 824g. NH3?
2NO + 3H2 --> 2NH3 + O2
824g. NH3 X
1 mole NH3
17g. NH3
2 moles NO
X
2 moles NH3
= 48.5 moles NO
2NO + 3H2 --> 2NH3 + O2
1 mole O2 3 moles H2
312g. O2 X
X
32g. O2
1 mole O2
= 29.3 moles H2
$100,000 Pyramid
Pencil
Cougars
Purple
Chemistry
Safety Shower
Mole
$100,000 Pyramid
Stoichiometry
Coefficient
Periodic Table
Molar Mass
Balance
Ratio
Turn
in:
 Worksheet #1 & Worksheet #2
Our Plan:
 Conversions Table Review
 Notes – Mass to Mass Conversions
 Worksheet #3
Homework (Write in Planner):
 Worksheet #3 due next class
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
As
a group of 3 or 4, complete
the problems on the team
conversion review. Each
student should have a
different colored writing
utensil and you should take
turns doing the work.
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
Given
the mass of
one substance and
asked to determine
the mass of another
substance.
molar
Mass mass
Given PT
molar
molar
Moles mass
Moles ratio
Given equation Unknown
PT
Mass
Unkn
2Na
+ Cl2 -->2NaCl
Find
the mass of
table salt
produced from
18.6g Na.
2Na + Cl2 --> 2NaCl
18.6g. Na X 1 mole Na X 2 moles NaCl X 58.5g NaCl
23g. Na
2 mole Na
1 mole NaCl
=47.3g NaCl
2Al
+ Fe2O3 --> Al2O3 + 2Fe
Find
the mass of
aluminum oxide
produced from 2.3g of
aluminum.
2.3g. Al X
1 mole Al
27g. Al
X 1 mole Al2O3 X 102g Al2O3
2 mole Al
1 mole Al2O3
= 4.3 g Al2O3
2Al + Fe2O3 --> Al2O3 + 2Fe
Methane
burns in air by
the following reaction:
CH4 + 2O2 --> CO2 + 2H2O
What
mass of water is
produced by burning
500. g of methane?
1 mole CH4 X 2 mole H2O
18g H2O
500.g CH4 X
X
16g CH4
1 mole CH4
1 mole H2O
= 1,130 g H2O
Complete
Worksheet #3 by
next class!
Turn
in:
 Worksheet #3
Our Plan:
 Questions on WS 1-3
 Stoichiometry Team Review
 Quiz WS 1 – 3
 Lab Day 2
Homework (Write in Planner):
 Finish missing worksheets
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
1.14.78
2.2,822
3.8.4528
Turn
in:
Any Missing Worksheets
Our Plan:
Lab Day 3
Turn in Lab Today
Homework (Write in Planner):
Lab, if not done
#1 – 1000 mg = 1 g, 1000 mL = 1 L
#3 – Convert both the 1g AgNO3 and
your g of Cu to g of Ag. The limiting
reactant is the reactant that produces
the least product. You will do 2 three
step conversions to find the answer.
# 5 & #6 – Percent yield = (actual/theoretical) x 100
WASH ALL SUPPLIES, RETURN TO SIDE
TABLE, WIPE DOWN COUNTER, WASH
HANDS!
Clicker
Review
Turn
in:
 Metal in Water Lab
Our
Plan:
 Relay
Race Review
 Daily Challenge
 Notes – Stoichiometry of Gases and
Molecules
 Worksheet #4
Homework (Write in Planner):
 WS#4 due next class
1.
2.
3.
How many liters are in 1 mole of
a gas?
How many molecules are in 1
mole of any compound?
For the reaction below, what
mass of water can be produced
from 1.5 moles of hydrogen?
(27g)
2H2 + O2 --> 2H2O
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
Given
the volume
of one substance
and asked to
determine the
volume of another
substance.
molar
Vol 22.4 L
Given
Moles 22.4 L
Moles ratio
Given equation Unknown
Vol
Unkn
Conversion
factor: 22.4
L/mole
You
can only use the
conversion factor 22.4 under
very specific conditions.
It must be a gas and it must
be at STP (Standard
Temperature and Pressure)
STP is 0 degrees Celsius and 1
atm
How
many
moles of Helium
are in a 3.7 L
balloon at STP?
3.7 L x 1 mole = 0.17 mole
22.4 L
2 CO + O2 --> 2 CO2
What
volume of carbon
dioxide will be
produced from 2.6 L of
oxygen at STP?
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
2 CO + O2 --> 2 CO2
2.6 L O2 X 1 mole O2 X 2 mole CO2 X 22.4 L CO2 =
22.4 L O2
1 mole O2
5.2 L CO2
1 mole CO2
Using
the same
equation, what
volume of carbon
dioxide will be
produced at STP from
45.3 grams of carbon
monoxide?
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
2 CO + O2 --> 2 CO2
45.3 g CO X 1 mole CO X 2 mole CO2 X 22.4 L CO2 =
28 g CO
2 mole CO
36.2 L CO2
1 mole CO2
Using
the same
equation, what volume
of oxygen is used to
produce 19.3 grams of
carbon dioxide at STP?
(4.91 L)
2 CO + O2 --> 2 CO2
19.3 g CO2 X 1 mole CO2 X 1 mole O2 X 22.4 L O2 =
44 g CO2
2 mole CO2
4.91 L O2
1 mole O2
You
can do the
same type of
problems using
molecules. Just use
Avogadro’s Number!
In
the reaction:
2H2 + O2 --> 2H2O
if you have 0.50
grams of hydrogen,
how many molecules
of water will form?
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
0.50 g H2 X
1 mole H2
2 g H2
=
X
2 mole H2O
2 mole H2
X
6.022 x 1023
molecules
1 mole H2O
1.5 x 1023 molecules H2O
2H2 + O2 --> 2H2O
Cooking
Planning
outcomes
of reactions
Race cars/
mechanics
Treating
an
upset stomach
Airbags
Artists
Launching
space shuttle
Baking bread
Engineering
Complete
Worksheet
#4 by next class
period!
 Turn
in:
 Get Worksheet #4 Out to Check
 Our Plan:
 Balloon Races
 Limiting Reactant Notes
 Finish Balloon Races Handout
 Worksheet #5
 Chemistry Cartoon
 Homework (Write in Planner):
 Worksheet 5 & Cartoon Due Next Class
Molecules
Molecules
6.022 x 1023
Mass
Volume
Molar Mass
PT
22.4
L/mole
6.022 x 1023
Moles
Given
Molar
Ratio
Equation
Moles
Unknown
Molar Mass
PT
Mass
22.4
L/mole
Volume
Pre Lab Review – The balanced equation for
the reaction of baking soda and vinegar is:
HC2H3O2 + NaHCO3 --> NaC2H3O2 + CO2 + H2O.
How many liters of carbon dioxide can form
from 3.00 g of baking soda (NaHCO3)? Show
your work at the top of your lab.
What
did you predict
would happen?
What did happen? Why?
Smore’s Recipe
2 Graham Crackers
3 Pieces of Chocolate
1 Marshmallow
1.If
you have 4
graham crackers, 9
pieces of chocolate,
and 7 marshmallows,
how many smores
can you make?
2.Which
ingredient
determined
this?
3.If
you have 22
graham crackers, 27
pieces of chocolate,
& 13 marshmallows,
how many smores
can you make?
4.Which
ingredient
determined
this?
Even
in abundance,
the reactants must still
combine in proportion
They follow
stoichiometry.
Is
the reactant
that limits the
amount of product
formed in a
chemical equation.
Is
the substance
that is not used
up completely in
a reaction.
The
quantities of
products formed in a
reaction are always
determined by the
quantity of the
limiting reactant.
1.Write a balanced
equation
2.Convert the mass
of reactants to the
mass of the product.
3.Whichever
produces less
product is the
limiting reactant.
Identify
the L.R
when 10.0 g of water
react with 4.50 g of
sodium to produce
sodium hydroxide &
hydrogen gas.
2H2O + 2Na --> 2NaOH + H2
1 mol H2O 2 mol NaOH
40g NaOH
10.0 g
X
X
X
H2O
18g H2O 2 mol H2O
1 mol NaOH
= 22.2 g NaOH
1 mol Na
2 mol NaOH
40g NaOH
4.50 g
X
X
X
Na
23g Na
2 mol Na
1 mol NaOH
= 7.83 g NaOH
Na= L.R.
Identify
the L.R when
8.90 g HF are combined
with 14.56g SiO2 in the
following reaction:
SiO2 + 4HF --> SiF4 + 2H2O
SiO2 + 4HF -->SiF4 + 2H2O
2
mol
H
O
1
mol
HF
18g H2O
2
8.9 g HF
X
X
X
20g HF
4 mol HF 1 mol H2O
= 4.0 g H2O
1 mol SiO2
2 mol H2O
14.50g SiO2 X
X
60g SiO2
1 mol SiO2
= 8.700 g H2O
X
18g H2O
1 mol H2O
HF= L.R.
Finish
Balloon Races
Questions
Worksheet #5
Chemistry Cartoon
Turn
in:
 Get Worksheet #5 Out to Check
 Turn in Chemistry Cartoon - Basket
Our Plan:
 L.R. Review Problem
 Percent Yield Notes
 Worksheet #6
Homework (Write in Planner):
 Worksheet #6 Due Tuesday/Wednesday
Acrylonitrile, C3H3N, is an important
ingredient in the production of various
fibers and plastics. Acrylonitrile is
produced from the following reaction:
C3H6 + NH3 + O2 --> C3H3N + H2O
If 850 g of C3H6 is mixed with 300. g of NH3
and unlimited O2, to produce 850. g of
acrylonitrile, what is the percent yield?
You must first balance the equation. 91 %
1.Babe
Ruth played in
2503 baseball games in
his career. He had 2873
hits in 8399 at bats
what was his batting
average?
 www.pbs.org
2.Hank
Aaron played in
3298 baseball games in
his career. He had 3771
hits in 12,364 at bats.
What was his batting
average?
 www.pbs.org
Theoretical
Yield is based
on
calculations
Actual
Yield is
based on the
actual chemical
reaction (the
lab)
%
yield = Actual yield
X 100
Theoretical yield
1.Write equation
2.Calculate the mass
of the product that
should have been
formed. (theoretical
yield)
3.Plug
numbers into
the equation
Determine
the % yield
for the reaction
between 2.80 g Al(NO3)3
& excess NaOH if 0.966g
Al(OH)3 is recovered.
Write
the
equation:
Al(NO3)3 + 3NaOH --> Al(OH)3 + 3NaNO3
 Calculate
theoretical yield
78g
Al(OH)3
2.80 g
1 mol Al(NO3)3
1 mol Al(OH)3
X
X
Al(NO3)3 X
213g Al(NO3)3
1 mol Al(NO3)3
1 mol
Al(OH)3
= 1.03 g Al(OH)3
Al(NO3)3 + 3NaOH --> Al(OH)3 + 3NaNO3
 Plug
numbers into equation
0.966g
% yield =
X 100
1.03g
= 93.8%
Determine
the
percent yield for the
reaction between
15.0 g N2 & excess H2
if 10.5g NH3 is
produced.
2 mol NH3
1 mol N2
17g NH3
15.0g N2 X
X
X
28g N2
1 mol NH3
1 mol N2
= 18.2 g NH3
%yield =
10.5g NH3
18.2 g NH3
X 100 = 57.7%
You
have to find the
limiting reactant first.
Calculate amount of
product for both reactants
& whichever is less, use as
theoretical yield.
Complete
Worksheet
#6 by next class!
Turn
in:
 Worksheet #6
Our Plan:
 Stoichiometry Lab Test
 Test Review
Homework (Write in Planner):
 Test Review due next class
 UNIT 8 TEST NEXT CLASS!
Turn
in:
 Get out Test Review
Our Plan:
 Questions on Test Review?
 Test
 Larry the Lawnchair Guy Reading
Homework (Write in Planner):
 Larry Reading due next class
 Given
the following
reaction:
C3H8 + 5O2 -------> 3CO2
+
4H2O
If you start with 14.8 g of C3H8 and 3.44 g of
O2, determine the limiting reagent
 Given
the following equation:
H3PO4 + 3 KOH ------> K3PO4 + 3 H2O
If 49.0 g of H3PO4 is reacted with excess
KOH, determine the percent yield of K3PO4 if
you isolate 49.0 g of K3PO4.
 Given
the following equation:
Al(OH)3 +3 HCl --> AlCl3 + 3 H2O
If you start with 50.3 g of Al(OH)3 and you
isolate 39.5 g of AlCl3, what is the
percent yield?
 Given
the following equation:
H2SO4 + Ba(OH)2 ----> BaSO4 + H2O
If 98.0 g of H2SO4 is reacted with excess
Ba(OH)2, determine the percent yield of if
213.7 g of BaSO4 is formed.
 http://www.youtube.com/watch?v=NC-
Km7aUjgc