In almost all reactions, matter is conserved: it is not created or destroyed, only transformed from one type into another. Therefore, you must have the same amount of matter on one side of a reaction as you have on the other. The molecules may be formed, destroyed, or changed, but the individual atoms remain. There is an exception to these rules, nuclear reactions, but these will not be discussed for a while yet.
Take a look at the reaction below. It is one we have referred to often, the combustion of hydrogen in oxygen to generate water. However, there is something wrong with this reaction. Can you see what it is?
H2 (g) + O2 (g) --> H2O (g)
If we consider each half of the reaction separately, the problem becomes apparent. There are two moles (or atoms, if you prefer to count this way) of both hydrogen and oxygen atoms on the left side; two moles of hydrogen also inhabit the right side, but only one mole of oxygen is present. It seems as if the extra mole oxygen just disappeared! Well, not really; the reaction is simply balanced incorrectly. In a proper reaction, the same number of moles (or atoms) of each element will appear on both sides. Think of a correct reaction as a scale; when equal amounts are on both sides, the reaction is in balance, much as a scale with equal masses on both pans is balanced. Now, can you find a way to fix the problem?
Since there are two moles of O atoms on the right but one on the left, two solutions are possible. First, we could change the coefficient of the O2 molecule so only one O is present on the left side. We do this by adding a 1/2 to the molecule; one-half mole of oxygen gas will contain one mole of oxygen atoms. The correct equation is shown below:
H2 (g) + 1/2 O2 (g) --> H2O (g)
Or, we could change the reaction so two moles of hydrogen are required and two moles of water are produced. Therefore, the two moles of oxygen atoms will be used up in the reaction:
2 H2 (g) + O2 (g) --> 2 H2O (g)
Either of these solutions is acceptable. Note that when balancing equations, the coefficient of each species should be as small as possible while avoiding fractions. In other words, the following two reaction equations are correctly balanced, but are not in "standard" form. The first could have each coefficient divided by two, to give a simpler equation; the second could be multiplied by two or four to minimize the number of fractions.
4 H2 (g) + 2 O2 (g) --> 2 H2O (g)
1/4 H2 (g) + 1/2 O2 (g) --> 1/2 H2O (g)
The process of balancing equations is called stoichiometry. It is an exercise you will do many, many times during your studies in chemistry. Knowing how to balance reactions quickly and accurately is crucial, since an error here can throw off the rest of a problem. The best way to balance a large equation is to tackle the biggest molecule first; set its coefficient to one (for now) and change the other coefficients as needed. Then, move on to progressively smaller molecules. Since water, carbon dioxide, and other common products are simple, save them for last. Note that in equations with ionic charges, the charges must be equal on both sides (for example, if two "-" charges exist on the left side, two must also exist on the right). In the paragraphs below, we will balance a fairly typical equation as an example.
FeCl3 (aq) + H2S (g) --> Fe2S3 + HCl
We will walk you through the process of balancing the reaction above. If you can handle this reaction, you should be able to follow the rest of this chapter.
The most complex molecule in this reaction is probably the Fe2S3, so let's give it a coefficient of one:
FeCl3 (aq) + H2S (g) --> 1 Fe2S3 + HCl
Since we have two iron (Fe) atoms on the right, we need two on the left, giving FeCl3 a coefficient of two:
2 FeCl3 (aq) + H2S (g) --> 1 Fe2S3 + HCl
We also have 3 sulphur atoms on the right, giving H2S a coefficient of 3 as well:
2 FeCl3 (aq) + 3 H2S (g) --> 1 Fe2S3 + HCl
Finally, we have 6 chlorine atoms on the left, giving a coefficient of 6 to the hydrochloric acid, HCl:
2 FeCl3 (aq) + 3 H2S (g) --> 1 Fe2S3 + 6 HCl
Let's check our work by calculating the number of each kind of atom on each side: 2 iron, 6 chlorine, 3 sulphur, and 6 hydrogen. Since the atoms are equivalent on each side, the reaction is now balanced. The practice problems below should give you a good handle on balancing equations. If you run into a dead end while balancing an equation, don't worry: just start over and try a different method.
Example Problem 1
Balance each reaction equation below. Use scratch paper if necessary to count up atoms
A. C2H6 + O2 --> H2O + CO2
B. Ag + O2 + H2S --> Ag2S + H2O
C. Ga2(CO3)3 + HCl --> GaCl3 + H2O + CO2
D. Cr2(SO4)3 + KOH --> Cr(OH)3 + K2SO4
E. Al + Fe3O4 --> Fe + Al2O3
A. C2H6 + 7/2 O2 --> 3 H2O + 2 CO2
B. 2 Ag + O2 + H2S --> Ag2S + 2 H2O
C. Ga2(CO3)3 + 6 HCl --> 2 GaCl3 + 3 H2O + 3 CO2
D. Cr2(SO4)3 + 6 KOH --> 2 Cr(OH)3 + 3 K2SO4
E. 8 Al + 3 Fe3O4 --> 9 Fe + 4 Al2O3