Stoichiometry is the study of the chemical ratios that cause chemical reactions to work the way they do. To really understand stoichiometry, let's start by examining how to "balance" a chemical equation. If I was studying a reaction, and I figured out that my reactants were C2H6 and O2, and my products are H2O and C2O, I need to balance this equation. Another way of thinking about this is, "How can I get the same number of atoms of each element on both sides of the equation?" Essentially, I have to realize that this equation produces three water molecules and two carbon dioxide molecules, and requires two oxygen molecules and one ethane molecule. When the molecules are present in this ratio, they complete this reaction without any difficulty.
So what would happen if one of these elements was not sufficient to complete the reaction--for example, if there was only one oxygen molecule instead of two for every ethane molecule? Well, this is what is known as a "limiting reagent" or a "limiting reactant."
To understand this idea a little better, let's do an example problem. (I can't claim credit for this one; I found it here.) If we have two grams of ammonia react with four grams of oxygen, then what will happen? Well, let's start by balancing the equation.
4 NH3 + 5 O2 ----> 4 NO + 6 H2O
Let's examine one of the products and see which reactant is the limiting reactant. A molar-mass conversion gives us
2 g NH3 x (1 mol NH3/17 g NH3) x (4 mol NO/4 mol NH3) x (30 g NO/1 mol NO) = 3.53 g NO
4 g O2 x (1 mol O2/32 g O2) x (4 mol NO/5 mol O2) x (30 g NO/1 mol NO) = 3 g NO
Here, we can see that oxygen is the limiting reagent, because it produces less NO than the 2 g of ammonia by itself would have.
We can also use these techniques to analyze how much of one element is needed to produce a different compound.
For example, let's consider calcium. How much calcium nitrate will be produced from 2 g of calcium?
As always, start with a balanced equation.
3Ca + N2 = Ca3Na2
Then, using molar mass techniques:
2 g Ca x (1 mol Ca/40 g Ca) x (1 mol Ca3N2/3 mol Ca) x (148 g Ca3N2/1 mol Ca3N2)
which gives us about 2.4 g of Ca3Na2.
Finally, these techniques can also be applied to find the percent yield of an equation. Considering the following example, I should expect to get 2.4 g of Ca3Na2 if I mix 2 g of calcium with nitrogen. However, even in a perfectly executed experiment, this won't be the case, for various reasons. If I recieved, say, 2 g of calcium nitrate, then I can say that my percent yield was about 83%, because 2/2.4 is .83333333....