Balancing Equations
Balancing combustion reactions, where a compound
is burned in the presence of O2 to make H2O and
CO2 sometimes presents its own set of problems.
Let’s start with the example below:
C3H8(g)
( ) + O2(g)
( ) = CO2(g)
( ) + H2O(g)
O( )
? C3H8(g) + ? O2(g) = ? CO2(g) + ? H2O(g)
C
3
0
1
0
In combustions reactions, dealing with the O is tricky
because there are two different products that contain O.
As a result I always save O for last when I balance
a combustion reaction. Therefore let’s start with the C.
We have 3 C’s on the left, so we will need 3 CO2’s
on the
th right.
i ht
1 C3H8(g) + ? O2(g) = 3 CO2(g) + ? H2O(g)
C
H
1(3)
1(8)
( )
0
0
3(1)
0
0
2
Since we are saving O for last, that makes H next.
We have 8 H’s on the left, so we will need 4 H2O’s
on the
th right.
i ht
1 C3H8(g) + ? O2(g) = 3 CO2(g) + 4 H2O(g)
C
H
O
1(3)
1(8)
( )
0
0
0
2
3(1)
0
3(2)
0
4(2)
( )
4(1)
And finally the O’s.
We have a total of 6+ 4 O’s on the left, so we will need
10 O’s
O’ on the
th right,
i ht or 5 O2’s.
’
1 C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g)
C
H
O
1(3)
1(8)
( )
0
0
0
5(2)
3(1)
0
3(2)
0
4(2)
( )
4(1)
Well that wasn’t so bad. Our final answer is:
1 C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g)
That was nice and easy.
g a little harder.
Now let’s do something
This is a similar combustion that is a little trickier:
? C4H10(g) + ? O2(g) = ? CO2(g) + ? H2O(g)
C
4
0
1
0
As before, start with C.
4 C’s on the left,, so we need 4 CO2’s on the right.
g
1 C4H10(g) + ? O2(g) = 4 CO2(g) + ? H2O(g)
C 1(4)
0
4(1)
0
H 1(10)
0
0
2
Next move on to H’s.
10 H’s on the left,, so we need 5 H2O’s on the right.
g
1 C4H10(g) + ? O2(g) = 4 CO2(g) + 5 H2O(g)
C 1(4)
0
4(1)
0
H 1(10)
0
0
5(2)
O
0
2
4(2)
5(1)
Last, but
b not least,
l
the
h O’s
’
There are 8 + 5 or 13 O’s on the right, so we need 13 O’s
on the
h left.
l f But how
h
do
d you get O2 to give you 13 O’s?
’
What we will do is a cross-multiplication.
p
We will multiply
py
anything on the right side of the equation that has O’s by
the number of O’s on the left side, and multiply anything
that has O’s
O s on the left hand
hand, by the number of O’s
O s on the
right.
1 C4H10(g) + ? O2(g) = 4 CO2(g) + 5 H2O(g)
O
0
2
2 O’s in 1O2
O
0
4(2)
5(1)
13 O’s total this side
Cross-multiplication
Cross
multiplication
1(2)
4(2)
[[1(2)]x13
( )] 3 [[4(2)
( )
5(1)
5(
5(1)]x2
)]
After the cross multiplication
p
our equation
q
looks like this:
? C4H10(g) + 13 O2(g) = 8 CO2(g) + 10 H2O(g)
(g). Once you
y
Notice how I stuck a ? In front of the C4H10(g)
have done a cross multiplication, you have to start all over!
That’s what we’ll do next.
? C4H10(g) + 13 O2(g) = 8 CO2(g) + 10 H2O(g)
O
C
0
4
13(2)
0
8(2)
8(1)
10(1)
0
Now we have to start all over with the C’s
8C
C’ss on the right
right, so we need two molecules of C4H10(g)
2 C4H10(g) + 13 O2(g) = 8 CO2(g) + 10 H2O(g)
O
C
H
0
2(4)
2(10)
13(2)
0
0
8(2)
8(1)
0
10(1)
0
10(2)
Now try the H’s again
Hooray, it finally worked. Our final answer is:
2 C4H10(g) + 13 O2(g) = 8 CO2(g) + 10 H2O(g)
(g)
Now try the last balancing tutorial to see where
cross-multiplication is used in other problems.