LIMITING REACTANTS

Definitions Limiting reactant Excess reactant The reactant that gets used up completely during a chemical reaction. The reactant that is used to determine the amount of products that will form and the amount of other reactants that will get used up. Excess reactant The reactant that doesn’t get used up. Some of this reactant remains after the reaction is complete.

N2 (g) + 3 H2 (g)  2 NH3 (g) What is the limiting reactant? What is the excess reactant?

Simple analogy Bike factory 1 F + 2 T  1 B A bike factory has 300 tires and 160 frames. 1. What is the limiting reactant? 2. How many bikes can be made?

Strategies for determining the limiting reactant Strategy 1 – Pick a product Determine the amount of that product that each reactant will form. The reactant that makes the smallest amount of that product is the limiting reactant.

Bike example (300 tires, 160 frames) Determine how many bikes can form from each reactant. 300 tires  150 bikes 160 frames  160 bikes The tires produce a smaller number of bikes than the frames. Therefore, the tires are the limiting reactant and only 150 bikes can be made.

Chemistry Example What is the limiting reactant when 3.85 g of hydrogen gas reacts with 19.2 g of oxygen gas to form water? 2 H2 (g) + O2 (g)  2 H2O (g)

What is the limiting reactant when 3 What is the limiting reactant when 3.85 g of hydrogen gas reacts with 19.2 g of oxygen gas to form water? Strategy: Change both amounts of reactants to moles of water. Because the O2 makes the smaller amount of H2O, it is the limiting reactant.

What is the limiting reactant when 3 What is the limiting reactant when 3.25 L of propane are reacted with 33.4 g of oxygen? C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) What is our strategy? 1. Change both amounts of reactants to moles of carbon dioxide, OR 2. Change both amounts of reactants to moles of water. WHY MOLES?

Strategies for determining the limiting reactant Strategy 2 – Compare the reactants Start with one reactant. Determine the amount of the second reactant that is required to react all of the first reactant. Compare this answer to how much of the second reactant is available. If there is more of the second reactant available than is needed, then the second reactant is the excess reactant. If there is not enough of the second reactant available, then the second reactant is the limiting reactant.

Bike example (300 tires, 160 frames) Determine how many frames are needed to use up all 300 tires. 300 tires  150 frames 150 frames are needed and 160 frames are available. There are more than enough frames to use up all of the tires. Therefore, the tires are the limiting reactant.

Chemistry Example What is the limiting reactant when 3.85 g of hydrogen gas react with 19.2 g of oxygen gas to form water? 2 H2 (g) + O2 (g)  2 H2O (g)

What is the limiting reactant when 3 What is the limiting reactant when 3.85 g of hydrogen gas react with 19.2 g of oxygen gas to form water? Strategy: Determine the mass of oxygen gas needed to completely use up the 3.85 g of hydrogen. More oxygen gas is needed (30.8 g) to use up all of the H2 than is available (19.2 g). Not all of the hydrogen will react. Therefore, the oxygen runs out first and is the limiting reactant.

What is the limiting reactant when 3 What is the limiting reactant when 3.25 L of propane are reacted with 33.4 g of oxygen? C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) What is our strategy? 1. Convert the volume of propane to grams of oxygen. This tells us how much oxygen is needed to use up all of the propane. 2. Compare that answer to 33.4 g of oxygen. 3. If there is enough oxygen, then the propane is the limiting reactant.

Most limiting reactant problems don’t mention the words “limiting reactant” or even ask you to specifically determine the limiting reactant. It is important that you realize that any problem where you are given amounts of two or more reactants is a limiting reactant problem. Whether the question asks or not, you will need to determine the limiting reactant in these problems.

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) What mass of water is formed when 5.12 L of methane are reacted with 7.85 g of oxygen gas? What strategy? Why? Pick a product (grams of H2O). If both reactants are converted to grams of H2O, we know the smaller number is the answer.

CO(g) + H2O(g)  CO2(g) + H2(g) 3.25 g of CO reacts with 3.00 g of H2O. What is the limiting reactant? What mass of excess reactant remains? Strategy? Why? Compare reactants. We might be able to use our calculations in part 1 to answer part 2.

5 C(s) + 2 SO2(g)  CS2(g) + 4 CO(g) 40.5 grams of carbon react with 7.25 L of sulfur dioxide. What is the limiting reactant? How much excess reactant remains (use the same units given in the problem). Strategy? Why? Compare reactants. We might be able to use our calculations in part 1 to answer part 2.

C5H12(g) + 8 O2(g)  5 CO2(g) + 6 H2O(g) 14.2 grams of pentane react with 31.5 grams of oxygen gas. What volume of CO2 forms? What mass of H2O forms? What mass of excess reactant remains? Strategy? Use any method to solve one of the problems. Then, use the limiting reactant to answer the remaining questions.