Stoichiometry: Limiting Reagents
At the conclusion of our time together, you should be able to:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction

OK, you guys, you really need to follow my lead on this!!!

Limiting Reactant: Cookies
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed and 4 extra eggs, could we make 4 dozen cookies? If not, what limits us??
What if we only had one egg, could we make 3 dozen cookies???

Limiting Reactant



Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.
That reactant is said to be in excess (there is too much).
The other reactant limits how much product we get. Once it runs out, the reaction ’s. This is called the limiting reactant .


Limiting
Reactant
Limiting Reactant: Example
10.0 g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl
2
Start with Al:
 2 AlCl
3
10.0 g Al 1 mol Al 2 mol AlCl
3
133.33 g AlCl
3
= 49.4 g AlCl
3

26.98 g Al 2 mol Al 1 mol AlCl
3
Now Cl
2
:
35.0 g Cl
2
1 mol Cl
2
70.90 g Cl
2
2 mol AlCl
3
133.33 g AlCl
3
= 43.9 g AlCl
3
3 mol Cl
2
1 mol AlCl
3

LR Example Continued

We get 49.4 g of aluminum chloride from the given amount of aluminum, but only 43.9 g of aluminum chloride from the given amount of chlorine.
Therefore, chlorine is the limiting reactant. Once the
35.0 g of chlorine is used up, the reaction comes to a complete .

Because It Scares The Dog.

Limiting Reactant Practice #5


3.00 g of magnesium reacts with 2.20 g of oxygen.
Calculate which reactant is in excess and how much product is made.
A little quicker way to do this is to pick one reactant and determine what amount of the other reactant is needed to completely react the first.

Calculate the mass in grams of oxygen required to react completely with 3.00 g of magnesium.
Mg+ O
2
 MgO
2 Mg+ O
2
 2 MgO
3.00 g Mg x 1 mol Mg
24.31 g Mg x 32.00 g O
2
1 mol O
2 x 1mol O
2
2 mol Mg
=
1.97 g O
2

Finding the Amount of Excess


To completely react 3.00 g of magnesium, we would need 1.97 g of oxygen. Since we have
2.20 g of oxygen, the magnesium limits the amount of product we can make. Oxygen is in excess.
By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Finding the Amount of Excess




Can we find the amount of excess oxygen in the previous problem?
To completely react 3.00 g of magnesium, we would need 1.97 g of oxygen. Since we have
2.20 g of oxygen, the magnesium limits the amount of product we can make. Oxygen is in excess.
2.20 g O
2 available – 1.97 g O
2 used =
0.23 g O
2 excess

Calculate the mass in grams of magnesium oxide produced when 3.00 g of magnesium (the limiting reagent) is burned in excess oxygen.
Mg+ O
2
 MgO
2 Mg+ O
2
 2 MgO
3.00 g Mg x 1 mol Mg
24.31 g Mg x 40.31 g MgO
1 mol MgO x 2 mol MgO
=
2 mol Mg
4.97 g MgO

Gee, I wonder what’s going to happen next??

Stoichiometry: Limiting Reagents
At the conclusion of our time together, you should be able to:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction

1.
2.
3.
4.
5.
6.
7.
Limiting Reactant: Recap
You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN
AMOUNT.
Pick a reactant (A), convert grams to moles, and compare it to the grams needed of the other reactant (B).
If the answer for B is higher than what you have in the problem,
You don’t have enough B, therefore B is limiting.
If the answer for B is lower than what you have in the problem,
You have enough B, therefore A is limiting.
If you have enough B, subtract your answer for B from what you have in the problem = excess B.

Stoichiometry: Limiting Reagents
Let’s see if you can:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction

What mass of ZnO is formed when 20.0 g of
MoO
3 is reacted with 10.0 g of Zn?
3 Zn + 2 MoO
3
 Mo
2
O
3
+ 3 ZnO
20.0 g MoO
3 x 1 mol MoO
3
143.94 g MoO
3 x 3 mol Zn
2 mol MoO
3 x 65.39 g Zn = 13.6 g Zn
1 mol Zn

What mass of ZnO is formed when 20.0 g of
MoO
3 is reacted with 10.0 g of Zn?
We need 13.6 g of Zn so Zn is limiting!
3 Zn + 2 MoO
3
 Mo
2
O
3
+ 3 ZnO
10.0 g Zn x 1 mol Zn
65.39 g Zn x 3 mol ZnO x 81.39 g ZnO
3 mol Zn 1 mol ZnO
= 12.4 g ZnO

Stoichiometry: Limiting Reagents
Let’s see if you can:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction

Another Limiting Reagent Worksheet #1
3 NH
4
NO
3
+ Na
3
PO
4
 (NH
4
)
3
PO
4
+ 3 NaNO
3 a. Which is limiting?
30.0 g NH
4
NO
3 x 1 mol NH
4
NO
3
80.06 g
NH
4
NO
3 x 1 mol
Na
3
PO
4
3 mol
NH
4
NO
3 x 163.94 g
Na
3
PO
4
1 mol
Na
3
PO
4
=
20.5 g Na
3
PO
4

a. Which is limiting? NH
4
NO
3
3 NH
4
NO
3
+ Na
3
PO
4
 (NH
4
)
3
PO
4
+ 3 NaNO
3 b. Maximum amount of each product?
30.0 g NH
4
NO
3 x 1 mol NH
4
NO
3
80.06 g
NH
4
NO
3 x 1 mol
(NH
4
)
3
PO
4
3 mol
NH
4
NO
3 x 149.12 g
(NH
4
)
3
PO
4
1 mol
(NH
4
)
3
PO
4
=
18.6 g (NH
4
)
3
PO
4

a. Which is limiting? NH
4
NO
3
3 NH
4
NO
3
+ Na
3
PO
4
 (NH
4
)
3
PO
4
+ 3 NaNO
3 b. Maximum amount of each product?
30.0 g NH
4
NO
3 x 1 mol NH
4
NO
3
80.06 g
NH
4
NO
3 x 3 mol
NaNO
3
3 mol
NH
4
NO
3 x 85.00 g
NaNO
3
1 mol
NaNO
3
=
31.9 g NaNO
3

c. How much of the other reagent is left over??
3 NH
4
NO
3
+ Na
3
PO
4
 (NH
4
)
3
PO
4
+ 3 NaNO
3
50.0 g
Na
3
PO
4
- 20.5 g NH
3
PO
4
=
29.5 g NH
3
PO
4

Stoichiometry: Limiting Reagents
Let’s see if you can:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction

FeS Lab Problem:
In the lab where you combined iron (II) with sulfur, if 5.00 g of iron is combined with 5.00 g of sulfur, what is the limiting reagent and how much excess reagent is left?
Fe + S  FeS
5.00 g Fe x 1 mol Fe
55.85 g Fe x 1 mol S
I mol Fe x 32.07 g S
1 mol S
= 2.87 g S

When 5.00 g of Fe is used, I would need 2.87 grams of S to completely react the Fe. Since I have 5.00 g of S, I have plenty of S and therefore, the Fe is limiting.
2.87 g S used
5.00 g S
– 2.87 g S used =
2.13 g excess S

Stoichiometry: Limiting Reagents with
Percent Yield
At the conclusion of our time together, you should be able to:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction
4. Determine the percent yield for a problem

What is the Percent Yield of a Chemical
Reaction?
Actual Product Produced
Theoretical Product that Should be Produced
X
100 =
Percent Yield

Limiting Reactant Practice

15.0 g of aluminum reacts with 15.0 g of iodine.
Calculate which reactant is limiting, how much excess reactant is available and how much product is made.

Determine the percent yield if 10.5 g of aluminum iodide is actually produced.

Calculate the mass in grams of iodine required to react completely with 15.0 g of aluminum.
Al + I
2
 AlI
3
2 Al + 3 I
2
 2 AlI
3
15.0 g Al x 1 mol Al
26.98 g Al x 253.80 g I
2
1 mol I
2 x 3 mol I
2
2 mol Al
=
212 g I
2

Finding the Amount of Excess


To completely react 15.0 g of aluminum, we would need 212 g of iodine. Since we only have
15.0 g of iodine, the iodine limits the amount of product we can make.
By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Finding Excess Practice

15.0 g of aluminum reacts with 15.0 g of iodine.
Calculate the excess of aluminum.
2 Al + 3 I
2
 2 AlI
3
Always start with the limiting reactant:
15.0 g I
2
1 mol I
2
2 mol Al 26.98 g Al
253.80 g I
2
3 mol I
2
1 mol Al
= 1.06 g Al
USED!
15.0 g Al – 1.06 g Al = 13.9 g Al EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only use it!

Calculate the mass in grams of aluminum iodide that would be produced in this reaction.
Al + I
2
 AlI
3
2 Al + 3 I
2
 2 AlI
3
15.0 g I
2 x 1 mol I
2
253.80 g
I
2 x 407.68 g AlI
3
1 mol I
2 x 2 mol AlI
3
3 mol
I
2
=
16.1 g
AlI
3


Determine the percent yield if 10.5 g of aluminum iodide is produced.
16.1 g of AlI
3 should have been produced. What is the percent yield?
2 Al + 3 I
2
 2 AlI
3
10.5 g AlI
3
16.1 g AlI
3 x 100
=
65.2 % Yield

What's The Difference Between a Bad
Golfer And a Bad Skydiver?
A Bad Golfer Goes, Whack, Dang!
A Bad Skydiver Goes Dang! Whack.

Stoichiometry: Limiting Reagents with
Percent Yield
Let’s see if you can:
1. Determine the limiting reagent
2. Use the limiting reagent to determine the amount of product produced in a reaction
3. Determine the excess amount(s) in the reaction
4. Determine the percent yield for a problem

Percent Yield #1


How many grams of antimony(III) iodide would be produced?
Determine the percent yield if 118.00 g of antimony(III) iodide is actually produced.

How many grams of antimony(III) iodide would be produced?
Sb + I
2
 SbI
3
2 Sb + 3 I
2
 2 SbI
3
98.60 g I
2 x 1 mol I
2
253.80 g I
2 x 502.46 g SbI
3
1 mol SbI
3 x 2 mol SbI
3
3 mol I
2
=
130.1 g
SbI
3


Determine the percent yield if 118.00 g of antimony
(III) iodide is produced.
130.1 g of Sbl
3 should have been produced. What is the percent yield?
2 Sb + 3 I
2
 2 SbI
3 x 118.00 g SbI
3
130.1 g SbI
3 x 100
=
90.70 % Yield