Limiting Reagent Introduction The concept of the limiting reagent can be demonstrated by an analogy: Jerry works in the Purchasing Department at a specialty car manufacturing plant in Detroit. The plant gets an order for 100 limited edition cars for Macy’s parade. Jerry purchases 100 car bodies, 100 steering wheels, and 100 tires. Sadly, Jerry gets fired. Do you know why? You probably realized that Jerry did not order enough tires to make the 100 cars required. With his order Jerry could only make 25 cars, and he would have 75 leftover car bodies and steering wheels. In this story the tires are the “limiting reagent” and the bodies and steering wheels are “in excess.” Now let’s apply this concept to chemistry. Consider the balanced combustion reaction of methane gas: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) According to the reaction one mole of methane gas will react with two moles of oxygen gas to produce one mole of carbon dioxide gas and two moles of steam. Worked example: When a reaction involves a solution, the concentration of the solution is generally reported as molarity (shown in units of molar, M, which is equal to moles of solute per liter of solution)—e.g. a 1.50M sodium chloride solution has 1.50 moles of NaCl per liter of 1.50 mol NaCl solution or . Given the volume and molarity of any solution, we can calculate the L number of moles of solute present. For example, given 50.0 mL of a 1.50M sodium chloride solution, we can calculate the number of moles of NaCl present as follows: 50.0 mL 1L 1000 mL 1.50 mol NaCl L 0.0750 mol NaCl What mass of precipitate could be produced if 10.0 mL of 1.500M aluminum chloride reacts with 10.0 mL of 1.500M sodium hydroxide? First, we write the balanced chemical equation: AlCl3(aq) + 3 NaOH(aq) Al(OH)3(s) + 3 NaCl(aq) The precipitate formed is aluminum hydroxide. In this example we must first calculate moles for each reactant, then continue with the molar ratio step, then convert moles of the precipitate back to grams. 10.0 mL 10.0 mL 1L 1000 mL 1L 1000 mL 1.500 mol AlCl3 L 1 mol Al(OH)3 1 mol AlCl3 78.0 g Al(OH)3 1 mol Al(OH)3 1.500 mol NaOH L 1 mol Al(OH)3 3 mol NaOH 78.0 g Al(OH)3 1 mol Al(OH)3 = 1.17 g Al(OH)3 = 0.390 g Al(OH)3 page 1 of 2 Notice that the two product amounts are very different. Even though there is enough aluminum chloride to make 1.17 grams of Al(OH)3 precipitate, there is only enough sodium hydroxide to make 0.390 grams of Al(OH)3. Once that amount of precipitate is made, we run out of sodium hydroxide. Thus, in this example, sodium hydroxide is the limiting reagent, aluminum chloride is in excess, and 0.390 grams of aluminum hydroxide precipitate can be produced. While these calculations may not be as exciting as manufacturing specialty cars, if a chemist miscalculates limiting reagent problems he or she could get fired, cause a serious accident, and (in the extreme case) kill someone due to an explosion or other serious accident! In this lab you will conduct several trials of a reaction, determine the limiting reagent and the reactant in excess for each trial, and record your observations. page 2 of 2