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					It’s time to learn about . . .
What's The Difference Between Roast
Beef And Pea Soup?




 Anyone Can Roast Beef.
Stoichiometry
          Stoichiometry : Mole Ratios to
                Determining Grams of Product
          At the conclusion of our time together, you
          should be able to:


1. Review the conversion of particles or grams to
  moles
2. Determine mole ratios from a balanced chemical
  equation
3. Determine the amount of product produced when
  given the amount of reactants
    Review the Molar Mass of Compounds

   The molar mass (MM) of a compound is determined
    by adding up all the atomic masses for the molecule
    (or compound)
                                                 20
                                                 Ca
    ◦   Ex. Molar mass of CaCl2                 40.08
    ◦   Avg. Atomic mass of Calcium = 40.08g
    ◦   Avg. Atomic mass of Chlorine = 35.45g
    ◦   Molar Mass of calcium chloride =
        40.08 g/mol Ca + (2 X 35.45) g/mol Cl    17
          110.98 g/mol CaCl2                    Cl
                                                35.45
Practice


   Calculate the Molar Mass of
    calcium phosphate
    ◦ Formula = Ca3(PO4)2
    ◦ Masses elements:

    ◦ Molar Mass =
                     310.18 g
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Middle Digit.
Chapter 14 – The Best Time to Let a
Foursome Play through Your Twosome.
  Calculations


    molar mass    Avogadro’s number
Grams        Moles                Particles


   Everything must go through Moles!!!
Atoms or             Flowchart
Molecules

                Divide by 6.02 X 1023
  Multiply by
  6.02 X 1023                           Multiply by
                     Moles              atomic/molar
                                        mass from
                                        periodic
                   Divide by            table
                   atomic/molar
                   mass from
                   periodic table       Mass
                                        (grams)
Chocolate Chip Cookies!!
              1 cup butter                 2 eggs
              1/2 cup white sugar          1 teaspoon salt
              1 cup packed brown sugar
              1 teaspoon vanilla extract
              2 1/2 cups all-purpose flour
              1 teaspoon baking soda
              2 cups semisweet chocolate chips
              Makes 3 dozen
How many eggs are needed to make 3 dozen cookies?
How much butter is needed for the amount of
chocolate chips used?
How much brown sugar would I need if I had 1½ cups
white sugar?
    Cookies and Chemistry…Huh!?!?
 Just like chocolate chip cookies
  have recipes, chemists have
  recipes as well
 Instead of calling them recipes,
  we call them chemical equations
 Furthermore, instead of using
  cups and teaspoons, we use
  moles
 Lastly, instead of eggs, butter,
  sugar, etc. we use chemical
  compounds as ingredients
     Chemistry Recipes

 Looking at a reaction tells us how much of
  something you need to react with something else to
  get a product (like the cookie recipe)
 Be sure you have a balanced reaction before you
  start! Balance #1 on your worksheet!!
     Example: 2 Ag+ Cl2  2 AgCl
     This reaction tells us that by mixing 2 moles of
      silver with 1 mole of chlorine we will get 2 moles
      of silver chloride
     What if we wanted 4 moles of AgCl?
     10 moles?
     50 moles?
Let’s sit down and figure this out!!
    Practice

   Write the balanced reaction for hydrogen gas reacting
    with oxygen gas.
              2 H2 + O2  2 H2O
    ◦ How many moles of reactants are needed?
    ◦ What if we wanted 4 moles of water?
    ◦ What if we had 3 moles of oxygen, how much
      hydrogen would we need to react, and how much
      water would we get?
    ◦ What if we had 50 moles of hydrogen, how much
      oxygen would we need, and how much water would
      be produced?
Mole Ratios

 These mole ratios can be used to calculate
  the moles of one chemical from the given
  amount of a different chemical
 Example: How many moles of chlorine are
  needed to react with 5 moles of silver
  (without any silver left over)?
 2 Ag +      Cl2  2 AgCl
 5 Ag + 2.5 Cl2  5 AgCl
Question Answer
2 Ag+ Cl2  2 AgCl


# moles of Cl2 =


5 mol Ag   x 1 mol Cl2
            2 mol Ag


                   = 2.5 mol Cl2
Mole-Mole Conversions
   How many moles of silver chloride will be
    produced if you react 2.6 moles of chlorine gas
    with an excess (more than you need) of silver
    metal?
           2 Ag + Cl2  2 AgCl
# moles of AgCl =

2.6 mol Cl2 x 2 mol AgCl
                   1 mol Cl2


                   = 5.2 mol AgCl
        Mass-Mole
 We can also start with mass and convert to moles of
  product or another reactant
 We use molar mass and the mole ratio to get to moles
  of the compound of interest
  ◦ Calculate the number of moles of the combustion of
    ethane (C2H6) needed to produce 10.0 g of water

    ◦   2 C2H6 + 7 O2  4 CO2 + 6 H20
       Mass-Mole
      2 C2H6 + 7 O2  4 CO2 + 6 H20

10.0 g H2O   x     1 mol H2O   x    2 mol C2H6
                 18.02 g H2O        6 mol H2O



                          = 0.185 mol C2H6
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Stoichiometry Continued:
          Stoichiometry : Mole Ratios to
                Determining Grams of Product
          At the conclusion of our time together, you
          should be able to:


1. Review the conversion of particles or grams to
  moles
2. Determine mole ratios from a balanced chemical
  equation
3. Determine the amount of product produced when
  given the amount of reactants
     Mass-Mass Conversions


 Most often we are given a starting mass and want
  to find out the mass of a product we will get (called
  theoretical yield) or how much of another reactant
  we need to completely react with it (no leftover
  ingredients!)
 Now we must go from grams to moles, to mole
  ratios, and back to grams of the compound we are
  interested in.
  Mass-Mass Conversion

    Ex. Calculate how many grams of ammonia are
     produced when you react 2.00 g of nitrogen
     with excess hydrogen.
    N2 + 3 H2  2 NH3



2.00 g N2    1 mol N2    2 mol NH3   17.06 g NH3
            28.02 g N2   1 mol N2    1 mol NH3

                 = 2.4 g NH3
    Practice Mass-Mole-Mass


   How many grams of calcium nitride are produced
    when 2.00 g of calcium reacts with an excess of
    nitrogen?

   Ca + N2  Ca3N2

   3 Ca + N2  Ca3N2
         Mass-Mole-Mass
        3 Ca + N2  Ca3N2

2.00 g Ca       x     1 mol Ca    x   1 mol Ca3N2
                    40.08 g Ca        3 mol Ca



      x 148.26 g Ca3N2
                             = 2.47 g Ca3N2
         1 mol Ca3N2
Some Things Just Take a Lot of Work!!!
          Stoichiometry : Mole Ratios to
               Determining Grams of Product
          Let’s see if you can:



1. Review the conversion of particles or grams to
  moles
2. Determine mole ratios from a balanced chemical
  equation
3. Determine the amount of product produced when
  given the amount of reactants
      Mass-Mole-Mass

    NaN3  Na + N2
    Synthesis
     2 NaN3  2 Na + 3 N2
      65.02 g 22.99 g 28.02 g

    94.5 g NaN3 x     1 mol NaN3   x   3 mol N2
                    40.08 g NaN3       2 mol NaN3
    x 28.02 g N2
                            = 99.1 g N2
       1 mol N2
Don’t you just love these types of
          problems???
     Review Mass-Mole-Molecules: Determine
     the number of molecules in 73 g of water



   # H2O molecules =


73 g H2O   x 1 mol H2O    x   6.02x1023 molecules
            18.02 g H2O          1 mol H2O


               = 2.4 x 1024 molecules H2O
       Try this one:

    Calculate the mass in grams of iodine required to react
          completely with 0.50 moles of aluminum.
    Al + I2  AlI3
                      2 Al + 3 I2  2 AlI3


             0.50 mol Al       x 3 mol I2
                                2 mol Al
            x    253.80 g I2
                    1 mol I2
                               = 190 g I2
   Try this one:

Calculate the mass in grams of iodine required to react
      completely with 0.50 g of aluminum.
Al + I2  AlI3
                  2 Al + 3 I2  2 AlI3


 0.50 g Al    x 1 mol Al        x 3 mol I2
                26.98 g Al        2 mol Al

         x   253.80 g I2
                               = 7.1 g I2
                1 mol I2

				
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posted:4/8/2013
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