1. Section 12.3: Limiting Reagent and Percent Yield
Stoichiometry III
Section 12.3: Limiting Reagent and Percent Yield

2. Objectives Upon completion of this presentation, you will be able to
identify the limiting reagent in a reaction.
calculate the theoretical yield, actual yield, or percent yield of a chemical reaction given appropriate information.

3. Introduction Let’s say that we are going to open up a sandwich shop.
Eventually, we are going to make a variety of sandwiches to sell.
But now, we are only going to offer one sandwich: a roast beef sandwich.
A single roast beef sandwich is made up of –
¼ lb. of beef
2 slices of bread
1 leaf of lettuce
½ tomatoes, sliced
2 Tsp. of mayonnaise
1 Tsp. of mustard

4. The next thing we need to do is to get our supplies.
We buy the following:
50 lb of roast beef
100 loaves of bread
10 heads of lettuce
50 tomatoes
10–16 oz. jars of mayonnaise
5–8 oz. jars of mustard

5. How many sandwiches can we make?
We can make 200 sandwiches from 50 lb of roast beef (50 lb./¼ lb. per sandwich = 200 sandwiches).
We can make 1,000 sandwiches from 100 loaves of bread (each loaf has 20 slices; that’s 10 sandwiches per loaf; 10×100 = 1,000 sandwiches).
We can make 120 sandwiches from 10 heads of lettuce (each head has about 12 good leaves; 12×10 = 120 sandwiches).
We can make 100 sandwiches from 50 tomatoes (we use ½ a tomato per sandwich; 2×50 = 100 sandwiches).
We can make 160 sandwiches from 10–16 oz. jars of mayonnaise (2 Tsp. is 1 oz.; 10×16 = 160 sandwiches).
We can make 80 sandwiches from 5–8 oz. jars of mustard (2 Tsp. is 1 oz.; 2×8×5 = 80 sandwiches).

6. Let’s look at this in another form
Let’s look at this in another form. The maximum number of sandwiches we can make is 80 sandwiches. The amount of mustard limits the number of sandwiches we can make. In chemistry, we call this the limiting reagent.
ingredient
quantity
sandwiches
Roast beef
50 lb.
200
Bread
100 loaves
1,000
Lettuce
10 heads
120
Tomatoes
50 tomatoes
100
Mayonnaise
10–16 oz. jars
160
Mustard
5–8 oz. jars 80

7. Limiting and Excess Reagents
In any chemical reaction, the amount of product is limited by the amount of reactants.
The more reactants we have, the more product we can get.
When we have a reaction with more than one reactant, the reactant that runs out first is called the limiting reagent.
The reactant that does not run out is call the excess reagent.

8. Limiting and Excess Reagents
In order to determine the amount of product we should get from a reaction, we need to know the limiting reagent.
We use stoichiometry to do this.

9. Limiting and Excess Reagents
For example, 48.0 g of methane, CH4, is burned with g of oxygen gas. What is the limiting reagent (which will run out faster, CH4 or O2)?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
M (g/mol)
n (mol) ____ ____ ① ②
m (g)
3.00
5.00
First, we calculate the number of mols of each reactant.
① nCH4 = mCH4/MCH4
= (48.0 g)/(16.0 g/mol)
= 3.00 mol
② nO2 = mO2/MO2
= (160. g)/(32.0 g/mol)
= 5.00 mol

10. Limiting and Excess Reagents
For example, 48.0 g of methane, CH4, is burned with g of oxygen gas. What is the limiting reagent (which will run out faster, CH4 or O2)?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
M (g/mol)
m (g)
n (mol)
3.00
5.00
Next, we calculate the number of mols of O2 needed to react with all of the CH4.
nCH4
nO2 = 1 2
⇒ 1 × nO2 = 2 × nCH4
⇒ nO2 = 2 × 3.00 mol
= 6.00 mol
But, we only have 5.00 mols of O2.
We will run out of O2 before we run out of CH4.
O2 is the limiting reagent.

11. Limiting and Excess Reagents
For example, 56.0 g of nitrogen reacts with 18.0 g of hydrogen to form ammonia. What is the limiting reagent (which will run out faster, N2 or H2)?
N2(g) + 3 H2(g) → 2 NH3(g)
M (g/mol)
n (mol) ____ ____ ① ②
m (g)
2.00
9.00
First, we calculate the number of mols of each reactant.
① nN2 = mN2/MN2
= (56.0g)/(28.0 g/mol)
= 2.00 mol
② nH2 = mH2/MH2
= (18.0 g)/(2.02 g/mol)
= 9.00 mol

12. Limiting and Excess Reagents
For example, 56.0 g of nitrogen reacts with 18.0 g of hydrogen to form ammonia. What is the limiting reagent (which will run out faster, N2 or H2)?
N2(g) + 3 H2(g) → 2 NH3(g)
M (g/mol)
n (mol) ____ ____ ① ②
m (g)
2.00
9.00
Next, we calculate the number of mols of H2 needed to react with all of the N2.
nN2
nH2 = 1 3
⇒ 1 × nH2 = 3 × nN2
⇒ nH2 = 3 × 2.00 mol
= 6.00 mol
But, we have 9.00 mols of H2.
We will run out of N2 before we run out of H2.
N2 is the limiting reagent.

13. Limiting and Excess Reagents
Sample problems. Find the limiting reagent for each of the following reactions.
N2(g) + 3 H2(g) → 2 NH3(g) [28.0 g N2 & 2.50 g H2]
CaO(s) + CO2(g) → CaCO3(s) [112 g CaO & 66.0 g CO2]
2 H2(g) + O2(g) → 2 H2O(l) [12.0 g H2 & 16.0 g O2]
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) [32.0 g CH4 & 168 g O2]
2 Mg(s) + O2(g) → 2 MgO(s) [121.5 g Mg & 64.0 g O2]
H2 is the limiting reagent.
CO2 is the limiting reagent.
O2 is the limiting reagent.
CH4 is the limiting reagent.
O2 is the limiting reagent.

14. Limiting and Excess Reagents
We can also use our knowledge of the limiting reagent to determine the quantity of a product in a reaction.
Again, we use stoichiometry to do this.

15. Limiting and Excess Reagents
For example, what is the maximum amount of CO2 that will form from 64.0 g of methane, CH4, and 224 g of oxygen gas?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
M (g/mol)
n (mol) ____ ____ ① ②
m (g)
4.00
7.00
First, we calculate the number of mols of each reactant.
① nCH4 = mCH4/MCH4
= (64.0 g)/(16.0 g/mol)
= 4.00 mol
② nO2 = mO2/MO2
= (224 g)/(32.0 g/mol)
= 7.00 mol

16. Limiting and Excess Reagents
For example, what is the maximum amount of CO2 that will form from 64.0 g of methane, CH4, and 224 g of oxygen gas?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
M (g/mol)
m (g)
n (mol)
4.00
7.00
Next, we calculate the number of mols of O2 needed to react with all of the CH4.
nCH4
nO2 = 1 2
⇒ 1 × nO2 = 2 × nCH4
⇒ nO2 = 2 × 4.00 mol
= 8.00 mol
But, we only have 7.00 mols of O2.
We will run out of O2 before we run out of CH4.
O2 is the limiting reagent and it will control the amount of CO2 produced.

17. Limiting and Excess Reagents
For example, what is the maximum amount of CO2 that will form from 64.0 g of methane, CH4, and 224 g of oxygen gas?
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
M (g/mol)
m (g)
154 ④
n (mol)
4.00
7.00
3.50 ③
Now, we use the amount of O2 to calculate the maximum amount of CO2 produced.
nO2
nCO2 = 2 1 ③
⇒ 2 × nCO2 = 1 × nO2
⇒ nCO2 = ½ × 7.00 mol
= 3.50 mol
mCO2 = nCO2 × MCO2 ④
= (3.50 mol) × (44.0 g/mol)
= 154 g

18. Limiting and Excess Reagents
Find the limiting reagent and determine the maximum amount of product that will result.
N2(g) + 3 H2(g) → 2 NH3(g) [42.0 g N2 & 10.0 g H2; find NH3]
CaO(s) + CO2(g) → CaCO3(s) [112 g CaO & 132 g CO2; find CaCO3]
2 H2(g) + O2(g) → 2 H2O(l) [12.0 g H2 & 16.0 g O2; find H2O]
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) [64.0 g CH4 & 288 g O2; find CO2]
2 Mg(s) + O2(g) → 2 MgO(s) [121.5 g Mg & 96.0 g O2; find MgO]
N2 is the limiting reagent; 51.0 g of NH3 produced.
CaO is the limiting reagent; 200. g of CaCO3 produced.
O2 is the limiting reagent; 18.0 g of H2O produced.
CH4 is the limiting reagent; 176 g of CO2 produced.
Mg is the limiting reagent; 202 g of MgO produced.

19. Percent Yield
When we do a stoichiometric calculation, we predict the maximum yield of products from the limiting reagents.
For example, we calculate the maximum mass that we could expect from using the masses of reactants we used.
This maximum yield is called the theoretical yield.

20. Percent Yield
When we actually do the reaction in the laboratory, the amount of product we get is usually less than that theoretical yield.
Why might that be so?
We might have had trouble transferring the product to the balance.
We might have lost some during heating or filtration or any other activity that involves handling the product.

21. Percent Yield
We want to have a measure of the efficiency of the reaction carried out in the laboratory.
We want something that tells us how close we came to the theoretical yield.
We use percent yield.
Percent yield is the ratio of the actual yield to the theoretical yield as a percentage.
Percent yield = × 100%
actual yield
theoretical yield

22. Sample Problem 12.9
Calcium carbonate, which is found in sea shells, is decomposed by heating. What is the theoretical yield of CaO if 24.8 g of CaCO3 is heated? The balanced chemical equation is: CaCO3(s) → CaO(s) + CO2(g)
M (g/mol) ∆
m (g)
13.9
n (mol) ① ③
0.248 ②
0.248
① nCaCO3 =
mCaCO3
MCaCO3 =
24.8 g
100.1 g/mol
= mol
② =
nCaO
nCaCO3 1
⇒ nCaO = nCaCO3
= mol
③ mCaO = nCaO × MCaO
= (0.248 mol)(56.1 g/mol)
= 13.9 g

23. Sample Problem 12.10
What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated? CaCO3(s) → CaO(s) + CO2(g) From Sample Problem 12.9, the theoretical yield is 13.9 g. The actual yield is 13.1 g. ∆
Percent yield = × 100%
theoretical yield
actual yield
Percent yield = × 100%
13.9 g
13.1 g
= 94.2%

24. Summary
In any chemical reaction, the amount of product is limited by the amount of reactants.
When we have a reaction with more than one reactant, the reactant that runs out first is called the limiting reagent.
The reactant that does not run out is call the excess reagent.
We can also use our knowledge of the limiting reagent to determine the quantity of a product in a reaction.

25. Summary
We want to have a measure of the efficiency of the reaction carried out in the laboratory.
We want something that tells us how close we came to the theoretical yield.
We use percent yield.
Percent yield is the ratio of the actual yield to the theoretical yield as a percentage.