Limiting Reagents and Percent Yield

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Limiting Reagents and Percent Yield
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What Is a Limiting Reagent?

• Many cooks follow a recipe when making a new
  dish.
• When a cook prepares to cook he/she needs to know
  that sufficient amounts of all the ingredients
  are available.
• Lets look at a recipe for the formation of a
  double cheeseburger

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1 hamburger bun
1 tomato slice
1 lettuce leaf
2 slices of cheese
2 burger patties
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• If you want to make 5 double cheese burgers

• How many hamburger buns do you need?

5

• How many hamburger patties do you need?

10

• How many slices of cheese do you need?

10

• How many slices of tomato do you need?

5
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• How many double cheeseburgers can you make if you
  start with

• 1 bun, 2 patties, 2 slices of cheese, 1 tomato
  slice

1

• 2 buns, 4 patties, 4 slices of cheese, 2 tomato
  slices

2

• 1 mole of buns, 2 moles of patties, 2 moles of
  cheese, 1 mole of tomato slices

1 mole

• 10 buns, 20 patties, 2 slices of cheese, 10
  tomato slices

1
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• We cant make anymore than 1 double cheeseburger
  with our ingredients.
• The slices of cheese limits the number of
  cheeseburgers we can make.
• If one of our ingredients gets used up during our
  preparation it is called the limiting reactant
  (LR)
• The LR limits the amount of product we can form
  in this case double cheeseburgers.
• It is equally impossible for a chemist to make a
  certain amount of a desired compound if there
  isnt enough of one of the reactants.

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• As weve been learning, a balanced chemical rxn
  is a chemists recipe.
• Which allows the chemist to predict the amount of
  product formed from the amounts of ingredients
  available

• Lets look at the reaction equation for the
  formation of ammonia

N2(g) 3H2(g) ? 2NH3(g)

• When 1 mole of N2 reacts with 3 moles of H2, 2
  moles of NH3 are produced.
• How much NH3 could be made if 2 moles of N2 were
  reacted with 3 moles of H2?

2 mols of ammonia
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N2(g) 3H2(g) ? 2NH3(g)

• The amount of H2 limits the amount of NH3 that
  can be made.
• From the amount of N2 available we can make 4
  moles of NH3
• From the amount of H2 available we can only make
  2 moles of NH3.
• H2 is our limiting reactant here.
• It runs out before the N2 is used up.
• Therefore, at the end of the reaction there
  should be N2 left over.
• When there is reactant left over it is said to be
  in excess.

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• How much N2 will be left over after the reaction?
• In our rxn it takes 1 mol of N2 to react all of 3
  mols of H2, so there must be 1 mol of N2 that
  remains unreacted.
• We can use our new stoich calculation skills to
  determine 3 possible types of LR type
  calculations.
• Determine which of the reactants will run out
  first (limiting reactant)
• Determine amount of product
• Determine how much excess reactant is wasted

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Limiting Reactant Problems
Given the following reaction 2Cu S ? Cu2S

• What is the limiting reactant when 82.0 g of Cu
  reacts with 25.0 g S?
• What is the maximum amount of Cu2S that can be
  formed?
• How much of the other reactant is wasted?

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• Our 1st goal is to calculate how much S would
  react if all of the Cu was reacted.
• From that we can determine the limiting reactant
  (LR).
• Then we can use the Limiting Reactant to
  calculate the amount of product formed and the
  amount of excess reactant left over.

82g Cu?
mol Cu?
mol S?
g S
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2Cu S ? Cu2S
1mol S
1molCu
32.1g S
82.0gCu
2molCu
1mol S
63.5gCu
20.7 g S

• So if all of our 82.0g of Copper were reacted
  completely it would require only 20.7 grams of
  Sulfur.
• Since we initially had 25g of S, we are going to
  run out of the Cu, the limiting reactant) end
  up with 4.3 grams of S

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• Copper being our Limiting Reactant is then used
  to determine how much product is produced.
• The amount of Copper we initially start with
  limits the amount of product we can make.

1molCu2S
1molCu
159gCu2S
82.0gCu
2molCu2S
1molCu2S
63.5gCu
103 g Cu2S
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• So the reaction between 82.0g of Cu and 25.0g of
  S can only produce 103g of Cu2S.
• The Cu runs out before the S and we will end up
  wasting 4.7 g of the S.

Ex 2 Hydrogen gas can be produced in the lab by
the rxn of Magnesium metal with HCl according to
the following rxn equation Mg 2HCl ? MgCl2
H2

• What is the LR when 6.0 g HCl reacts with 5.0 g
  Mg? What is the maximum amount of H2 that can be
  formed? And how much of the other reactant is
  wasted?

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5.0g Mg ?
mol Mg ?
2mol HCl ?
g HCl
36.5gHCl
1molMg
2molHCl
5.0g Mg
1molHCl
1molMg
24.3gMg
15.0g HCl

• So if 5.0g of Mg were used up it would take 15.0g
  HCl, but we only had 6.0g of HCl to begin with.
• Therefore, the 6.0g of HCl will run out before
  the 5.0g of Mg, so HCl is our Limiting Reactant.

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6.0g HCl?
2mol HCl ?
1mol H2 ?
g H2
2.0gH2
1molHCl
1molH2
6.0g HCl
1molH2
2molHCl
36.5gHCl
0.164 g H2 produced
6.0g HCl?
2mol HCl ?
1mol Mg ?
g Mg
24.3gMg
1molHCl
1molMg
6.0g HCl
1molMg
2molHCl
36.5gHCl
1.997 g Mg
- 5.0 g Mg
3.01g Mg extra
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Calculating Percent Yield

• In theory, when a teacher gives an exam to the
  class, every student should get a grade of 100.
• Your exam grade, expressed as a perc-ent, is a
  quantity that shows how well you did on the exam
  compared with how well you could have done if you
  had answered all questions correctly

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• This calc is similar to the percent yield calc
  that you do in the lab when the product from a
  chemical rxn is less than you expected based on
  the balanced eqn.
• You might have assumed that if we use stoich to
  calculate that our rxn will produce 5.2 g of
  product, that we will actually recover 5.2 g of
  product in the lab.
• This assumption is as faulty as assuming that all
  students will score 100 on an exam.

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• When an equation is used to calculate the amount
  of product that is possible during a rxn, a value
  representing the theoretical yield is obtained.
• The theoretical yield is the maximum amount of
  product that could be formed from given amounts
  of reactants.
• In contrast, the amount of product that forms
  when the rxn is carried out in the lab is called
  the actual yield.
• The actual yield is often less than the
  theoretical yield.

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• The percent yield is the ratio of the actual
  yield to the theoretical yield as a percent
• It measures the measures the efficiency of the
  reaction

actual yield
Percent yield
x 100
theoretical yield

• What causes a percent yield to be less than 100?

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• Rxns dont always go to completion when this
  occurs, less than the expected amnt of product is
  formed.
• Impure reactants and competing side rxns may
  cause unwanted products to form.
• Actual yield can also be lower than the
  theoretical yield due to a loss of product during
  filtration or transferring between containers.
• If a wet precipitate is recovered it might weigh
  heavy due to incomplete drying, etc.

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Calcium carbonate is synthesized by heating,as
shown in the following equation CaO CO2 ?
CaCO3

• What is the theoretical yield of CaCO3 if 24.8 g
  of CaO is heated with 43.0 g of CO2?
• What is the percent yield if 33.1 g of CaCO3 is
  produced?

Determine which reactant is the limiting and
then decide what the theoretical yield is.
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24.8gCaO?
molCaO?
mol CO2?
gCO2
24.8 g CaO
1molCaO
1mol CO2
44 g CO2
56g CaO
1mol CaO
1molCO2
LR
19.5gCO2
1mol CaO
100g CaCO3
24.8 g CaO
1molCaCO3
56g CaO
1mol CaO
1molCaCO3
44.3 g CaCO3
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• CaO is our LR, so the reaction should
  theoretically produce 44.3 g of CaCO3 (How
  efficient were we?)
• Our percent yield is

33.1 g CaCO3
Percent yield
x 100
44.3 g CaCO3
Percent yield 74.7