Chapter 12 Stoichiometry

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Chapter 12 Stoichiometry
Pioneer High School Mrs. Julia V. Bermudez
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Chapter 12 Stoichiometry

• OBJECTIVES
• To predict how much is used or formed in a
  chemical reaction

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Chapter 12 Stoichiometry

• Stoichiometry is Greek for Measuring Elements
• It starts with a balanced equation
• 2H2 O2 2H2O, means
• 2 moles of hydrogen reacts with
• 1 mole of oxygen
• Forming 2 moles of water.

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Chapter 12 Stoichiometry

• The coefficients tell us how many moles of each
  substance. Not Grams!
• For example
• 2H2 O2 2H2O
• 2 g of H2 1 g of O2 2 g of H2O
• 3 g of reactants cant make only 2 g of products

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Chapter 12 Stoichiometry

• The Law of Conservation of Mass applies
• Convert the moles to grams and the equation does
  work.
• 2H2 O2 2H2O

2.02 g H2
2 moles H2

4.04 g H2
1 mole H2
32.00 g O2
1 mole O2

32.00 g O2
1 mole O2
36.04 g H2O2
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Chapter 12 Stoichiometry

• 2H2 O2 2H2O

18.02 g H2O
2 moles H2O
36.04 g H2O

1 mole H2O
2H2 O2 2H2O
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Chapter 12 Stoichiometry

• Try 1a-e on page 356
• Only check to see if the mass of reactants equals
  the mass of the products.
• Row 1 (by the clock) do 1a
• Row 2 do 1b
• Row 3 do 1c
• Row 4 do 1d
• Row 5 do 1e

• Row 6 do 1a
• Row 7 do 1b
• Side lab tables do 1c
• Back Lab tables do 1e

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Chapter 12 Stoichiometry

• Try 1a-e on page 356
• Only check to see if the mass of reactants equals
  the mass of the products.
• 28g N2 6.06g H2 ? 34.1g NH3
• 36.5g HCl 56.1g KOH ? 74.6g KCl 18.0g H2O
• 262g Zn 630g HNO3 ? 758gZn (NO3)2 44.0g N2O
  90.1g H2O
• 48.6g Mg 32.0g O2 ? 80.6g MgO
• 46.0g Na 36.0g H2O ? 80.0g NaOH 2.02g H2

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Chapter 12 Stoichiometry

• OBJECTIVES
• Construct mole ratios from balanced chemical
  equations
• Apply these ratios in stoichiometric calculations.

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Chapter 12 Stoichiometry

• OBJECTIVES
• Calculate stoichiometric quantities from balanced
  chemical equations using units of moles mass.

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Mole to Mole conversions

• 2 Al2O3 4Al 3O2
• What are the mole ratios between Al2O3 O2?

2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
Both are possible conversion factors
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Mole to Mole conversions

• How many moles of O2 are produced when 3.34 moles
  of Al2O3 decompose?
• 2 Al2O3 4Al 3O2

mol O2
3
3.34 mol Al2O3

5.01 mol O2
2
mol Al2O3
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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
  of O2 are needed?

(9.6 mol O2)
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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• How many moles of C2H2 are needed to produce
  8.95 mole of H2O?

(8.95 mol C2H2)
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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• If 2.47 moles of C2H2 are burned, how many moles
  of CO2 are formed?

(4.94 mol CO2)
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How do you get good at this?
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Classwork

• Page 357 2-3
• Page 359 9-10
• Page 379 61-63
• 9 2SO2 O2 2H2O ? 2H2SO4

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Mass-Mass Calculations

• In the lab, you can not measure a mole directly.
• You always start with grams and you want answers
  in grams
• Since you start with mass and end the problem
  with mass
• this is called a Mass-Mass problem

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Mass-Mass Calculations

• There are 3 ways to do these problems
• Convert the Equation Method (1 step)
• Mole Space Method (2 steps)
• Dimensional Analysis Method (2 steps)
• Ill teach the easy one so you wont get
  confused.

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Mass-Mass Calculations

• If 10.1 g Fe are added to a solution of Copper
  (II) Sulfate, how much solid copper would form?
• 2Fe 3CuSO4 Fe2(SO4)3 3Cu

Mole Ratio
2 3 1
3
Mass Ratio for Fe and Cu
55.8 x 2 111.6
63.5 x 3 190.5
g Cu
190.5
g Fe
111.6
Answer 17.2 g Cu
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2Na Cl2 ? 2NaCl

• How many grams of sodium are needed to react with
  .071g of chlorine gas?

Mole Ratio
2 1 2
Mass Ratio
35.5 x 2 x 1 71
23 x 2 46
58.5 x 2 117
Answer .046g Na
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Practice Time

• 2Na Cl2 ? 2NaCl
• If 2.47 g of NaCl are dissolved, how many grams
  of Cl2 are formed?

Answer 1.50 g Cl2
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Classwork

• Page 362 13,14
• Page 380 69-70, 74a, 75

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Section 12.3Limiting Reagent

• OBJECTIVES
• Identify and use the limiting reagent to
  calculate the max amount of product(s) produced.

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Limiting Reagent

• If you are given a loaf of bread, a bottle of
  mustard, and four slices of salami, how many
  salami sandwiches can you make?

ONLY 4!
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Limiting Reagent

• The limiting reagent is the reactant you run out
  of first.
• The excess reagent is the one you have left over.
• The limiting reagent determines how much product
  you can make

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How do you find out?

• Do as many stoichiometry problems as there are
  givens.
• The one that makes the least product is the
  limiting reagent.

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• If 10.6 g of copper reacts with 3.83 g S. How
  many grams of product will be formed?
• 2Cu S Cu2S

Mole Ratio
2 1 1
63.5 x 2 127
32.1 x 1 32.1
159.1 x 1 159.1
Mass Ratio
10.6 g Cu
159.1 g Cu2S
13.3 g Cu2S
127 g Cu
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• If 10.6 g of copper reacts with 3.83 g S. How
  many grams of product will be formed?
• 2Cu S Cu2S

Mole Ratio
2 1 1
63.5 x 2 127
32.1 x 1 32.1
159.1 x 1 159.1
Mass Ratio
3.83 g S
159.1 g Cu2S
19.0 g Cu2S
32.1 g S
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If 10.6 g of copper reacts with 3.83 g S. How
many grams of product will be formed?2Cu S
Cu2S

• 10.6 g of Cu produces 13.3 g of Cu2 S.
• 3.83 g of S produces 19.0 g Cu2 S.
• ? Cu is the Limiting Reagent and only 13.3 g of
  Cu2S are produced.

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Another example

• AgCl NaNO3 ? AgNO3 NaCl
• If 10.1 g of AgCl and 2.87 g of NaNO3 are
  reacted, how many grams of NaCl will be produced?

1.98 g NaCl
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Still another example

• 4KBr CS2 ? 2K2S CBr4
• If 14.3 g of KBr are reacted with 21.7 g of CS2
  how much CBr4 will be produced? Which reactant is
  the Limiting Reagent?

9.96g CBr4 KBr is the L.R
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Homework

• Pg 368 20 a-c, 21
• Pg 380 77, 79 ,82 ,83

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Section 12.3Limiting Reagent Percent Yield

• OBJECTIVES
• Calculate theoretical yield, actual yield, or
  percent yield, given appropriate information.

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Chemical Yield

• The amount of product made in a chemical
  reaction.
• There are three types
• 1.Theoretical yield- the amount of Product that
  should be made (from calculations)
• 2. Actual yield- the amount of Product formed in
  the laboratory (always given)

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Chemical Yield

• 3. Percent yield a percentage/ratio between the
  actual yield and the theoretical yield.
• Yield
• yield tells us how efficient a reaction is.
• yield can not be bigger than 100 .

x 100
Theoretical Yield
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Example 1

• According to your calculations the theoretical
  yield for the production of NaCl is 13.6 grams.
  In the laboratory your actual yield is 11.8 grams
  of NaCl. What is the percent yield?

x 100
13.6 g NaCl
Answer 86.7
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Example

• In the laboratory 6.78 g of copper is actually
  produced when 3.92 g of Al are reacted with
  excess copper (II) sulfate.
• 2Al 3 CuSO4 Al2(SO4)3 3Cu
• What is the actual yield?
• What is the theoretical yield?
• What is the percent yield?

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Classwork

• Pg 372 27-29
• Homework
• Pg 381 84-85, 87 a,c,d, 88