Visualizing the balancing of
chemical equations
By Daniel R. Barnes
Init: 1/9/2007 from previous work
You may hit the “End” key at any time to jump to the hyperTable of Contents.
NOTE: This presentation is meant to go along with Mr. Barnes’ cardboard disc reactions
worksheet, which is meant to be used with a set of specially-colored cardboard discs,
each disc representing an atom of an element indicated by the color of the disc.
NOTE: As always, some of the images in this presentation have been taken from the
world wide web without permission of their owners. Copying and distribution of this
presentation may be, therefore, illegal. In fact, its mere existence may be illegal.
SWBAT . . .
. . . balance chemical equations.
. . . explain why some reactions are
endothermic and other reactions are
exothermic.
. . . define “activation energy” and
give real life examples.
reactants
1.
C+
turn into
O2 
products
CO2
Skip to answer
1.
C+
O2 
CO2
Skip to answer
1.
C+
O2 
CO2
Skip to answer
1.
C+
O2 
CO2
Please note: charcoal briquette
ashes are not made of carbon
dioxide. CO2 is a colorless, odorless
gas. Charcoal ashes are made of
alkali metal compounds and other
stuff. Charcoal is not pure carbon. If
it were, there would be no ashes left
after charcoal burned.
Skip to answer
1.
C= 1
O= 2
C+
O2 
CO2
C= 1
O= 2
Before you go to your lab tables and start working with the discs,
we’d better make sure we know the definitions of . . .
Chemicals absorb energy  Surroundings get colder.
Chemicals release energy  Surroundings get hotter.
All chemical reactions, whether they are endothermic or
exothermic, typically involve endothermic bond
breakage AND exothermic bond formation.
If forming new bonds releases more energy than the
activation energy required to break the old bonds . . .
. . . The reaction will be EXOTHERMIC
If forming new bonds releases less energy than the
activation energy required to break the old bonds . . .
. . . The reaction will be ENDOTHERMIC
Sell $100
worth of
hot dogs
Sell $400
worth of
hot dogs
Spend $200
for materials
Spend $200
for materials
$100
LOSS
$200
PROFIT
Yields
money
Yields
energy
Sell $100
worth of
hot dogs
Making
new bonds
Sell $400
worth of
hot dogs
Spend $200
for materials
Breaking
old bonds
Spend $200
for materials
$100
LOSS
Costs
money
$200
PROFIT
Costs
energy
NOTE: The circled numbers are made up. They’re not real.
Relase 100 kJ
of energy
forming new
bonds
Absorb 200 kJ
of energy
breaking old
bonds
ENDOthermic
Making
new bonds
Breaking
old bonds
Release 400 kJ
of energy
forming new
bonds
Absorb 200 kJ
of energy
breaking old
bonds
EXOthermic
Q: What is the difference between an endothermic
process and an exothermic process?
A: Endothermic processes absorb heat, making their
surroundings colder, whereas exothermic processes give
off heat, making their surroundings get hotter.
Q: What is activation energy?
A: The energy input required to make a reaction happen,
even if it’s an exothermic reaction. It’s the energy
required to break old bonds.
Mr. Barnes, please
(1) Open the windows.
(2) Put goggles on the kids in the front two rows.
(3) Put a few drops of glycerine on a dimpled pile of potassium
permanganate.
This reaction is definitely exothermic, as shown by the fire coming
out of it.
The reaction was spontaneous at room temperature. No
matches, sparks, UV light, or other input of activation energy
was needed to get it going.
The apparently low activation energy may indicate weak bonds
somewhere in the reactants.
One of the main reactions that happens when these two
chemicals go crazy on each other is the following:
14KMnO4(s) + 4C3H5(OH)3(l) →
7K2CO3(s) + 7Mn2O3(s) + 5CO2(g) + 16H2O(g)
I’ve left the coefficients covered so that you can try to balance this
monster for extra credit when you’re all done with the worksheet.
A slide near the end of this presentation has the coefficients on it if
you want to find out what they are.
1.
C+
O2 
C= 1
O= 2
CO2
C= 1
O= 2
Your worksheet should already look like this for #1.
Mr. Barnes will now have you animate the reaction with cardboard
discs. At this point, Mr. Barnes may wish to turn off the projector.
If he does, the rest of this presentation can be viewed at
hhscougars.org, on Mr. Barnes’ Chemistry Power Points page.
2.
H2 +
O2 
H2O
formulas
subscripts
Skip to answer
2.
H2 +
O2 
H2O
Explosive chemical reactions
can blow up a vehicle and
destroy it, but they can also
make vehicles move in the first
place.
The big, orange external fuel
tank of the space shuttle
contains hydrogen and oxygen.
The two gasses are mixed
together and ignited in the three
engines under the tail fin of the
orbiter. The flames that come
Skip to answer
out of those engines are made
The white rockets on the side of
of water vapor – water vapor
the fuel tank use a solid fuel. I
that’s so hot that it glows.
don’t know what that fuel is.
2 H2 +
2.
H=2 4
O=2
O2 
2 H2O
H =2 4
O =1 2
coefficient
InIf ayou
chemical
reaction,
start out
with two
atoms
are
neither
oxygen
atoms,
you
created
nor destroyed.
should end
up with two
They
justatoms
change
oxygen
who they’re stuck to.
Okay. Let’s re-run the reaction from the beginning, but this time we’re going
to get it right the first time.
Skip to answer
2.
H=2 4
O=2
2 H2 +
O2 
2 H2O
H =2 4
O =1 2
Skip to answer
2.
H=2 4
O=2
2 H2 +
O2 
2 H2O
H =2 4
O =1 2
Skip to answer
2.
H=2 4
O=2
2 H2 +
O2 
2 H2O
H =2 4
O =1 2
3. CH4 +
C= 1
H= 4
O=2
O2
H 2O +
CO2
C= 1
H= 2
O= 3
First, we need to see if the equation is balanced or not.
We need to do an . . . atom count.
Skip to answer
3. CH4 +
C= 1
H= 4
O=2 4
2 O2
2 H 2O +
CO2
C= 1
H= 2 4
O= 3 4
Something is obviously wrong. Let’s animate the equation as it is
and see if we notice anything weird.
Skip to answer
3. CH4 +
2 O2
2 H 2O +
CO2
Skip to answer
3. CH4 +
C= 1
H= 4
O=2 4
2 O2
2 H 2O +
CO2
C= 1
H= 2 4
O= 3 4
Skip to answer
3. CH4 +
2 O2
C= 1
H= 4
O=2 4
atom
census
“before”
(reactants)
picture
2 H 2O +
CO2
C= 1
H= 2 4
O= 3 4
atom
census
“after”
picture (products)
Skip to answer
3. CH4 +
C= 1
H= 4
O=2 4
2 O2
2 H 2O +
CO2
C= 1
H= 2 4
O= 3 4
The famous “ammonia fountain” demonstration is
something we won’t explain until chapter 14
(gases). However, if you want to see it happen,
click the link below. If that doesn’t work, just go to
YouTube and type in “ammonia fountain.”
http://www.youtube.com/watch?v=4U-DgdWPKyo
4.
NH3 +
N= 1
H= 5
O=1
H2O
NH4OH
N= 1
H= 5
O= 1
Skip to answer
The famous “ammonia fountain” demonstration is
something we won’t explain until chapter 14
(gases). However, if you want to see it happen,
click the link below. If that doesn’t work, just go to
YouTube and type in “ammonia fountain.”
http://www.youtube.com/watch?v=4U-DgdWPKyo
4.
NH3 +
N= 1
H= 5
O=1
N
H2O
NH4OH
N
N= 1
H= 5
O= 1
Skip to answer
4.
NH3 +
N= 1
H= 5
O=1
N
H2O
NH4OH
N
N= 1
H= 5
O= 1
Skip to answer
4.
NH3 +
N= 1
H= 5
O=1
N
H2O
NH4OH
N
N= 1
H= 5
O= 1
When water and carbon dioxide mix, they produce
carbonic acid. Carbonic acid is a “weak” acid
found in soda and any other carbonated beverage.
It may be a “weak” acid, but it is acidic enough to
partially dissolve certain rocks over time. Carbonic
acid, therefore, is an agent of “chemical
weathering”.
5.
C= 1
O=3
H= 2
CO2 +
H2O
H2CO3
C= 1
O= 3
H= 2
Skip to answer
When water and carbon dioxide mix, they produce
carbonic acid. Carbonic acid is a “weak” acid
found in soda and any other carbonated beverage.
It may be a “weak” acid, but it is acidic enough to
partially dissolve certain rocks over time. Carbonic
acid, therefore, is an agent of “chemical
weathering”.
5.
C= 1
O=3
H= 2
CO2 +
H2O
H2CO3
C= 1
O= 3
H= 2
Skip to answer
5.
C= 1
O=3
H= 2
CO2 +
H2O
H2CO3
C= 1
O= 3
H= 2
Skip to answer
5.
C= 1
O=3
H= 2
CO2 +
H2O
H2CO3
C= 1
O= 3
H= 2
The reaction below is an example of an acid-base “neutralization”.
6.
HCl +
Ba(OH)2
H= 3
Cl = 1
Ba = 1
O= 2
H 2O +
H=
Cl =
Ba =
O=
BaCl2
2
2
1
1
You might be tempted to try to balance hydrogen first,
because it’s the first element in the equation, but I wouldn’t if
I were you.
Notice that hydrogen appears in three different formulas in
the equation.
Trust me when I say that you should balance first those
elements that appear in the least number of formulas.
Let’s save hydrogen for later . . .
Skip to answer
6.
HCl +
Ba(OH)2
H= 3
Cl = 1
Ba = 1
O= 2
H 2O +
H=
Cl =
Ba =
O=
BaCl2
2
2
1
1
The second element in the equation, chlorine, appears in only
two formulas (the smallest possible # of places).
Let’s balance chlorine first.
We balance an element by increasing whatever coefficient
will increase the amount of that element.
Remember: when you’re balancing an equation you can only
change the coefficients.
DON’T TOUCH THE SUBSCRIPTS!
Skip to answer
6.
HCl 2+
Ba(OH)2
H 2O +
BaCl2
H= 2
H= 3
Cl = 2
Cl = 1
Ba = 1
Ba = 1
O= 1
O= 2
EXAMPLE: Someone who didn’t know what they’re doing
might foolishly put a “2” as a subscript after the “Cl” in
“HCl”.
DON’T DO THAT! You’re not supposed to change subscripts
when you balance a chemical equation!
There’s no such chemical as HCl2, and even if there were, it
wouldn’t be hydrochloric acid anymore.
When you balance a chemical equation, you’re not changing
the identity of the chemicals in the reaction, just how much of
Skip to answer
each chemical there is.
6. 2HCl +
H= 3 4
Cl = 1 2
Ba = 1
O= 2
Ba(OH)2
H 2O +
H=
Cl =
Ba =
O=
BaCl2
2
2
1
1
The proper thing to do is to put a coefficient of “2” in front of
HCl on the left.
As soon as you change a coefficient, you need to adjust the
atom count for any element in that formula.
Doubling the amount of HCl increases the amount of H . . .
. . . and also the amount of Cl.
We’ve now balanced the # of Cl’s on the left and the right.
Skip to answer
6. 2HCl +
Ba(OH)2
H= 3 4
Cl = 1 2
Ba = 1
O= 2
2 H 2O +
H=
Cl =
Ba =
O=
BaCl2
2 4
2
1
1 2
Now chlorine is fixed, but there still isn’t enough oxygen on
the right.
Putting a coefficient of “2” in front of H2O on the right will
double the amount of oxygen, balancing it as well.
Of course, the “2” doubles not only the O’s on the right, but ,
but also the H’s.
Would you look at that?
The H’s got balanced by accident!
Skip to answer
6. 2HCl +
H= 3 4
Cl = 1 2
Ba = 1
O= 2
Ba(OH)2
2 H 2O +
Cl
Cl
Ba
Cl
Ba
H=
Cl =
Ba =
O=
BaCl2
2 4
2
1
1 2
Cl
And now, for the cartoon . . .
Skip to answer
6. 2HCl +
H= 3 4
Cl = 1 2
Ba = 1
O= 2
Ba(OH)2
Cl
Cl
Ba
Cl
2 H 2O +
Ba
Cl
H=
Cl =
Ba =
O=
BaCl2
2 4
2
1
1 2
Here’s an even more classic acid-base neutraliztion:
7.
HCl +
H= 2
Cl = 1
Na = 1
O= 1
NaOH
H 2O +
Na
Cl
Na
Cl
NaCl
H= 2
Cl = 1
Na = 1
O= 1
How boring. This one’s already balanced.
Cartoon, please!
Skip to answer
7.
HCl +
H= 2
Cl = 1
Na = 1
O= 1
NaOH
H 2O +
Na
Cl
Na
Cl
NaCl
H= 2
Cl = 1
Na = 1
O= 1
This reaction is the decomposition of nitrogen triiodide.
http://www.youtube.com/watch?v=2KlAf936E90
8.
2 NI3
N2 +
3 I2
N= 1 2
I= 3 6
N= 2
I= 2 6
There’s not enough nitrogen on the left, so we need to get
more by . . .
. . . putting a coefficient of “2” in front of NI3.
As always, any time you change a coefficient, you must
adjust the atom counts for every element in the formula you
put the coefficient in front of.
Now we need more iodine on the right, so . . .
Skip to answer
This reaction is the decomposition of nitrogen triiodide.
http://www.youtube.com/watch?v=2KlAf936E90
8.
2 NI3
N2 +
I
N= 1 2
I= 3 6
I
N
I
I
N N
I I
I
N
3 I2
I
N= 2
I= 2 6
I I
I I
Here comes the cartoon . . .
Skip to answer
8.
2 NI3
N2 +
I
N= 1 2
I= 3 6
I
N
I
I
N N
I I
I
N
3 I2
I
I I
N= 2
I= 2 6
I I
The following reaction is steel wool burning.
http://www.youtube.com/watch?v=39diUfWnbPY
9.
2 Fe +
3
2 O2
Fe = 1 2
O= 2 3
Fe2O3
Fe = 2
O= 3
There’s not enough iron on the left, so . . .
Now, we’ve got a problem. We want to end up with three
oxygen atoms on the left, but if we put a “2” in front of the
“O2” on the left, we’ll end up with four “O’s” on the left
instead of three. 1O2 is not enough, and 2O2 is too much.
The solution is a little crazy. We can put a 3/2 as a coefficent.
3/2 x 2 = 6/2 = 3. Now we’re balanced.
It’s actually standard procedure for chemists to use fractional
coefficients, but only if it’s an odd number over two, and only
Skip to answer
if it’s in front of a diatomic element.
The following reaction is steel wool burning.
http://www.youtube.com/watch?v=39diUfWnbPY
9.
4 2 Fe + 3
Fe = 1 2 4
O= 2 3 6
3
2 O2
2 Fe2O3
Fe = 2 4
O= 3 6
We’ve got a problem when it comes time to draw the cartoon
for the reaction, though.
Drawing 3/2 of an oxygen molecule is kind of like drawing 3/2
of a person or 3/2 of a tv.
The solution is to double all the coefficients in the equation.
Doubling all the coefficients doubles all the atom counts . . .
. . . and the equation is still just as balanced as before the
doubling.
Skip to answer
The following reaction is steel wool burning.
http://www.youtube.com/watch?v=39diUfWnbPY
9.
4 2 Fe + 3
Fe = 1 2 4
O= 2 3 6
Fe
3
2 O2
2 Fe2O3
Fe
Fe
Fe
Fe
Fe = 2 4
O= 3 6
Fe
Fe
Fe
Now that all the coefficients are whole numbers, drawing the
cartoon will be much easier. We won’t have to split any
circles in half or anything dumb like that.
Skip to answer
9.
4 2 Fe + 3
Fe = 1 2 4
O= 2 3 6
Fe
3
2 O2
2 Fe2O3
Fe
Fe
Fe
Fe
Fe = 2 4
O= 3 6
Fe
Fe
Fe
And now, we burn methanol, like a “top methanol dragster”.
10.
CH3OH +
O2
C= 1
H= 4
O= 3
CO2 +
H 2O
C= 1
H= 2
O= 3
Notice that oxygen occurs in all four formulas in this
equation.
We should save oxygen for last and balance something else
first.
Carbon is already balanced, and
hydrogen appears in only two
places, so hydrogen should be
our first element to balance.
Skip to answer
And now, we burn methanol, like a “top methanol dragster”.
10.
CH3OH +
O2
C= 1
H= 4
O= 3
CO2 +
2 H 2O
C= 1
H= 2 4
O= 3 4
We’re almost balanced now, but we need one more oxygen
atom on the left.
Unfortunately, if we add an O2, we get two more oxygen
atoms, not one more.
Don’t you worry! Oxygen is a
diatomic element, so we can
use fractional coefficients
with O2, as long as the
fraction is an odd number
divided by two.
Skip to answer
And now, we burn methanol, like a “top methanol dragster”.
10. 2 CH3OH + 3 23 O2
2 CO2 + 4 2 H2O
C= 1 2
C= 1 2
H= 4 8
H= 2 4 8
O= 3 4 8
O= 3 4 8
Now we’re balanced, and most chemists would say we’re
done.
However, we want to draw a cartoon, and half-circles look
stupid, so let’s get rid of that fraction by . . .
. . . multiplying all coefficients by
two.
This means multiplying all atom
counts by two.
Now we’re ready to draw the
cartoon.
Skip to answer
And now, we burn methanol, like a “top methanol dragster”.
10. 2 CH3OH + 3 23 O2
C= 1 2
H= 4 8
O= 3 4 8
2 CO2 + 4 2 H2O
C= 1 2
H= 2 4 8
O= 3 4 8
Skip to answer
10. 2 CH3OH + 3 23 O2
C= 1 2
H= 4 8
O= 3 4 8
2 CO2 + 4 2 H2O
C= 1 2
H= 2 4 8
O= 3 4 8
Dang.
You made it this
far?
E-mail Mr. Barnes
and tell him that
you saw the
sasquatch if you
want some XCR.
I bet you need more in-class practice.
You need to balance some more equations.
Your group leader needs to get
everyone in your group an “SP Book”.
YOUR ASSIGNMENT:
Section 9:2
pg 64:
Section 9:3
pg 65:
pg 66:
pg 67:
PROBLEMS 1-5
PROBLEMS 11 – 14
PROBLEMS 21 – 24
PROBLEMS 31 – 34
The answers are in the back of the book, on pages 273-274
(You can do the other problems in 9:2 and 9:3 for extra credit if
you’re really brave, but they might be hard.)
What do the following things mean?
(cr) = “crystal”
(s)
= “solid”
(l)
= “liquid”
(g)
= “gas”
(aq) = “aqueous” = dissolved in water
They all have something to do with . . . states of matter.
Please ignore them in this exercise. They don’t affect
equation-balancing at all.
EQUATION-COPYING TIPS:
(i) Don’t copy state of matter info
((cr), (s), (l), (g), or (aq)).
(ii) Leave plenty of room in front of each formula
so that you can put coefficients there.
Consider that you might have to change a
coefficient more than once, so leave plenty of
room.
For example, if the SP book says
“1. Mg(cr) + O2(g)  MgO(cr)”
You should write, on your paper,
“1.
LOTS
OF Mg
SPACE
LOTS
+ OF O2
SPACE

LOTS
OF MgO”
SPACE
One last thing . . . Please remember the following saying:
Give a man a
fish, and you
feed him for
a day.
Teach a man
to fish, and
you feed him
for a lifetime.
There’s only one teacher in the room, and I might be busy when
you need help.
If you run into trouble, ask the people around you for help.
If you do, don’t just ask for the answer.
Ask HOW you get the answer.
now
“SP Book Problems, 9:2 & 9:3”
YOUR ASSIGNMENT:
Section 9:2
pg 64:
Section 9:3
pg 65:
pg 66:
pg 67:
PROBLEMS 1-5
PROBLEMS 11 – 14
PROBLEMS 21 – 24
PROBLEMS 31 – 34
The answers are in the back of the book, on pages 273-274.
If you get one right, put a happy face. If you get it wrong but
understand it after you see the answer, put a check. If you still
don’t get it even after you see the answer, put a question mark
and get help from a neighbor or your teacher.
Have you tried balancing the equation below yet?
14KMnO4(s) + 4C3H5(OH)3(l) →
7K2CO3(s) + 7Mn2O3(s) + 5CO2(g) + 16H2O(g)
Don’t press any buttons until you’ve tried, or you’ll spoil the
surprise.
Did you get it right? If you couldn’t, you might ask Mr. Barnes to
show you his algebraic method for balancing difficult equations.
1.
C= 1
O= 2
C+
O2 
CO2
C= 1
O= 2
2.
H=2 4
O=2
2 H2 +
O2 
2 H2O
H =2 4
O =1 2
3. CH4 +
C= 1
H= 4
O=2 4
2 O2
2 H 2O +
CO2
C= 1
H= 2 4
O= 3 4
4.
NH3 +
N= 1
H= 5
O=1
N
H2O
NH4OH
N
N= 1
H= 5
O= 1
5.
C= 1
O=3
H= 2
CO2 +
H2O
H2CO3
C= 1
O= 3
H= 2
6. 2HCl +
H= 3 4
Cl = 1 2
Ba = 1
O= 2
Ba(OH)2
Cl
Cl
Ba
Cl
2 H 2O +
Ba
Cl
H=
Cl =
Ba =
O=
BaCl2
2 4
2
1
1 2
7.
HCl +
H= 2
Cl = 1
Na = 1
O= 1
NaOH
H 2O +
Na
Cl
Na
Cl
NaCl
H= 2
Cl = 1
Na = 1
O= 1
8.
2 NI3
N2 +
I
N= 1 2
I= 3 6
I
N
I
I
N N
I I
I
N
3 I2
I
I I
N= 2
I= 2 6
I I
9.
4 2 Fe + 3
Fe = 1 2 4
O= 2 3 6
Fe
3
2 O2
2 Fe2O3
Fe
Fe
Fe
Fe
Fe = 2 4
O= 3 6
Fe
Fe
Fe
10. 2 CH3OH + 3 23 O2
C= 1 2
H= 4 8
O= 3 4 8
2 CO2 + 4 2 H2O
C= 1 2
H= 2 4 8
O= 3 4 8
Title Page
5.
SP Book
SWBATS
6.
Demo EQ
answer
1.
7.
2.
8.
Just the
answers
3.
9.
4.
10.