General Chemistry/Balancing Equations

Balancing Equations
Chemical equations are useful because they give the relative amounts of the substances that react in a chemical equation.

$$\hbox{N}_{2} + 3{\hbox{H}_{2}} \to 2\hbox{NH}_{3}$$

In some cases, however, we may not know the relative amounts of each substance that reacts. Fortunately, we can always find the correct coefficients of an equation (the relative amounts of each reactant and product). The process of finding the coefficients is known as balancing the equation.

During a chemical reaction, atoms are neither created or destroyed. The same atoms are present before and after a reaction takes place; they are just rearranged. This is called the Law of Conservation of Matter, and we can use this law to help us find the right coefficients to balance an equation.

For example, assume in the above equation that we do not know how many moles of ammonia gas will be produced:

$$\hbox{N}_{2} + 3{\hbox{H}_{2}} \to {\color{Red}?}\hbox{NH}_{3}$$

From the left side of this equation, we see that there are 2 atoms of nitrogen gas in the molecule N2 (2 atoms per molecule x 1 molecule), and 6 atoms of hydrogen gas in the 3 H2 molecules (2 atoms per molecule x 3 molecules). Because of the Law of Conservation of Matter, there must also be 2 atoms nitrogen gas and 6 atoms of hydrogen gas on the right side. Since each molecule of the resultant ammonia gas (NH3) contains 1 atom of nitrogen and 3 atoms of hydrogen, 2 molecules are needed to obtain 2 atoms of nitrogen and 6 atoms of hydrogen.

Combustion
A combustion reaction is a reaction between a carbon chain (basically, a molecule consisting of carbons, hydrogen, and perhaps oxygen) with oxygen to form carbon dioxide and water, plus heat. Combustion reactions could get very complex: $$2\hbox{C}_6\hbox{H}_6 + 15\hbox{O}_2 \to 12 \hbox{CO}_2 + 6\hbox{H}_2\hbox{O}$$

Fortunately, there is an easy way to balance these reactions.

First, note that the carbon in C6H6 can only appear on the product side in CO2. Thus, we can write a coefficient of 6 in front of CO2.

Next, note that the hydrogen in C6H6 can only go to H2O. Thus, we put a 3 in front of H2O.

We have 15 oxygen atoms on the product side, so there are $$\frac{15}{2}$$ O2 molecules on the reactant side. To make this an integer, we multiply all coefficients by 2.