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Chemistry 11 Chapter 6

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									Chemistry 11
Chapter 6




  STOICHIOMETRY OF EXCESS
          QUANTITIES
Introduction: So far ...
 wehave
 assumed that a
 given reactant is
 completely used
 up during the
 reaction
In reality...
                   reactions are often
                    carried out in such
                    a way that one or
                    more of the
                    second reactants
                    actually are
                    present in EXCESS
                    amounts.
Definitions
   EXCESS REACTANT = the reactant in excess
   LIMITING REACTANT = the reactant that
    completely reacts
   THE LIMITING REACTANT determines the yield of
    the product (how much product(s) will form)
A Simple Analogy
              Imagine you work
               at McDonalds™ …
              You have 10
               hamburger buns
               and 5 beef patties
              How many regular
               hamburgers can
               you make?
Da Answer ...
   Indeed, you would
    get 5 regular
    hamburgers!

   And what was left
    over?
Connecting the Lingo …
                There would be 5
                 hamburger buns in
                 EXCESS!
                Therefore, the beef
                 patty is known as the
                 LIMITING ingredient
                 since it “limits” or
                 determines how many
                 regular buns can be
                 made!
Note that




   do not predict based on the
 we
 number of hamburger buns
Example 1
   If 20.0 g of
    hydrogen gas
    react with 100.0 g
    of oxygen, which
    reactant is present
    in excess and by
    how many grams?
Step 1 …


 The   balanced equation:




  2H2(g) + O2(g)         2H2O(l)
Step 2 …
              First PREDICT which
               reactant is limiting
               (it’s ok if you predict
               wrong)

              USUALLY the reactant
               with the least number
               of moles is limiting
               (but not always)
Convert masses to moles
Number of moles of H2 present
      = 20.0 g x 1 mol H2       =   10 mol H2
               2.0 g H2
Number of moles of O2 present
    = 100.0 g x 1 mol O2      = 3.125 mol O2
                32.0 g O2

Let’s make a prediction ...
Prediction: O2 is limiting
Mass of H2 that reacts with 100.0 g O2
 = 100.0 g O2 x 1 mol O2 x 2 mol H2 x 2.0 g H2
                  32.0 g O2 1 mol O2   1 mol H2
 = 12.5 g H2
Analysing the numbers …
   What we have: 100.0 g O2 and 20.0 g H2
   We predict O2 is limiting (i.e. all 100.0 g reacted)
   We calculated that we would need 12.5 g H2
   Is the prediction correct ?
Da Answer (again!)
                 Yes!! Prediction is
                  correct
                 only 12.5 g H2 is
                  required, so we have
                  an excess of 7.5 g H2
                  (20.0 g - 12.5 g)

                 so H2 is in EXCESS of
                  7.5 g.
The other side of the coin
So what if we predicted that H2 was
 limiting?

Mass of O2 that reacts with 20.0 g H2
= 20.0 g H2 x 1 mol H2 x 1 mol O2 x 32.0 g O2
               2.0 g H2    2 mol H2 1 mol O2
= 160.0 g O2
Therefore …
   If ALL 20.0 g of H2 were to completely react we
    would need 160.0 g of O2

   BUT we only have 100.0 g of O2

   So the prediction that H2 limiting is INCORRECT!
Example 2.
             If 79.1 g of Zn reacts
             with 1.05 L of 2.00 M
             HCl,

                a) Which reactant is
                  in excess and by
                  how much?
                b) What is the mass
                  of each product?
a) which reactant is excess?
The balanced equation:
     Zn + 2HCl            ZnCl2 + H2

     79.1 g 1.05 L, 2.00 M   xg    yg
                
     1.21 mol 2.10 mol
       (what we HAVE)
Prediction: Zn is limiting
Moles of HCl required
 = 1.21 mol Zn x 2 mol HCl = 2.43 mol HCl
                      1 mol Zn

Therefore 2.42 mol HCl would be required to react
with 1.21 mol Zn.
We ONLY have 2.10 mol HCl
So is our prediction correct?
Uh Oh! You’re wrong!
                 We would need more
                  HCl (2.42 mol) than
                  what we have (2.10
                  mol) if all the Zn were
                  to react
                 Thus: Zn is in excess,
                  and HCl is limiting!
To find how much in
excess:
             We must find how many
             moles of Zn is required
             to react with 2.10 molHCl
             Mol of Zn
             = 2.10 mol HCl x 1 mol Zn
                              2 mol HCl
             = 1.05 mol Zn
             Excess Zn = 1.21 - 1.05
                         = 0.16 mol Zn
b) mass of products?
Since HCl is limiting we MUST use this amount to
calculate the mass of products

x g ZnCl2 = 2.10 mol HCl x 1 mol ZnCl2 x 136.4 g
                           2 mol HCl   1 mol ZnCl2
         = 143 g ZnCl2
y g H2    = 2.10 mol HCl x 1 mol H2    x 2.0 g
                           2 mol HCl     1 mol H2
         = 2.1 g H2
Example 3:
   3.00 L of 0.1 M NaCl reacts with 2.50 L of 0.125 M
    AgNO3. Calculate the yield of solid AgCl (in
    grams) that will be produced.

   This problem requires us to determine how much
    product (AgCl) will form, so we will need to first
    determine which reactant is limiting.
The balanced equation:

 NaCl(aq) + AgNO3(aq)    NaNO3(aq) + AgCl(s)

 3.00 L     2.50 L                       ?g
 0.1M      0.125 M
              
 0.300 mol 0.325 mol
 Limiting    Excess         (since 1:1 ratio)
NaCl limiting ...
Therefore: mol NaCl = mol AgCl = 0.300 mol
             (also 1:1 ratio)

Mass of AgCl = 0.300 mol AgCl x 143.5 g AgCl
                                 1 mol AgCl
            = 43.1 g AgCl
Percent Yield
                   Often 100% of the
                    expected amount of
                    product cannot be
                    obtained from a
                    reaction
                   The term “Percent
                    Yield” is used to
                    describe the amount of
                    product actually
                    obtained as a
                    percentage of the
                    expected amount
Reasons for reduced yields
A) the reactants may not all
   react because:
   i) not all of the pure
   material
      actually reacts
   ii) the reactants may be
      impure
B) Some of the products are
   lost during procedures
   such as solvent extraction,
   filtration etc
The equation:
Percent Yield       = ACTUAL YIELD          x 100%
                    THEORETICAL YIELD

   Actual yield = amount of product obtained
    (determined experimentally)
   Theoretical yield = amount of product expected
    (determined from calculations based on the
    stoichiometry of the reaction)
   The amounts may be expressed in g, mol,
    molecules
Types of calculations
A) Find the percentage yield, given the mass of
  reactant used and mass of product formed
B) Find the mass of product formed, given the mass
  of reactant used and the percentage yield
C) Find the mass of reactant used, given the mass
  of product formed and percentage yield
 Note that the percentage yield must be less than
  100%
 But when calculating the theoretical yield assume

  a 100% yield
Example 1
            When 15.0 g of CH4 is
            reacted with an excess
            of Cl2 according to the
            reaction:
            CH4 + Cl2  CH3Cl + HCl

            a total of 29.7 g of CH3Cl
            is formed. Calculate the
            percentage yield.
The solution ...
The actual yield of CH3Cl = 29.7 g

To find the theoretical yield of CH3Cl: (assuming a
100% yield)

g of CH3Cl = 15.0 g CH4 x 1 mol CH4 x 1 mol CH3Cl x 50.5 g
                          16.0g CH4 1 mol CH4     1 mol CH3Cl
           = 47.34 g
Then:
Percentage yield = actual yield    x 100%
                 theoretical yield
                = 29.7 g x 100%       = 62.7 %
                  47.34 g
Example deux!




What mass of K2CO3 is produced when 1.50 g of
KO2 is reacted with an excess of CO2 if the reaction
has a 76.0% yield? The reaction is:
  4KO2(s) + 2 CO2(g)  2K2CO3(s) + 3O2(g)
 The solution:
We are looking for the actual yield (some idiot
forgot to weigh and record the mass of product!)
First calculate the mass of K2CO3 produced
(assuming a 100% yield) i.e. the theoretical yield

g of K2CO3 = 1.50 g KO2 x 1 mol KO2 x 2 mol K2CO3 x 138.2 g
                          71.1 g KO2 4 mol KO2 1 mol K2CO3
            = 1.458 g
actual yield = 76.0 % x 1.458 g = 0.760 x 1.458 = 1.11 g K2CO3
Last (but not least) example
...
What mass of CuO is required to make 10.0 g of Cu
according to the reaction
    2NH3 + 3CuO  N2 + 3Cu + 3H2O
if the reaction has 58.0 % yield?
Here we go again …
Actual yield = 10.0 g Cu
From the percentage yield equation, calculate the
theoretical yield of Cu.

Theoretical yield of Cu = Actual yield x 100%
                        Percentage yield
                      = 10.0 g x 100 %
                            58.0 %
                      = 17.24 g
Now find the mass of CuO:
Use this theoretical yield and find the mass of CuO
that would be needed:

g CuO = 17.24 g Cu x 1 mol Cu x 3 mol CuO x 79.5g
                    63.5 g Cu 3 mol Cu     1mol CuO


       = 21.6 g CuO

								
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