4.3 Limiting Reactant,

Theoretical Yield and Percent

Yield

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.

Actual Yield and Percent YieldIf 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

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:

3H2(g) + N2(g) 2NH3(g)

What is the theoretical yield of ammonia, in kg, that we can synthesize from 5.22 kg of H2 and 31.5 kg of N2?

Sample Problem 1 Worked Out

3H2(g) + N2(g) 2NH3(g)

Step 1 Find the limiting reagent:

1000 g H2 1 mol H2 2 mol NH3

5.22 kg H2 X--------------- X ------------ X -------------- = 1.72X103 mol NH3

1 kg H2 2.02 g H2 3 mol H2

1000 g N2 1 mol N2 2 mol NH3

31.5 kg N2 X -------------- X --------------- X -------------- = 2.25X103 mol NH3

1 kg N2 28.02 g N2 1 mol N2

H2 is the limiting reagent

Sample Problem 1 (Continued)

Step 2: Calculate the theoretical yield of ammonia in kg.

17.04 g NH3 1 kg NH3

1.72X103 mol NH3 X ------------------- X ---------------- = 29.3 kg NH3

1 mol NH3 1000 g NH3

29.3 kg NH3 is the theoretical yield!

Let’s Try Another!

Mining companies use this reaction to obtain iron from iron ore:

Fe2O3(s) + 3CO(g) 2Fe(s) +3CO2(g)

The reaction of 167 g Fe2O3 with 85.8 g CO produces 72.3 g Fe. Determine the limiting reactant, theoretical yield, and percent yield.

Sample Problem 2 Worked Out

Fe2O3(s) + 3CO(g) 2Fe(s) +3CO2(g)

Step 1: Find the limiting reagent.

1.00 mol Fe2O3 2 mol Fe

176 g Fe2O3 X ----------------------- X ---------------- = 2.20 mol Fe

159.7 g Fe2O3 1 mol Fe2O3

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 Fe2.04 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

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