# Balancing Burning Equations

Most burning reactions are the oxidation of a fuel material
with oxygen gas. Complete burning produces carbon dioxide
from all the carbon in the fuel, water from the hydrogen in the
fuel, and sulfur dioxide from any sulfur in the fuel. Methane
burns in air to make carbon dioxide and water.

_ CH4 + _
O
2 ===> _
H
2O + _ CO2

Easy. Put a two in front of the water to take care of all the
hydrogens and a two in front of the oxygen. Anything you have to
gather (any atom that comes from two or more sources in the
reactants or gets distributed to two or more products) should
be considered last.

CH4 + _
O
2 ===> 2 H2O +
CO
2

CH4 + 2
O
2 ===> 2
H
2O + CO2

What if the oxygen does not come out right? Let’s
consider the equation for the burning of butane, C4H10.

_ C4H10 + _
O
2 ===> _
CO
2 + _ H2O

Insert the coefficients for carbon dioxide and water.

_ C4H10 + _
O
2 ===> 4
CO
2 + 5 H2O

We now have two oxygens on the left and thirteen oxygens on
the right. The real problem is that we must write the oxygen
as a diatomic gas. The chemical equation is not any different
from an algebraic equation in that you can multiply both sides
by the same thing and not change the equation. Multiply both
sides by two to get the following.

2 C4H10 + _
O
2 ===> 8
CO
2 + 10 H2O

Now the oxygens are easy to balance. There are twenty-six
oxygens on the right, so the coefficient for the oxygen gas on
the left must be thirteen.

2 C4H10 + 13
O
2 ===> 8
CO
2 + 10 H2O

Now it is correctly balanced. What if you finally balanced the
same equation with:

4 C4H10 + 26
O
2 ===> 16
CO
2 + 20 H2O

or

6 C4H10 + 39
O
2 ===> 24
CO
2 + 30 H2O

Either equation is balanced, but not to the lowest integer.
Algebraically you can divide these equations by two or three to
get
the lowest integer coefficients in front of all of the materials
in the equation.

Now that we are complete pyromaniacs, let’s try burning
isopropyl alcohol, C3H7OH.

_ C3H7OH + _
O
2 ===> _
CO
2 + _ H2O

First take care of the carbon and hydrogen.

_ C3H7OH + _
O
2 ===> 3
CO
2 + 4 H2O

But again we come up with an oxygen problem. The same process
works here. Multiply the whole equation (except
oxygen) by two.

2 C3H7OH + _
O
2 ===> 6
CO
2 + 8 H2O

Now the number nine fits in the oxygen coefficient. (Do
you understand why?) The equation is balanced with six
carbons, sixteen hydrogens, and twenty oxygens on each
side.

2 C3H7OH + 9
O
2 ===> 6
CO
2 + 8 H2O

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