Limiting Reactant

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Limiting Reactant

• Theoretical and Percent Yield

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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

Limiting Reactant - The reactant in a chemical
reaction that limits the amount of product that
can be formed.  The reaction will stop when all
of the limiting reactant is consumed. Excess
Reactant - The reactant in a chemical reaction
that remains when a reaction stops when the
limiting reactant is completely consumed.  The
excess reactant remains because there is nothing
with which it can react.
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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.

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Limiting Reactant

• To find the correct answer, we have to try all of
  the reactants. We have to calculate how much of
  a product we can get from each of the reactants
  to determine which reactant is the limiting one.
• The lower amount of a product is the correct
  answer.
• The reactant that makes the least amount of
  product is the limiting reactant.
• Be sure to pick the same product!

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Limiting Reactant Example
LimitingReactant

• 10.0g 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 AlCl3
• Start with Al
• Now Cl2

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g
AlCl3 27.0 g Al 2 mol Al
1 mol AlCl3
49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g
AlCl3 71.0 g Cl2 3 mol Cl2
1 mol AlCl3
43.9g AlCl3
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Theoretical Yield

• We get 49.4g of aluminum chloride from the given
  amount of aluminum, but only 43.9g of aluminum
  chloride from the given amount of chlorine.
  Therefore, chlorine is the limiting reactant.
• Theoretical Yield The predicted amount of
  product is the theoretical yield. 43.9 g of
  AlCl3 is the calculated product, so that is the
  theoretical yield.

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Limiting Reactant Practice

• 15.0 g of potassium reacts with 15.0 g of iodine.
  Calculate which reactant is limiting and how
  much product is made.
• 2 K I2 ? 2 KI
• 15.0 g K x 1mole x 2 KI x 166 g 63.7 g KI
• 40 g 2 K 1 mole
• 15. 0 g I2 x 1 mole x 2 KI x 166g 19.8 g KI
• 252 g I2 1 mole

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Finding the Amount of 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.
• Can we find the amount of excess potassium in the
  previous problem?

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Finding Excess Practice

• 15.0 g of potassium reacts with 15.0 g of iodine.
  2 K I2 ? 2 KI
• We found that Iodine is the limiting reactant,
  and 19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K
254 g I2 1 mol I2 1
mol K
4.62 g K USED!
15.0 g K 4.62 g K 10.38 g K 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 start
with it!
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Percentage Yield

• Percentage yield is the percent of the
  theoretical yield that was made by our
  experiment. 
•                       actual amount of
  productpercentage yield -----------------------
  ----------------- x 100                          
    theoretical yield
• (the amount you expect to get from the reaction)

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Percentage Yield

• Example  A student conducts a single
  displacement reaction that produces 2.755 grams
  of copper.  If 3.150 grams of copper should have
  been produced what is the student's percentage
  yield?

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Percentage Yield

• 2.755gpercentage yield --------------- x
  100                           3.150g  
• percentage yield 87.46

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Sample problem

• Q - What is the yield of H2O if 138 g H2O is
  produced from 16 g H2 and excess O2?
• Step 1 write the balanced chemical equation
• 2H2 O2 ? 2H2O
• Step 2 determine actual and theoretical yield.
  Actual is given, theoretical is calculated

g H2O
16 g H2
Step 3 Calculate yield
yield
x 100
x 100
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Practice problem

• Q - What is the yield of NH3 if 40.5 g NH3 is
  produced from 20.0 mol H2 and excess N2?
• Step 1 write the balanced chemical equation
• N2 3H2 ? 2NH3
• Step 2 determine actual and theoretical yield.
  Actual is given, theoretical is calculated

g NH3
20.0 mol H2
Step 3 Calculate yield
yield
x 100
x 100