4 3 Limiting Reactant Theoretical Yield and Percent

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4. 3 Limiting Reactant, Theoretical Yield and Percent Yield

4. 3 Limiting Reactant, Theoretical Yield and Percent Yield

Limiting Reactant and Excess reactants and Theoretical Yield The reactant that makes the least

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 theoretical yield) All other reactants would be considered in excess.

Actual Yield and Percent Yield If after making the 3 sandwiches one dropped on

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 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

Limiting Reactant and Excess reactants and Theoretical Yield from Initial Reactant Masses To solve

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.

Let’s Try a Few Sample Problems Ammonia can be synthesized by the reaction: 3

Let’s Try a Few Sample Problems Ammonia can be synthesized by the reaction: 3 H 2(g) + N 2(g) 2 NH 3(g) What is theoretical yield of ammonia, in kg, that we can synthesize from 5. 22 kg of H 2 and 31. 5 kg of N 2?

Sample Problem 1 Worked Out 3 H 2(g) + N 2(g) � 2 NH

Sample Problem 1 Worked Out 3 H 2(g) + N 2(g) � 2 NH 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 -------------- = 1. 72 X 103 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 ------- = 2. 25 X 103 mol NH 3 1 kg N 2 28. 02 g N 2 1 mol N 2 H 2 is the limiting reagent

Sample Problem 1 (Continued) Step 2: Calculate theoretical yield of ammonia in kg. 17.

Sample Problem 1 (Continued) Step 2: Calculate theoretical yield of ammonia in kg. 17. 04 g NH 3 1 kg NH 3 1. 72 X 103 mol NH 3 X ---------- X -------- = 29. 3 kg NH 3 1 mol NH 3 1000 g NH 3 29. 3 kg NH 3 is theoretical yield!

Let’s Try Another! Mining companies use this reaction to obtain iron from iron ore:

Let’s Try Another! Mining companies use this reaction to obtain iron from iron ore: Fe 2 O 3(s) + 3 CO(g) 2 Fe(s) +3 CO 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.

Sample Problem 2 Worked Out Fe 2 O 3(s) + 3 CO(g) 2 Fe(s)

Sample Problem 2 Worked Out Fe 2 O 3(s) + 3 CO(g) 2 Fe(s) +3 CO 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!

Sample 2 (continued) Step 2: Calculate theoretical yield 55. 85 g Fe 2. 04

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 = 63. 4% theoretical yield 114 g Fe

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

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

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