Every chemical reaction can be described by a chemical equation. Consider the reaction that was responsible for the discovery of oxygen: the decomposition of mercuric oxide to metallic mercury and oxygen gas. This reaction is given in equation form as:
2 HgO → 2 Hg + O2
In the equation, the arrow should be read as "produces" or "goes to". The substances on the left side of the arrow are called the reactants or starting materials; the substances on the right are the products. As the name implies, every equation must be balanced; that is, have the same number of atoms of each element on each side of the equation. This fact, that matter is neither created nor destroyed during a chemical reaction, is called the Law of Conservation of Mass. Notice that the equation has two mercury atoms on the left because of the coefficient in front of the HgO and two mercury atoms on the right because of the coefficient in front of the Hg. The equation has two oxygen atoms on the left as a result of the coefficient in front of the HgO and two oxygen atoms on the right as a consequence of the fact that elemental oxygen exists as diatomic molecules.
As a second example, consider the reaction between elemental nitrogen and elemental hydrogen to form ammonia. This reaction is the extremely important nitrogen fixation reaction that describes how nitrogen in the air is converted in certain types of plants to ammonia, which is later incorporated into amino acids and proteins. In writing an equation for this reaction, we must first ask, "what are the molecular formulas for elemental nitrogen, hydrogen, and ammonia?" We then write these formulas along with the "goes to" arrow:
N2 + H2 → NH3
Now we ask, "Is this reaction balanced? Are there the same number of atoms of each element on both sides of the arrow?" The answer, of course, is no: the left side contains two nitrogens, while the right side has only one. Moreover, there are two hydrogens on the left, but three on the right. We must remedy this situtation by placing coefficients in front of at least some, perhaps all, of the substances. If we start with the nitrogens we can see that the nitrogens are easily balanced by placing a two in front of the ammonia. However, that coefficient of two means that there now are six hydrogens on the right and only two on the left. Fortunately, this can be easily taken care of by placing a three in front of the H2:
N2 + 3 H2 → 2 NH3
The equation is now balanced, and we can read it as "one nitrogen molecule reacts with three hydrogen molecules to produce two ammonia molecules." Or, if we prefer to think in moles (remember that chemists are accustomed to working with gram-size quantities, which means that they like to deal with moles), we can read it as "one mole of nitrogen molecules reacts with three moles of hydrogen molecules to give two moles of ammonia molecules."
Write an equation, balanced, of course, for the decomposition of CuCl to Cu and CuCl2
2 CuCl → Cu + CuCl2
Using the equation in the question above, determine how many moles of CuCl is required to produce one mole of CuCl2
The equation tells us the two moles of CuCl are required to produce one mole of CuCl2.
When you start thinking about chemical equations in terms of moles, you have entered the important arena of chemical stoichiometry. Stoichiometry is a strange word, derived from the Greek meaning "to measure the elements". Chemists are thoroughly trained in this area of chemistry because all chemical reactions must be described by an equation, and these equations tell the chemical engineer, for example, how many tons of ammonia will be produced in a reactor when a certain amount of nitrogen and hydrogen are mixed, given that the yield for the reaction is, say, 30%. The yield of a reaction is the percent of the product that is actually obtained under experimental conditions relative to the amount that would be produced if the reaction went 100% to completion. For example, if the preparation of ammonia at 400 °C and 300 atmospheres of pressure were started with 1.0 mole of N2 and the reaction went to completion, 2.0 moles of ammonia would be obtained. If the reaction does not go to completion under these conditions and only 1.0 mole of ammonia is produced, the yield would be only 50%.