C. Johannesson
I
II III IV V

1. Write the unbalanced equation.
2. Count atoms on each side.
3. Add coefficients to make #s equal.
Coefficient
 subscript = # of atoms
4. Reduce coefficients to lowest possible ratio, if necessary.
5. Double check atom balance!!!
C. Johannesson

 Balance one element at a time.
 Update ALL atom counts after adding a coefficient.
 If an element appears more than once per side, balance it last.
 Balance polyatomic ions as single units.
 “1 SO
4
” instead of “1 S” and “4 O”
C. Johannesson

Aluminum and copper(II) chloride react to form copper and aluminum chloride.
2 3
2

3 2
3
2

1 1

2
3
 
3
6
 
6
C. Johannesson

 Mr. Rapps slides next
C. Johannesson

 Compare the numbers of each kind of atom in the balanced equation with the numbers of each kind of atom in the sketched representation. Both the equation and the sketch have the same number of atoms in the reactants and in the products.

Practice


Al(s) + O
2
(g) ---> Al
2
O
3
(s)
Inventory: 1Al and 2O’s on left and
2Al’s and 3O’s on right.
 Start with Al
 Need 2 on left to balance right
 2 Al(s) + O
2
(g) ---> Al
2
O
3
(s)
 Al is balanced but O is not.
 Double everything but O
2

Practice
 4 Al(s) + O
2
(g) ---> 2 Al
2
O
3
(s)
 Al is still balanced, now we have
2O’s on left and 6O’s on right.
 How can we balance that?
 3x2 = 6 so we need 3 O
2 molecules


4 Al(s) + 3 O
2
(g) ---> 2 Al
2
O
3
(s)
Inventory: 4Al’s and 6O’s on left and
4Al’s and 6O’s on right

It’s balanced









Practice
Balance: FeCl
3
+ NaOH  Fe(OH)
3
+ NaCl
Inventory: 1Fe, 3Cl’s, 1Na and 1OH on left and 1Fe, 3OH’s, 1Na and 1Cl on right
Balance Cl first  need 3 on right
FeCl
3
+ NaOH  Fe(OH)
3
+ 3 NaCl
Na is now unbalanced  need 3 on left now
FeCl
3
+ 3 NaOH  Fe(OH)
3
+ 3 NaCl
Inventory again: 1Fe, 3Cl, 3Na, and 3OH on left and 1Fe, 3OH, 3Na, and 3Cl on right
It’s balanced!!!

Practice





Ba(NO
3
)
2
+ Al
2
(SO
4
)
3
 BaSO
4
+ Al(NO
3
)
3
Inventory: 1Ba, 2NO
3
’s, 2Al’s and 3SO
4 left and 1Ba, 1SO
4
, 1Al, and 3NO
3
’s on
’s on right
Balance SO
4
’s first  need 3 on right
Ba(NO
3
)
2
+ Al
2
(SO
4
)
3
 3 BaSO
4
Balance Ba next  need 3 on left
+ Al(NO
3
)
3


3 Ba(NO
3
)
2
Al(NO
3
)
3
+ Al
2
(SO
4
)
3
 3 BaSO
4
+
Balance NO3’s next  6 on left and 3 on right, use 2 as coefficient

Practice



3 Ba(NO
3
)
2
2 Al(NO
3
)
3
+ Al
2
(SO
4
)
3
 3 BaSO
4
+
That gives 6NO
3
’s on both sides
Inventory: 3Ba’s, 6NO3’s, 2Al’s and 3SO4’s on left and 3Ba’s, 3SO4’s, 2Al’s, and 6NO3’s on right.

It’s balanced

Practice


Ca(OH)
2
+ H
3
PO
4
 Ca
3
(PO
4
)
2
+ H
2
O
Inventory: 1Ca, 5H’s, 6O’s and 1P on left and 3Ca’s, 9O’s, 2H’s, and 2 P on right.
 Balance Ca first  need 3 on left




3 Ca(OH)
2
+ H
3
PO
4
 Ca
3
(PO
4
)
2
+ H
2
O
Balance PO
4
’s next  need 2 on left
3 Ca(OH)
2
+ 2 H
3
PO
4
 Ca
3
(PO
4
)
2
+ H
2
O
Inventory again for H’s and O’s:
6O’s+8O’s(14) and 6H’s+6H’s(12) on left and 8O’s+1O’s(9) and 2H’s on right

Practice

To get 12H’s on right we can use 6 as a coefficient for water


3 Ca(OH)
2
6 H
2
O
+ 2 H
3
PO
4
 Ca
3
(PO
4
)
2
+
Inventory: 3Ca’s, 2P’s, (6+6=12)H’s, and (6+8=14)O’s on left and 3Ca’s,
2P’s, (6x2=12)H’s and (8+6=14)O’s on right

It’s balanced

Practice
 ____C
3
H
8
(g) + _____ O
2
(g) ---->

>
_____CO
2
(g) + _____ H
2
O(g)
____B
4
H
10
___ B
2
O
3
2
(g) ----
2
O(g)

Practice
 NH
4
NO
3
---> N
2
O + 2 H
2
O
 Carbon monoxide reacts with nitrogen monoxide to form carbon dioxide and nitrogen gas (pairs)
 Equation: CO + NO  CO
2
2 CO + NO 
2 CO
2
+ N
2
+ N
2

Practice
 Phosphorus trichloride reacts with water to form hydrochloric acid and
 phosphorous acid.
Equation: PCl
3
+ H
2
O

HCl + H
3
PO
3
 PCl
3
+ H
2
O 
3 HCl + H
3
PO
3
 Magnesium nitride reacts with water to
 form magnesium hydroxide and ammonia
Equation: Mg
3
N
2
+ H
2
O

Mg(OH)
2
+ NH
3
Mg
3
N
2
6
2
O 
3
Mg(OH)
2
2
3