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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 + _ O2 ===> _ H2O + _ 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 + _ O2 ===> 2 H2O + CO2

CH4 + 2 O2 ===> 2 H2O + CO2

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

_ C4H10 + _ O2 ===> _ CO2 + _ H2O

Insert the coefficients for carbon dioxide and water.

_ C4H10 + _ O2 ===> 4 CO2 + 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 + _ O2 ===> 8 CO2 + 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 O2 ===> 8 CO2 + 10 H2O

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

4 C4H10 + 26 O2 ===> 16 CO2 + 20 H2O

or

6 C4H10 + 39 O2 ===> 24 CO2 + 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 + _ O2 ===> _ CO2 + _ H2O

First take care of the carbon and hydrogen.

_ C3H7OH + _ O2 ===> 3 CO2 + 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 + _ O2 ===> 6 CO2 + 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 O2 ===> 6 CO2 + 8 H2O

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