Lecture 11

Limiting Reagent and Percent Yield

A. A Simple Example of Limiting Reagent

In the last lecture, we made turkey sandwiches according to the following "equation" (the recipe):

2 Bd + 3 Tk ---> Bd₂Tk₃

That is, one sandwich (Bd₂Tk₃) is made by combining 3 slices of turkey with 2 slices of bread (Bd). Our equation is balanced: the ingredients (reactants) are used up to form the sandwich (product).

Now let's consider the concept of limiting reagent with this same example. Suppose you have 10 slices of of bread and 24 slices of turkey. We can ask these questions:

• Which ingredient is used up first (the limiting reagent)?
• Which ingredient is left over?
• How much of it remains after the other ingredient is used up?
• How many sandwiches can you make?

Solution: Let's consider how much turkey we need to go with 10 slices of bread. Our "stoichiometric factor" from the above equation is 3 slices Tk/2 slices Bd:

But we know that we have 24 slices of turkey, so we have more turkey than we need to go with the 10 slices of bread. Thus the bread is said to be the limiting reagent: it is used up first, and limits the number of sandwiches that can be made by the recipe given.

When we've made all the sandwiches we can, there will be 9 slices of turkey left over (24 available, 15 used in the sandwiches.

If we use all our bread, we can make 5 sandwiches:

 

B. A Simple Example of % Yield

You used enough bread and turkey to make 5 sandwiches. However, your dog grabs a sandwich while your back is turned. Now you have only 4 sandwiches, not the 5 you could have made. So your % yield is

C. A Chemical Example of Limiting Reagent: Production of Ammonia, NH₃

Ammonia is produced in a combination reaction between hydrogen and nitrogen:

3 H₂ + N₂ ---> 2 NH₃

Suppose we have 50.0 g H₂ and 200. g N₂. Let's answer these questions, taking the molar masses of reactants and products to 3 significant figures. (H₂ 2.02 g/mol; N₂ 28.0 g/mol; NH₃ 17.0 g/mol).

• Which reactant is used up first (the limiting reagent)?
• Which reactant is left over?
• How much of it remains after the other reactant is used up?
• How much ammonia (in grams) can you make?

Solution: Step 1. First we need to determine the moles of each reactant, since our balanced equation is in terms of moles (3 moles of H₂ react with 1 mole of N₂ to give 2 moles of NH₃). We'll carry one extra significant figure in our reported figures, and only round down at the end of the calculations.

moles H₂ = 50.0 g H₂ x (1 mol H₂ / 2.02 g H₂) = 24.75 mol H₂
moles N₂ = 200. g N₂ x (1 mol N₂ / 28.0 g N₂) = 7.143 mol N₂

Step 2. Now let's calculate how much N₂ is needed to react with all the hydrogen:

(24.75 mol H₂) x (1 mol N₂/3 mol H₂) = 8.251 mol N₂ needed

But we don't have 8.251 mol N₂! We only have 7.143 moles, so N₂ is limiting.

Step 3. If N₂ is limiting, there must be leftover H₂. Let's check and see:

(7.143 mol N₂) x (3 mol H₂ / mol N₂) = 21.43 mol H₂ needed to react with all the N₂. But we have 24.75 mol H₂, so we have the amount needed, plus some left over:

Have 24.75 mol H₂
Use up 21.43 mol H₂
Remaining: (24.75 - 21.43) mol H₂ = 3.32 mol H₂ left over

Step 4. Since N₂ is limiting, we have to use it to calculate the amount of NH₃ that can be made:

(7.143 mol N₂) x (2 mol NH₃ / mol N₂) x (17.0 g NH₃ / mol NH₃)

= 243 g NH₃ is the maximum possible yield (all N₂ is used up)

D. A Chemical Example of % Yield: Production of Ammonia, NH₃

In the example in part C, suppose we actually obtain 211 g NH₃. What is the % yield?

 

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/CHEM1P80_Lecture_11.html
Created October 3, 2000 by M. F. Richardson
© Brock University, 2000