1. 4.3 Limiting Reactant, Theoretical Yield and Percent Yield

2. Limiting Reactant and Excess reactants and Theoretical Yield The reactant that makes the least amount of product is the limiting reactant. For example, going back to Mr. Stevenson’s egg sandwich, if we started with 7 eggs, 5 slices of cheese and 8 english muffins, a tub of butter, and 40 slices of bacon, which would be the limiting reactant? As a reminder: 2 eggs + 4 bacon strips + 1 slice of cheese + 1 English muffin + butter  Mr. Stevenson’s egg sandwich The eggs would be the limiting reactant since we would only be able to make 3 sandwiches (this would be the theoretical yield) All other reactants would be considered in excess.

3. Actual Yield and Percent Yield If after making the 3 sandwiches one dropped on the floor…this would make me sad…but besides that we would end up with 2 sandwiches (this is the actual yield). Percent yield is the percentage of the theoretical yield actually attained. actual yield % yield = ------------------------ X 100 theoretical yield What is the percent yield if sandwiches made? 2 sandwiches % yield of sandwiches = --------------------- X 100 = 67% 3 sandwiches

4. Limiting Reactant and Excess reactants and Theoretical Yield from Initial Reactant Masses To solve one of these types of problems where we begin the problem being given the masses of each reactant, we must: Step 1: Convert g of each reactant to moles Step 2: Convert moles of each reactant to moles of the product being asked about. Step 3: That reactant that produces the least amount of product is the limiting reactant, the other is the excess reactant. Step 4: Use the least amount of moles attained and convert to mass of product. This is your theoretical yield. Step 5: If the actual yield is also given in the problem, or an actual yield is obtained in a lab, you can calculate the percent yield.

5. Let’s Try a Few Sample Problems Ammonia can be synthesized by the reaction: 3H 2 (g) + N 2 (g)  2NH 3 (g) What is the theoretical yield of ammonia, in kg, that we can synthesize from 5.22 kg of H 2 and 31.5 kg of N 2 ?

6. Sample Problem 1 Worked Out 3H 2 (g) + N 2 (g)  2NH 3 (g) Step 1 Find the limiting reagent: 1000 g H 2 1 mol H 2 2 mol NH 3 5.22 kg H 2 X--------------- X ------------ X -------------- = 1.72X10 3 mol NH 3 1 kg H 2 2.02 g H 2 3 mol H 2 1000 g N 2 1 mol N 2 2 mol NH 3 31.5 kg N 2 X -------------- X --------------- X -------------- = 2.25X10 3 mol NH 3 1 kg N 2 28.02 g N 2 1 mol N 2 H 2 is the limiting reagent

7. Sample Problem 1 (Continued) Step 2: Calculate the theoretical yield of ammonia in kg. 17.04 g NH 3 1 kg NH 3 1.72X10 3 mol NH 3 X ------------------- X ---------------- = 29.3 kg NH 3 1 mol NH 3 1000 g NH 3 29.3 kg NH 3 is the theoretical yield!

8. Let’s Try Another! Mining companies use this reaction to obtain iron from iron ore: Fe 2 O 3 (s) + 3CO(g)  2Fe(s) +3CO 2 (g) The reaction of 167 g Fe 2 O 3 with 85.8 g CO produces 72.3 g Fe. Determine the limiting reactant, theoretical yield, and percent yield.

9. Sample Problem 2 Worked Out Fe 2 O 3 (s) + 3CO(g)  2Fe(s) +3CO 2 (g) Step 1: Find the limiting reagent. 1.00 mol Fe 2 O 3 2 mol Fe 176 g Fe 2 O 3 X ----------------------- X ---------------- = 2.20 mol Fe 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 1.00 mol CO 2 mol Fe 85.8 g CO X -------------------- X -------------- = 2.04 mol Fe 28.01 g CO 3 mol CO CO is the limiting reagent!

10. Sample 2 (continued) Step 2: Calculate theoretical yield 55.85 g Fe 2.04 mol Fe X ---------------- = 114 g Fe  theoretical yield 1 mol Fe Step 3: Calculate percent yield actual yield 72.3 g Fe % yield Fe = ----------------------- X 100 = ------------- X 100 = 63.4% theoretical yield 114 g Fe

11. Chapter 4 pg. 187 #’s 38, 40, 44, & 50 (only the a’s) Study for Quiz Chapter 3