General Chemistry/Balancing Equations

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.

${\text{O}}_{2(g)}+{\text{H}}_{2(g)}\to {\text{H}}_2{O}_{(g)}$

This chemical equation shows the compounds being consumed and produced, however it does not appropriately deal with the quantities of the compounds. There appears 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. To make the equation balanced, add coefficients in front of each molecule as needed:

${\text{O}}_{2(g)}+2{\text{H}}_{2(g)}\to 2{\text{H}}_2{O}_{(g)}$

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. Lets consider a similar reaction between hydrogen and nitrogen:

${\text{N}}_{2(g)}+{\text{H}}_{2(g)}\to {\text{NH}}_{3(g)}$

Typically it is easiest to balance all pure elements last, especially hydrogen. First, by placing a two infront of ammonia, the nitrogens are balanced.

${\text{N}}_{2(g)}+{\text{H}}_{2(g)}\to 2{\text{NH}}_{3(g)}$

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.

${\text{N}}_{2(g)}+3{\text{H}}_{2(g)}\to 2{\text{NH}}_{3(g)}$

Remember when balancing a reaction that only coefficients can be changed, since changing a subscript would give a different reaction. As reactions become more complex they become more difficult to balance. For example the combustion of butane (lighter fluid):

${\text{C}}_4{H}_{10(g)}+{\text{O}}_{2(g)}\to {\text{CO}}_{2(g)}+{\text{H}}_2{O}_{(g)}$

Once again, it is better to leave pure elements till the end, so first we'll balance carbon and hydrogen. Oxygen can then be easily balanced. It is easy to see that one mole of butane will produce four moles of carbon dioxide and five moles of water.

${\text{C}}_4{H}_{10(g)}+{\text{O}}_{2(g)}\to 4{\text{CO}}_{2(g)}+5{\text{H}}_2{O}_{(g)}$

Now there are 13 oxygen atoms on the right and two on the left. The odd number of oxygens prevents blancing 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.

$2{\text{C}}_4{H}_{10(g)}+{\text{O}}_{2(g)}\to 8{\text{CO}}_{2(g)}+10{\text{H}}_2{O}_{(g)}$

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.

$2{\text{C}}_4{H}_{10(g)}+13{\text{O}}_{2(g)}\to 8{\text{CO}}_{2(g)}+10{\text{H}}_2{O}_{(g)}$