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Stoichiometry: Limiting Stoichiometry: Limiting ReagentsReagentsAt the conclusion of our time At the conclusion of our time together, you should be able to:together, you should be able to:

1. Determine the limiting reagent2. Use the limiting reagent to determine the

amount of product produced in a reaction3. Determine the excess amount(s) in the

reaction

Limiting Reactant: CookiesLimiting Reactant: Cookies1 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 chipsMakes 3 dozen

If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? If not, what limits us??

What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? If not, what limits us???

What if we only had one egg, could we make 3 dozen cookies?

Limiting ReactantLimiting 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 excessexcess (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 reactantlimiting reactant.

Limiting ReactantLimiting ReactantTo find the correct amount of product

produced, we have to try allall of the reactants. We have to calculate how much of aa product we can get from each each of the reactants to determine which reactant is the limiting one.

The The lowerlower amount of amount of aa product is the product is the correct answer. correct answer.

The reactant that makes the least amount of product is the limiting reactantlimiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!

Be sure to pick Be sure to pick aa product! You can’t product! You can’t compare to see which is greater and which compare to see which is greater and which is lower unless the product is the same!is lower unless the product is the same!

Limiting Reactant: ExampleLimiting Reactant: Example10.0 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 Cl2 2 AlCl3Start with Al:

Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.33 g AlCl3

26.98 g Al 2 mol Al 1 mol AlCl3

= 49.42 g AlCl3

35.0 g Cl2 1 mol Cl2 2 mol AlCl3 133.33 g AlCl3

70.90 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.88 g AlCl3

LimitingLimitingReactantReactant

LR Example ContinuedLR Example Continued

We get 49.42 g of aluminum chloride from the given amount of aluminum, but only 43.88 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 .

Limiting Reactant PracticeLimiting Reactant Practice

15.0 g of aluminum reacts with 15.0 g of iodine. Calculate which reactant is limiting 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 iodine required to react completely with 15.0 g of aluminum.

Al + I2 AlI32 Al + 3 I2 2 AlI3

= 212 g I2

x 3 mol I2 2 mol Al

x 253.80 g I2

1 mol I2

15.0 g Al x 1 mol Al

26.98 g Al

Finding the Amount of ExcessFinding 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.

Can we find the amount of excess aluminum in the previous problem?

Finding Excess PracticeFinding Excess Practice15.0 g of aluminum reacts with 15.0 g of

iodine. Calculate the excess of aluminum.2 Al + 3 I2 2 AlI3

Always start with the limiting reactant:

15.0 g I2 1 mol I2 2 mol Al 26.98 g Al

253.80 g I2 1 mol I2 1 mol Al= 3.19 g Al USED!

15.0 g Al – 3.19 g Al = 11.81 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!

Limiting Reactant: RecapLimiting Reactant: Recap1. You can recognize a limiting reactant

problem because there is MORE THAN ONE GIVEN AMOUNT.

2. Convert ALL of the reactants to the SAME product (pick any product you choose.)

3. The lowest answer is the correct answer.4. The reactant that gave you the lowest

answer is the LIMITING REACTANT.5. The other reactant(s) are in EXCESS.6. To find the amount of excess, subtract

the amount used from the given amount.7. If you have to find more than one

product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

Stoichiometry: Limiting Stoichiometry: Limiting ReagentsReagentsLet’s see if you can:Let’s see if you can:

1. Determine the limiting reagent2. Use the limiting reagent to determine the

amount of product produced in a reaction3. Determine the excess amount(s) in the

reaction

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?

5.00 g Fe x 1 mol Fe

55.85 g Fe

= 0.0895 mol Fe

Fe + S FeS

5.00 g S x 1 mol S

32.07 g S

= 0.156 mol S

0.0895 mol Fe + 0.156 mol S FeS

The ratio must be 1:1, therefore Fe is limiting and will combine with 0.0895 mol S to produce 0.0895 mol of FeS

0.0895 mol S x 32.07 g S

1 mol S

= 2.87 g S used

5.00 g S – 2.87 g S used =

2.13 g excess S