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STOICHIOMETRY

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Title: STOICHIOMETRY


1
Stoichiometric Calculations Stoicheion
(element) metron (measurement) Relating mass or
moles of reactants in a chemical reaction to mass
or moles of products.
2
USING EQUATIONS
  • Nearly everything we use is manufactured from
    chemicals.
  • Soaps, shampoos, conditioners, cds, cosmetics,
    medications, and clothes.
  • For a manufacturer to make a profit the cost of
    making any of these items cant be more than the
    money paid for them.
  • Chemical processes carried out in industry must
    be economical, this is where balanced equations
    help.

3
USING EQUATIONS
  • Equations are a chemists recipe.
  • Equations tell chemists what amounts of reactants
    to mix and what amounts of products to expect.
  • When you know the quantity of one substance in a
    reaction, you can calculate the quantity of any
    other substance consumed or created in the
    reaction.
  • Quantity meaning the amount of a substance in
    grams, liters, molecules, or moles.

4
USING EQUATIONS
  • When you bake cookies you probably use a recipe.
  • A cookie recipe tells you the amounts of
    ingredients to mix together to make a certain
    number of cookies.
  • If you need more cookies than the yield of the
    recipe, the amounts of ingredients can be doubled
    or tripled.

5
USING EQUATIONS
  • In a way, a cookie recipe provides the same kind
    of information that a balanced chemical equation
    provides
  • Ingredients are the reactants
  • Cookies are the products.

6
USING EQUATIONS
  • Imagine you are in charge of manu-facturing for a
    Bicycle Company.
  • The business plan for Rugged Rider requires the
    production of 128 custom-made bikes each day.
  • One of your responsibilities is to be sure that
    there are enough parts available at the start of
    each day.

7
USING EQUATIONS
  • Assume that the major components of the bike are
    the frame (F), the seat (S), the wheels (W), the
    handlebars (H), and the pedals (P).
  • The finished bike has a formula of FSW2HP2.
  • The balanced equation for the production of 1
    bike is.

F S2WH2P ? FSW2HP2
8
(No Transcript)
9
USING EQUATIONS
  • Now in a 5 day workweek, the bike company is
    scheduled to make 640 bikes. How many wheels
    should be in the plant on Monday morning to make
    these bikes?
  • What do we know?
  • Number of bikes 640 bikes
  • 1 FSW2HP22W
  • What is unknown?
  • of wheels ? wheels

10
  • The connection between wheels and bikes is 2
    wheels per bike. We can use this information as
    a conversion factor to do the calculation.

2 W
640 FSW2HP2
1 FSW2HP2
11
  • A CONVERSION FACTOR IS SIMPLY A THIS PER THAT
    EQUATION
  • FOR EVERY TWO WHEELS (THIS) YOU GET ONE BIKE
    (THAT)
  • ANOTHER WAY TO THINK OF A CONVERSION WOULD BE
  • FOR EVERY 12 INCHES YOU GET 1 FOOT
  • 1 FOOT
  • 12 INCHES

12
  • Another way to think of a conversion would be a
    chemical equation.
  • 2 H2 (g) O2 (g) ? 2 H20 (l)
  • This is the only recipe that makes water.
  • You need 2 moles of H2 and 1 mole of O2 to make 2
    moles of water. As usual the 1 is understood with
    coefficients and of atoms.
  • 2 mol H2 2 mol H20
  • As a conversion you would write
  • 2mol H2 or 1 mol H2O
  • 1 mol H2O 2 mol H2

13
  • You can also use the reactant oxygen in the
    formula to set up a conversion.
  • Try that at your seat.
  • If you wrote
  • 1 mol O2 2 mol H2O
  • 1 mol O2 or 2 mol H2O
  • 2mol H2O 1 mol O2
  • Good Job!
  • You just did a stoichiometric conversion!
  • Big word, but you did it!

14
  • Take the following sentence
  • Nitrogen gas and hydrogen gas react to form
    ammonia gas.
  • It would read in a chemical equation like this
  • N2(g) H2(g) ? NH3(g)
  • Is it balanced?
  • no

15
  • In order for it to be balanced, the first step
    would be change the coefficients. A 2 in front of
    the product will give the correct number of
    nitrogen atoms.
  • Placing the coefficient 3 in front of the number
    of hydrogen atoms in the reaction will balance
    the number of hydrogen atoms.

N2(g) 3H2(g) ? 2NH3(g)
  • What can we learn from this equation?

16
  • INTERPRETING CHEMICAL EQUATIONS
  • Open up your text book to page 239 and look at
    figure 9.4.
  • You can interpret this balanced equation in two
    ways
  • (1.) the number of particles (1 molecule of
    nitrogen will react with 3 molecules of hydrogen
    to produce 2 molecules of ammonia.)
  • (2.) or by the number of moles

17
  • The coefficients of the balanced chemical
    equation indicate the numbers of moles of
    reactants and products in a chemical reaction.
  • 1 mole of N2 reacts with 3 moles of H2 to produce
    2 moles of NH3.
  • N2 and H2 will always react to form ammonia in
    this 132 ratio of moles.
  • N2(g) 3H2(g) ? 2NH3(g)

18
  • Using the coefficients, to make connections
    between reactants and products, is the most
    important information that a chemical equation
    provides.

19
  • Using the formula for water provided you with 2
    possible conversions if you wanted to work with
    the molar ratio (how many moles of H2 and moles
    of O2 produce how many moles of H2O)
  • 2 H2 (g) O2 (g) ? 2 H20 (l)
  • 2 H2 2 H2O and 1 O2 2 H2O
  • 2H2 1 O2
  • 2H2O 2 H2O
  • What two conversions can you use for this
    formula?
  • N2(g) 3H2(g) ? 2NH3(g)

20
  • N2(g) 3H2(g) ? 2NH3(g)
  • 1 N2 2 NH3
  • 1 N2
  • 2 NH3
  • and
  • 3 H2 2 NH3
  • 3 H2
  • 2 NH3

21
  • The mass of 1 mol of N2 molecules is 28 g the
    mass of 3 mols of H2 molecules is 6 g for a
    total mass of reactants of 34 g.
  • The mass of 2 moles of NH3 molecules is 2 X 17g
    or 34 g.
  • Reactants mass must products mass

22
  • Review
  • Where did I get those values?
  • 1 mol N 14.0 g there are 2 moles28.0 g
  • There is 1 mole of N2 so 1 X 28.0 g 28.0 g
  • 1mol H 1.0g there are 2 moles 2 g
  • The coefficient tells you there are 3 X 2g
  • 3 X 2 6g in all
  • For NH3
  • 14.0g3g 17.0g there are 2 mol 34.0 g

N2(g) 3H2(g) ? 2NH3(g)
23
  • N2(g) 3H2(g) ? 2NH3(g)
  • 28.0g 6.0 g 34.0g 34.0 g
  • mass of reactants mass of products
  • The law of conservation of mass is obeyed.
  • Note the law of conservation of mass is always
    obeyed.
  • A different mass would mean that the equation was
    not properly balanced.

24
  • Using molar information, you can calculate
    amounts of reactants and products in a number of
    ways.
  • For example
  • You can make a similar connections with volume of
    a gas at STP .
  • What is that relationship?

25
  • 1 mole of a gas at STP is equal to 22.4 L
  • This is referred to as molar volume
  • That means, that a mole of any gas will occupy
    that amount of space as long as the temperature
    and pressure in the environment is at STP.

26
  • It also means that if you counted how many
    molecules are present when you have a mole of
    that gas it would add up to
  • 6.02 X 1023 molecules

27
  • The following slide is similar to the picture on
    page 239 of your text book.
  • You can open to that page while we talk about the
    different relationships.

28
22.4 L
29
  • Remember that 1 mole of any gas at STP occupies
    22.4 L of space.
  • It follows that 22.4 L of N2 reacts with 67.2 L
    of H2 to form 44.8 L of Ammonia gas. (notice
    volumes dont balance)

30
  • Unlike mass and atoms, - molecules, formula
    units, moles, and volumes of gases will not
    necessarily be conserved- although in some
    situations they may.
  • For example in the formation of hydrogen iodide
    all are conserved
  • H2(g) I2 (g) ? 2HI (g)

31
  • NOTE VERY IMPORTANT CONCEPT
  • Only mass and atoms are conserved in every
    chemical reaction.

32
  • Using the coefficients of balanced reaction
    equations and our knowledge of mole conversions
    we can perform powerful calculations. A.K.A.
    stoichiometry.
  • A balanced equation is essential for all
    calculations involving amounts of reactants and
    products.

33
  • If you know the number of moles of 1 substance,
    the balanced equation allows you to calculate the
    number of moles of all other substances in the
    equation.

34
9.2 MOLE MOLE CALCULATIONS
  • The following reaction shows the synthesis of
    aluminum oxide.

3O2(g) 4Al(s) ? 2Al2O3(s)
3O2(g) 4Al(s) ? 2Al2O3(s)
  • How many moles of aluminum are needed to form 3.7
    mol Al2O3?

Given 3.7 moles of Al2O3
Uknown ____ moles of Al
35
  • Solve for the unknown

3O2(g) 4Al(s) ? 2Al2O3(s)
4 mol Al
3.7 mol Al2O3
7.4 mol Al
2 mol Al2O3
36
  • Mathmatically, the 7.4 mol of Al was solved by
    the following
  • 3.7 mol X 4 mol 14.8 mol
  • That number is then divided by 2
  • 14.8 2 7.4 mol.
  • The given, (3.7) is always first multiplied by
    the numerator (4) and then divided by the divisor
    (2)
  • You always put the given first when solving
    stoiciometric problems)
  • The units you need to end up with should be in
    the numerator

37
  • 3O2(g) 4Al(s) ? 2Al2O3(s)
  • Suppose you were asked how many moles of Al2O3
    would be produced with 4 mol of O2?
  • What is the given?
  • The given is 4 mol O2
  • What do you want to find out?
  • mol of Al2O3

38
  • 3O2(g) 4Al(s) ? 2Al2O3(s)
  • 4 mol O2 X 2 mol Al2O3 2.6 mol Al2O3
  • 3 mol O2


39
  • IMPORTANT NOTE
  • MOLE TO MOLE CALCULATIONS DO NOT REQUIRE YOU TO
    REFER TO THE PERIODIC TABLE.
  • YOU SIMPLY SET UP THE PROBLEM BY LOOKING AT THE
    MOLAR RATIO OF THE BALANCED CHEMICAL EQUATION

40
  • 3O2(g) 4Al(s) ? 2Al2O3(s)
  • If asked how many moles of O2 are required to
    produce 5 moles of Al2O3?
  • Mol of given X mol O2
  • mol of Al2O3
  • No need to determine molar mass
  • How would you set this up?

41
  • 3O2(g) 4Al(s) ? 2Al2O3(s)
  • 5 mol Al2O3 X 3 O2
  • 2 Al2O3
  • 7.5 mol O2

42
MASS MASS CALCULATIONS
  • No lab balance measures moles directly, instead
    the mass of a substance is usually measured.
  • From the mass of 1 reactant or
  • product, the mass of any other reactant or
    product in a given chemical equation can be
    calculated
  • As in mole-mole calculations, the unknown can be
    either a reactant or a product.

43
  • In mass to mass conversions you use the molar
    mass ratio of what 1 mol of a compound is
    (periodic table)
  • For example the ratio conversion for H2O
  • from mol to H2O to g H2O would be
  • 1 mol H2O 18.0 g
  • 1 mol H2O
  • 18.0 g H2O
  • Where did I get 18.0 g ?
  • 2 X 1.0g 16.0 g 18.0 g

44
  • 3O2(g) 4Al(s) ? 2Al2O3(s)
  • NOTE The molar mass is calculated by
  • (1.) determine what 1 mole is equal to first
  • (2.) then multiply that number by the coefficient
  • Example using the above
  • 3O2
  • (1.) O2 16.0 g X 2 since there are 2 atoms of
    oxygen in the molecule
  • (2.) 3 X 32.0g 96.0 g

45
MASS MASS CALCULATIONS
Acetylene gas (C2H2) is produced by adding water
to calcium carbide (CaC2).
CaC2 2H2O ? C2H2 Ca(OH)2
How many grams of C2H2 are produced by adding
water to 5.00 g CaC2?
46
CaC2 2H2O ? C2H2 Ca(OH)2
  • What do we know?
  • Given mass 5.0 g CaC2
  • Mole ratio 1 mol CaC2 1 mol C2H2
  • Molar mass of CaC2 64.0 g CaC2
  • How do we know that?
  • By calculating molar mass
  • Molar mass of C2H2 26.0g C2H2
  • What do we want to know?
  • grams of C2H2 produced

47
5.0CaC2
26.0 g C2H2
2.03 g C2H2
64.0 g CaC2
48
  • It can also be set up like this

5.0 g CaC2
1 mol CaC2
1 mol C2H2
64.0 g CaC2
1mol CaC2
26.0 g C2H2
1mol C2H2
2.03 g C2H2
49
  • Answer to 19
  • a. 15.3 mol O2
  • b. 10.2 mol CO2, 13.6 mol H2O
  • Solve for a.
  • 3.40 mol C3H7OH X 9 mol O2
  • 2 mol C3H7OH
  • 15.3 mol O2
  • Solve for b
  • 3.4 mol C3H7OH X6 molCO2 10.2 mol CO2
  • 2 mol C3H7OH

50
  • For mol H2O
  • 3.40 C3H7OH X 8 mol H2O
  • 2 C3H7OH
  • 13.6 mol H2O
  • Now try 20 0n page 250

51
  • Open to page 250
  • At your seat,
  • work on problem 20 a
  • and 21. a, b, and c
  • For 21 convert every reactant and product to
    molar mass

52
PRACTICE TIME Problem 20 a on and 21 a, b, c on
page 250
53
16.0 g CH4 1 mol CH4
54
  • 21.
  • a. 176 g CO2, 36.0 g H2O
  • 2C2H2(g) 5O2(g) ? 4CO2(g) 2H20(g)
  • 52 g 160g ? 176g
    36 g
  • b. 160g 0f O2
  • c. 212 g total of reactants 212 g total of
    products

55
  • A balanced reaction equation indicates the
    relative numbers of moles of reactants and
    products.
  • We can expand our stoichiometric calculations to
    include any unit of measure that is related to
    the mole.
  • The given quantity can be expressed in numbers of
    particles, units of mass, or volumes of gases at
    STP.
  • The problems can include mass-volume,
    volume-volume, and particle-mass calculations.

56
  • Sn(s) 2HF(g) ? SnF2(s) H2(g)
  • What does this balanced equation tell me in
    words?
  • Tin reacts with hydrogen fluoride gas to yield
    Tin(II)fluoride and hydrogen gas.

57
  • Sn(s) 2HF(g) ? SnF2(s) H2(g)
  • What does it tell me in terms of moles
  • 1 mol Sn 2 mol HF ? 1 mol SnF2 1 mol H2
  • 1 mol Sn 1 mol SnF2
  • 1 mol Sn 1 mol H2
  • 2 mol HF 1 mol SnF2
  • 2 mol HF 1 mol H2
  • 1 mol Sn 2 mol HF (for reactants ratio)
  • 1 mol SnF2 1 mol H2 (for product ratio)
  • There are 3 mol of reactants and 2 mol of products

58
  • Sn(s) 2HF(g) ? SnF2(s) H2(g)
  • What does it tell me in terms of volume at STP?
  • 22.4 L Sn 1 mol SnF2
  • 22.4 L Sn 22.4 L H2
  • 44.8 L HF 1 mol SnF2
  • 44.8 L HF 22.4 L H2
  • 22.4 L Sn 44.8 L HF for reactant ratio
  • 1 mol SnF2 22.4 L H2 for product ratio
  • 67.2 L(22.4 44.8) on reactant side 22.4 L on
    product side

59
  • Sn(s) 2HF(g) ? SnF2(s) H2(g)
  • In terms of mass
  • 118.7 g Sn 40.0 g HF ? 156.7 SnF2 2.0g H2
  • 118.7 g Sn 156.7 g SnF2
  • 118.7 g Sn 2.0 g H2
  • 40.0 g HF 156.7 g SnF2
  • 40.0 g HF 2.0 g H2
  • 158.7 g reactants(118.7 40.0) 158.7 g
    products(156.7 2.0)

60
  • Sn(s) 2HF(g) ? SnF2(s) H2(g)
  • In terms of of molecules
  • . 6.02 X 1023 molecules Sn 6.02 X 1023
    molecules SnF2
  • 6.02 X 1023 molecules Sn 6.02 X 1023 molecules
    H2
  • 1.2 X 1024 molecules HF 6.02 X 1023 molecules
    SnF2
  • 1.2 X 1024 molecules HF 6.02 X 1023 molecules H2

61
  • Now try 20 b and c on page 250

62
PRACTICE TIME Problem 20 b and c on page 250
63
  • b. 1 mol CH4
  • 22.4 L CH4
  • c. 1 mol CH4
  • 6.02 X 1023 molecules CH4

64
  • Sn(s) 2HF(g) ? SnF2(s) H2(g)
  • How many liters of H2 will be produced with 12.0
    L HF?
  • Given 12.0 L HF
  • Unknown ? L of H2
  • Conversion 22.4 L Sn / 22.4 L H2
  • 12.0 L Sn X 22.4 L H2 6.0 L H2
  • 44.8 L HF

65
  • Now try 22 a on page 250
  • at your seat

66
22.a Sn(s) 2HF(g) ? SnF2(s) H2(g) 9.40
L H2 X 44.8 L HF 18.8 L HF
22.4 L H2
67
  • 22 b would require you to combine two different
    equalities
  • 44.8 L HF 6.02 X 1023 molecules H2
  • Given 20.L HF
  • ? How many molecules of H2
  • 20.0 L HF X 6.02 X 1023 molecules of H2
  • 44.8 L HF
  • 2.69 X 1023 molecules H2

68
  • c.
  • 7.42 X 1024 molecules of HF
  • X 156.7 X 100 g SnF2
  • 1.2 X 1024 molecules
    of HF
  • 969 g SnF2

69
  • NOTE
  • Mole to mole and volume to volume problems do not
    require you to refer to the periodic table.
  • For mol to mol look at coefficients
  • For volume to volume(STP) know that 1 mol is
    equal to 22.4 L
  • The coefficient tells you what to multiply 22.4 L
    by
  • No coefficient always means 1 so you multiply
    22.4 L by 1

70
  • OPEN UP TEXT TO PAGE 262 PROBLEM 37
  • 2.7 mol C X 1mol CS2
  • 5 mol C
  • .54 mol CS2
  • 5.44 mol SO2 X 5 mol C
  • 2 mol SO2
  • 13.6 mol C

71
  • c. .246 mol CS2 X 4 mol CO
  • 1 mol CS2
  • .984 mol
  • End of chapter 9 for period 2

72
  • In any of these problems, the given quantity is
    first converted to moles.
  • Then the mole ratio from the balanced equation is
    used to convert from the moles of given to the
    number of moles of the unknown
  • Then the moles of the unknown are converted to
    the units that the problem requests.
  • The next slide summarizes these steps for all
    typical stoichiometric problems

73
  • 20.
  • a. 16.0 g CH4
  • 1 mol CH4
  • b. 1 mol CH4
  • 22.4 L CH4
  • c. 1 mol CH4
  • 6.02 X 1023 molecules CH4
  • Try 21 on page 250 hint convert reactants and
    products to molar mass(g)

74
Limiting Reactants Suppose you do not
have enough of one reactant to carry the
reaction to completion? To determine the
moles or masses of products, we
need to know which reactant will run
out first and which is in excess. We
cannot use coefficients or masses to make the
determination. As usual, everything is
measured in moles.
75
Limiting Reactants 1. Write the balanced
equation. 2. Convert masses to moles. 3. The
reactant producing the lowest moles of product
will be consumed (limiting reactant).
76
Take the following balanced chemical
equation Zn(s) 2HCl(aq) ? ZnCl2(aq)
H2(g) If 0.30 mol Zn is added to 0.52 mol HCl,
how many moles of H2 will be produced?
77
Zn(s) 2HCl(aq) ? ZnCl2(aq) H2(g)
0.30 mol Zn x 1 mol H2 0.30 mol H2
1 mol Zn 0.52 mol HCl x 1
mol H2 0.26 mol H2 2 mol
HCl The mole ratio conversions come directly
from coefficients in the balanced equation.
0.26 mol is the maximum amount of H2
produced when 0.30 mol of Zn and 0.52 mol
of HCl are mixed.
78
The HCL is the limiting reagent and the Zn will
be in excess. The limiting reagent determines the
amount of product possible for a reaction and is
used up entirely during a reaction. Excess
reagent will be what was not used up entirely
during the reaction
79
  • What is the limiting reagent when 4.3 mol of SO2
    react with 2.9 mol O2 according to the equation
  • 2 SO2(g) O2(g) ? 2SO3(g)
  • Solution
  • 4.3 mol SO2 X 2 mol SO3 4.3 mol SO3
  • 2 mol SO2
  • 2.9 mol O2 X 2 mol SO3 5.8 mol SO3
  • 1 mol O2
  • SO2 is the limiting reagent O2 will be in excess

80
  • Step two
  • Solve for percent yield by using the following
    formula
  • Percent yield actual yield X 100
  • theoretical yield
  • 4.65 g of Cu X 100
  • 6.61 g of Cu
  • 70.5

81
  • OPEN UP YOUR TEXT AND WORK ON PROBLEM 32 ON PAGE
    259
  • 32 SOLUTION
  • step one compare weight ratio of reactants in
    the balanced equation
  • the weight ratio is 480 g to 480 g 11
  • The available O2 is the limiting reagent and
  • FeS2 will be in excess, only 16 g will react.

82
How much of the reactant O2 will be in
excess? Solution First solve for how much O2
will react with 4.3 mol SO2 4.3 mol SO2 X 1 mol
O2 2.2 mol O2 2 mol
SO2 5.8mol - 2.2mol 3.6 mol O2 in excess
83
Period 5 only, period 7 and 8 will get to this
after the test
84
  • 1. 2Na(s) Cl2 ?2NaCl(s),
  • If 7.4 mol Na and 3.2 mol HCl are allowed to
    react,
  • a. what is the limiting reagent?
  • b. How many moles of NaCl are produced?
  • c. How many moles of excess reagent remain?
  • 2. C0(g) 2H2(g) ? CH3OH(g)
  • CH3OH(g) is called mehtanol
  • Interpret the above equation using the following
    1. name, including physical properties. 2. volume
    at STP 3. moles ratio 4. molar mass 5. any other
    combination

85
Percent Yield In most reactions not all reactants
go forward to form products. Some are lost to
byproducts and some do not react. The actual
yield is usually less than the anticipated,
theoretical yield. Percent yield actual
yield x 100 theoretical yield
86
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
87
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
88
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
89
  • Calcium carbonate is decomposed by heating, as
    shown in the following equation
  • CaCO3(s)? CaO(s) CO2
  • What is the theoretical yield of 24.8 g CaCo3 is
    heated?
  • B. What is the percent yield if 13.0 g CaO is
    produced?

90
knowns Mass of CaCO3 24.8g CaCO3 1mol CaCO3
1mol CaO 1 mol CaCO3 100.1 g CaCO3 (molar
mass) 1 mol CaO 56.1 g (molar mass) Unknowns
theoretical yield CaO with using 24.8 g
CaCO3 mass A ? mol A ? mol B ? mass B 24.8 g
CaCO3 X 1 mol CaCO3 X 1mol CaO X
56.1 g CaO 100.1 g
CaCO3 1 mol CaCO3 1 mol CaO
91
13 .9 g CaO unknowns for b is percent
yield KnownActual yield 13.0 g
CaO KnownTheoretical yield 13.9 g CaO Percent
yield actual X 100
theoretical 13.1 g CaO X 100
94.2 13.9 g CaO
92
  • OPEN UP YOUR TEXT AND WORK ON PROBLEM 30 ON PAGE
    259
  • SOLUTION
  • Step 1 determine theoretical yield
  • from the balanced chemical equation
  • Solution 1.87g of Al X 190.62 g of CuSO4
  • 53.96 g of Al
  • 6.61 g CuSO4

93
  • 20.0 g FeS2 X 640.5 g of SO3
  • 480g of FeS2
  • 26.7 g SO3
  • 16 g O2 X 640.5 g of SO3
  • 480g of O2
  • 21.4 g SO3

94
  • 6.6 g of Cu produced (theoretical yield)
  • Where did I get my gram amounts?
  • What do I know from the balanced equation?
  • That 2 moles of Aluminum will equal 3 Moles of
    Copper.
  • 2 moles of Aluminum53.96 g (2 X 23.96)
  • 3 moles of Copper 190.62 g (3 X 63.54)
  • Use that ratio to determine the expected yield
    for only 1.87 g of aluminum.

95
  • 29.
  • A limiting reagent determines the amount of
    product that can be produced from a reaction.
  • Excess reagent is not used up by a reaction.
  • 30. Step 1 determine theoretical yield
  • Solution 1.87g of Al X 190.62 g of Cu
  • 53.96 g of Al

96
  • 6.6 g of Cu produced (theoretical yield)
  • Where did I get my gram amounts?
  • What do I know from the balanced equation?
  • That 2 moles of Aluminum will equal 3 Moles of
    Copper.
  • 2 moles of Aluminum53.96 g (2 X 23.96)
  • 3 moles of Copper 190.62 g (3 X 63.54)
  • Use that ratio to determine the expected yield
    for only 1.87 g of aluminum.

97
Calcium carbonate is decomposed by heating,
as shown in the following equation. CaCO3(s) ?
CaO (s) CO2 (g) a. What is the theoretical
yield of CaO if 24.8g CaCO3 is heated? b. What
is the percent yield if 13.1 g CaO is
produced? List the known and unknowns for a
(theoretical yield)
98
  • Step two
  • Solve for percent yield by using the following
    formula
  • Percent yield actual yield X 100
  • theoretical yield
  • 4.65 g of Cu X 100
  • 6.6 g of Cu
  • 70.5

99
  • 32.
  • step one compare weight ratio of reactants in
    the balanced equation
  • the weight ratio is 480 g to 480 g 11
  • The available O2 is the limiting reagent and
  • FeS2 will be in excess, only 16 g will react.
  • 640.5 g of SO3 are produced with 480 g of each
    reactant

100
16.0 g O2 X 640.5 g of SO3
480 g of O2 21.4 g SO3 20.0 g FeS2 X
640.5 g of SO3 480 g of
FeS2 26.7 g SO3
101
Grams of A
Use molar mass as a conversion factor
Moles of A
Use coefficients from balanced equation as a
conversion factor
Moles of B
Use molar mass as a conversion factor
Grams of B
102
  • SOLUTIONS TO WORKSHEET

103
  • If you start with 10 grams of lithium hydroxide,
    how many grams of lithium bromide will be
    produced?
  • LiOH HBr ? LiBr H2O
  • What do you know?
  • 1mol LiOH 23.9g ( gfm)
  • 1 mol LiOH 1 mol LiBr
  • 1 mol LiBr 86.8 g (gfm)

104
  • 1. Mass A ? Moles A ? Moles B ? Mass B
  • 10g LiOH 1mol LiOH 1mol LiBr
  • 23.9 g LiOH 1mol LiOH

  • 86.6g LiBr

  • 1mol LiBr
  • 36.23g LiBr

105
  • 2. C2H4 3 O2 ? 2 CO2 2 H2O
  • 1 mol C2H428.0g 2 mol CO2 2(44.0g)
  • 28.0 g of C2H4 88.0g of CO2
  • 45g C2H4 X 88.0 g CO2 (molar mass) 141 g CO2
  • 28.0 g C2H4 (molar mass)
  • 4. 2 HCl Na2SO4 ? 2 NaCl
    H2SO4
  • 2(36.5 g) HCl is 73.0 g 98.0 g H2SO4
  • 20 g HCl X 98.0 g H2SO4 26.84 g H2SO4
  • 73.0 g HCl
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