Unit 8 –
Chemical Equations and Reactions
• Notes p.1-2
• These ones are in order!
Format of Chemical Equations:
2H2 + O2  2H2O
Reactants/Reagents
Format of Chemical Equations:
2H2 + O2  2H2O
Product(s)
Format of Chemical Equations:
2H2 + O2  2H2O
Yield(s)
Format of Chemical Equations:
2H2 + O2  2H2O
Subscript(s)
Format of Chemical Equations:
2H2 + O2  2H2O
Coefficients(s)
Format of Chemical Equations:
• Coefficients
• Large numbers in front
• Give how many molecules of each
substance are in the equation
• Subscripts
• Small numbers after an element
• Give how many of each atom are needed
in the formula
Now for some practice
• These are not in the notes but are essential skills to have
What is the coefficient for …
• 2Al + 3CuSO4  Al2(SO4)3 + 3Cu
• Aluminum: 2
• Copper sulfate: 3
• Aluminum sulfate: 1 (when there is no number written, it is 1)
• Copper: 3
What is the coefficient for …
• 2H2O  2H2 + O2
• Water: 2
• Hydrogen gas: 2
• Oxygen gas: 1
How many ___ are there in ___
• Mg(OH)2 + 2HCl  MgCl2 + 2H2O
• How many H are in Mg(OH)2?
• 2 (distribute the subscript through the parentheses)
• How many H are in HCl?
• 2 (the coefficient means there are 2 H)
• How many H are in the reactants?
• 4 total: 2 from HCl and 2 from Mg(OH)2
• How many H are in the products?
• 4 (all from H2O)
• Law of Conservation of mass
• Matter cannot be created nor destroyed, only
change form
• Balancing Equations
• Must be the same number of each type of atom
on the reactant side and the product side
Even though a reaction has taken place,
the mass at the beginning and the mass at
end are the same. No mass (or matter) is
gained or lost in the reaction.
• Never change SUBSCRIPTS in
balancing!!
• Only change COEFFICIENTS!!!
• To Balance an Equation:
1. Total the number of each element for both sides
2. Check if each element is the same on both sides
3. Change the total on the inventory list so that the
number of atoms matches.
4. Add a coefficient to the substance in the equation
5. Adjust the inventory list to reflect any changes in other
atoms
6. Move on to each atom until the lists match for EVERY
element
• Tricks:
• Balance polyatomic ions as a group
• If an element is present twice on one side, leave
it for last
• If the list isn’t working because you have an odd
number on one side and an even number on the
other, try doubling everything
• Simplify: Check after balancing for any factor
that could divide through all the coefficients
EXAMPLE: WRITE A BALANCED EQUATION
• Chlorine + sodium bromide  bromine + sodium chloride
Cl2 +
NaBr

Br2 +
NaCl
Cl: 2
Na: 1
Br: 1
Cl: 1
Na: 1
Br: 2
(You might need to brush
up on writing chemical
formulas if you are only
given the name.)
• 1. Total the number of each element for both sides
• Look at subscripts to tell how many are in the unbalanced equation
• All coefficients in an unbalanced equation are 1
EXAMPLE: WRITE A BALANCED EQUATION
• Chlorine + sodium bromide  bromine + sodium chloride
Cl2 +
NaBr

Br2 +
NaCl
Cl: 2
Na: 1
Br: 1
≠
=
≠
Cl: 1
Na: 1
Br: 2
2. Check if each element is the same on both sides
- Chlorine and bromine are unbalanced
EXAMPLE: WRITE A BALANCED EQUATION
• Chlorine + sodium bromide  bromine + sodium chloride
Cl2 +
NaBr

Br2 +
NaCl
Cl: 2
Na: 1
Br: 1
≠
=
≠
Cl: 1 2
Na: 1
Br: 2
3. Change the total on the inventory list so that the number of atoms
matches.
- I like to start with the first element on my inventory list. It doesn’t really
matter which one you start with.
EXAMPLE: WRITE A BALANCED EQUATION
• Chlorine + sodium bromide  bromine + sodium chloride
Cl2 +
NaBr

Br2 +
2 NaCl
Cl: 2
Na: 1
Br: 1
≠
=
≠
Cl: 1 2
Na: 1
Br: 2
4. Add a coefficient to the substance in the equation
- 1 x coefficient = 2, therefore the coefficient is 2
EXAMPLE: WRITE A BALANCED EQUATION
• Chlorine + sodium bromide  bromine + sodium chloride
Cl2 +
NaBr

Br2 +
2 NaCl
Cl: 2
Na: 1
Br: 1
=
≠
≠
Cl: 1 2
Na: 1 2
Br: 2
5. Adjust the inventory list to reflect any changes in other atoms
- The new coefficient applies to both Na and Cl. According to the
equation, there are now 2 Na in the products.
EXAMPLE: WRITE A BALANCED EQUATION
• Chlorine + sodium bromide  bromine + sodium chloride
Cl2 +
2 NaBr

Br2 +
2 NaCl
Cl: 2
Na: 1 2
Br: 1 2
=
=
=
Cl: 1 2
Na: 1 2
Br: 2
6. Move on to each atom until the lists match for EVERY element
- Na is now unbalanced. The new coefficient in front of NaBr should
be 2. That means there are now also 2 Br. The equation is balanced!
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ CaCl2

AlCl3
+ CaSO4
Al: 2
Al: 1
Trick: Because SO4 is present as a
SO4: 3
SO4: 1
polyatomic ion on both sides, it
(sulfate) can be treated as one unit
Ca: 1
Ca: 1
instead as S and O separately.
Cl: 2
Cl: 3
• 1. Total the number of each element for both sides
• Look at subscripts to tell how many are in the unbalanced equation
• All coefficients in an unbalanced equation are 1
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ CaCl2

AlCl3
+ CaSO4
Al: 2
≠
Al: 1
SO4: 3
≠
SO4: 1
Ca: 1
=
Ca: 1
Cl: 2
≠
Cl: 3
2. Check if each element is the same on both sides
- Aluminum, sulfate, and chlorine are unbalanced
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ CaCl2

AlCl3
+ CaSO4
Al: 2
=
Al: 1 2
SO4: 3
≠
SO4: 1
Ca: 1
=
Ca: 1
Cl: 2
≠
Cl: 3
3. Change the total on the inventory list so that the number of atoms matches.
- I like to start with the first element on my inventory list. It doesn’t really
matter which one you start with – but do one at a time!
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ CaCl2

2 AlCl3
+ CaSO4
Al: 2
=
Al: 1 2
SO4: 3
≠
SO4: 1
Ca: 1
=
Ca: 1
Cl: 2
≠
Cl: 3
4. Add a coefficient to the substance in the equation
- 1 x coefficient = 2, therefore the new coefficient is 2. Now Al is balanced.
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ CaCl2

2 AlCl3
+ CaSO4
Al: 2
=
Al: 1 2
SO4: 3
≠
SO4: 1
Ca: 1
=
Ca: 1
Cl: 2
≠
Cl: 3 6
5. Adjust the inventory list to reflect any changes in other atoms
- The new coefficient applies to both Al and Cl. According to the equation, there
are now 6 Cl in the products (2 from coefficient x 3 from subscript).
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ 3 CaCl2

2 AlCl3
+ CaSO4
Al: 2
=
Al: 1 2
SO4: 3
≠
SO4: 1
Ca: 1
=
Ca: 1
Cl: 2 6
=
Cl: 3 6
6. Move on to each atom until the lists match for EVERY element
- Cl in the reactants is now unbalanced. There need to be 6. 2 x coefficient = 6,
so the coefficient for CaCl2 needs to be 3. Now Cl is balanced.
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ 3 CaCl2

2 AlCl3
+ CaSO4
Al: 2
=
Al: 1 2
SO4: 3
≠
SO4: 1
Ca: 1 3
=
Ca: 1 3
Cl: 2 6
=
Cl: 3 6
6. Move on to each atom until the lists match for EVERY element
- Changing the coefficient of CaCl2 means there are now 3 Ca. To balance
calcium we need to change the number in the products.
EXAMPLE: WRITE A BALANCED EQUATION
aluminum sulfate+ calcium chloride  aluminum chloride +calcium sulfate
Al2(SO4)3
+ 3 CaCl2

2 AlCl3
+ 3 CaSO4
Al: 2
=
Al: 1 2
SO4: 3
=
SO4: 1 3
Ca: 1 3
=
Ca: 1 3
Cl: 2 6
=
Cl: 3 6
6. Move on to each atom until the lists match for EVERY element
- Calcium now balances! The new coefficient means there are three sulfates.
Now sulfate balances!
Practice Problems
1 C3H8 + ___
5 O2 
___
HINTS
3 CO2
___
4 H2O
+ ___
Balance oxygen LAST!
3 O2
2 NaBr + ___
2 NaBrO3  ___
___
1 NH4NO3  ___
1 N2O
___
1
___ H
2
2 H2O
+ ___
Balance oxygen LAST! Make sure to
account for both N in the reactants.
1 CO + ___
1 O2  ___
1 H2CO3
+ ___
It was already balanced! Yay!
1 La2O3 + ___
3 H2O  ___
2 La(OH)3
___
Balance oxygen LAST!
2 NaI + ___
1 Pb(NO3)2  ___
2 NaNO3 + ___PbI
1 2
___
Treat the nitrate ion as one unit.