Write a balanced stoichiometric equation for the reaction in question. Consider, for example, the reaction between sulfuric acid and sodium hydroxide. Acids react with bases to make water and a salt, in this case sodium sulfate. Each sulfuric acid supplies two hydrogen ions to the reaction, but sodium hydroxide can only supply one hydroxide ion. So, twice as many moles of sodium hydroxide are required to react with sulfuric acid; 0.75 moles of sulfuric acid requires 1.5 moles of sodium hydroxide to react completely.
Calculate the number of moles of reactant you have. If the reactants are pure, divide the mass (in grams) of the reactant by the reactants molar mass (grams per mole). If the reactants are in solution, multiply the concentration (in moles per liter) by the volume of solution you have (in liters). For example, if you were mixing 10 grams of pure sulfuric acid with 10 grams of sodium hydroxide you would have [10 grams / 98.08 grams per mole = 0.102 moles] of sulfuric acid and [10 grams / 40 grams per mole = 0.25 moles] of sodium hydroxide.
Calculate the number of moles required if a particular reaction were to react completely using the balanced stoichiometric equation and the actual amount of that reactant you have. Keeping in line with the previous example, if 0.102 moles of sulfuric acid were to react completely, [2 x 0.102 = 0.204 moles] of sodium hydroxide would be required.
Consider how many moles of reactant are present and how many would be required for one of them to react completely. In Step 3 it was calculated that 0.204 moles of sodium hydroxide are needed for all of the sulfuric acid to react. We have 0.25 moles of sodium hydroxide, so there is more than enough to react with all of the sulfuric acid. There will be sodium hydroxide left over. Since the sulfuric acid will run out first, it is the limiting reagent. The will be 0.046 moles of sodium hydroxide left over.
Consider the reaction of hydrochloric acid with sodium hydroxide to make water and sodium chloride, as shown in balanced equation. If the reaction began with 0.3 liters of 2 molar (moles per liter) hydrochloric acid and 0.2 liters of 2 molar sodium hydroxide, what will be the limiting reagent? First, calculate the number of moles of reactant: [0.3 liters x 2 mole per liter = 0.6 moles] hydrochloric acid and [0.2 x 2 moles per liter = 0.4 moles] sodium hydroxide. The reactants react one to one, so if all of the hydrochloric acid were to react, it would require 0.6 moles of sodium hydroxide. But, there isn't 0.6 moles of sodium hydroxide, there is only 0.4. So, in this case, sodium hydroxide is the limiting reagent. There will be 0.2 moles of hydrochloric acid left over.