# General Chemistry/Balancing Equations

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## 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 N$_2$ (2 atoms per molecule x 1 molecule), and 6 atoms of hydrogen gas in the 3 H$_2$ 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 (NH$_3$) 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.

### An Example

| ${\hbox{O}}_{2}+{{\hbox{H}}_{2}}\to {\hbox{H}}_{2}{\hbox{O}}$ | This chemical equation shows the compounds being consumed and produced; however, it does not appropriately deal with the quantities of the compounds. There appear to be two oxygen atoms on the left and only one on the right. But we know that there should be the same number of atoms on both sides. This equation is said to be unbalanced, because the number of atoms are different. |
| ${\hbox{O}}_{\color {Blue}2}+{\color {Blue}2}{{\hbox{H}}_{2}}\to {\color {Blue}2}{\hbox{H}}_{2}{\hbox{O}}$ | To make the equation balanced, add coefficients in front of each molecule as needed. The 2 in front of hydrogen on the left indicates that twice as many atoms of hydrogen are needed to react with a certain number of oxygen atoms. The coefficient 1 is not written, since it assumed in the absence of any coefficient. |
| ${\hbox{N}}_{2}+{{\hbox{H}}_{2}}\to {\hbox{NH}}_{3}$ | Now, let's consider a similar reaction between hydrogen and nitrogen. |
| ${\hbox{N}}_{\color {Blue}2}+{{\hbox{H}}_{2}}\to {\color {Blue}2}{\hbox{NH}}_{3}$ | Typically, it is easiest to balance all pure elements last, especially hydrogen. First, by placing a two in front of ammonia, the nitrogens are balanced. |
| ${\hbox{N}}_{\color {Blue}2}+{\color {Red}3}{{\hbox{H}}_{\color {Blue}2}}\to {\color {Blue}2}{\hbox{NH}}_{\color {Red}3}$ | This leaves 6 moles of atomic hydrogen in the products and only two moles in the reactants. A coefficient of 3 is then placed in front of the hydrogen to give a fully balanced reaction. |

## Tricks in balancing certain reactions
### 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 C$_6$H$_6$ can only appear on the product side in CO$_2$. Thus, we can write a coefficient of 6 in front of CO$_2$.
Next, note that the hydrogen in C$_6$H$_6$ can only go to H$_2$O. Thus, we put a 3 in front of H$_2$O.
We have 15 oxygen atoms on the product side, so there are ${\frac {15}{2}}$ O$_2$ molecules on the reactant side. To make this an integer, we multiply all coefficients by 2.

### Another Example

| ${\hbox{C}}_{4}{\hbox{H}}_{10}+{{\hbox{O}}_{2}}\to {\hbox{CO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}$ | As reactions become more complex, they become more difficult to balance. For example, the combustion of butane (lighter fluid). |
| ${\hbox{C}}_{\color {Red}4}{\hbox{H}}_{\color {Blue}10}+{{\hbox{O}}_{2}}\to {\color {Red}4}{\hbox{CO}}_{2}+{\color {Blue}5}{\hbox{H}}_{\color {Blue}2}{\hbox{O}}$ | Once again, it is better to leave pure elements until the end, so first we'll balance carbon and hydrogen. Oxygen can then be balanced after. It is easy to see that one mole of butane will produce four moles of carbon dioxide and five moles of water. |
| ${\color {OliveGreen}2}{\hbox{C}}_{4}{\hbox{H}}_{10}+{{\hbox{O}}_{\color {OliveGreen}2}}\to {\color {OliveGreen}8}{\hbox{CO}}_{2}+{\color {OliveGreen}10}{\hbox{H}}_{2}{\hbox{O}}$ | Now there are 13 oxygen atoms on the right and two on the left. The odd number of oxygens prevents balancing with elemental oxygen. Because elemental oxygen is diatomic, this problem comes up in nearly every combustion reaction. Simply double every species except for oxygen to get an even number of oxygen atoms in the product. |
| $2{\hbox{C}}_{4}{\hbox{H}}_{10}+{\color {Blue}13}{{\hbox{O}}_{\color {Blue}2}}\to {\color {Blue}8}{\hbox{CO}}_{\color {Blue}2}+{\color {Blue}10}{\hbox{H}}_{2}{\hbox{O}}$ | The carbon and hydrogens are still balanced, and now there are an even number of oxygens in the product. Finally, the reaction can be balanced. |