Topic 8.3 Stoichiometry Limiting and Excess Reagent Calculations by dfhdhdhdhjr


									Topic 8.3 Stoichiometry:
   Limiting and Excess
  Reagent Calculations
              Page 320-327

                         By Kirsten
              What is Stoichiometry?
              Lets take a closer look.
   A method of predicting
    or analyzing the
    quantities of the
    reactants and products
    participating in a
    chemical process
   Three forms including
    gas stoichiometry,
    solution stoichiometry,
    and gravimetric
         Example of Stoichiometry
   In a precipitation reaction, KOH(aq)
    reacts with excess Sn(NO3)2(aq) to
    produce a precipitate. If the mass of
    precipitate is 2.57 g, what mass of
    KOH(s) was present in the original
                                  First Step:
Balance Equation:

2KOH(aq) + Sn(NO3) 2(aq)          Sn(OH)2(s) + 2KNO3(aq)

 Moles              n2                n1

 Mass               ?                 2.57g

 Molar Mass         56.11 g/mol       152.71 g/mol
                         Second Step:
n1= 2.57 g x (1 mol/ 152.71 g)
  = 0.0168 mol

n2= 0.0168 mol x (2/1)
  = 0.0337 mol

m= 0.0337 mol x (56.11 mol/g)
 = 1.89g
               Understanding Chemical
   Conservation of Mass in a Chemical Reaction
       In chemical reactions the mass is conserved,
        meaning that the mass of the products equals the
        mass of the reactants.
           Mass Products = Mass Reactants
       There will always be an equal number of moles of
        each element before and after a reaction takes
                    Identifying Reagents
   Limiting Reagents
       Completely consumed in a chemical reaction
   Excess Reagents
       More is present than is necessary to react with
        the limiting reagent
   The limited reagent is the reagent that is
    being analyzed in a quantitative analysis
    where limiting and excess reagents are
     How to find Limiting Regent:
   Ensure that all masses given are in moles.
   Take the molar amounts from each
    substance and divide by the coefficient of that
    substance in the balanced equation.
   The smaller number will be the limiting
                               Example 1:
   300 mL of 0.100 mol/L of BaCl2(aq) and 200
    mL of 0.110 mol/L of Na2CO3(aq) are mixed.
    What is the limiting reagent in the reaction?

             Answer: Na2CO3
BaCl2(aq) + Na2CO3(aq)      BaCO3(s) + 2NaCl(aq)
With a 1:1 mole ratio of reactants he species
  present in least amount is the limiting
nBaCl2= 300mL x (0.100mol/L)
         =30.0 mmol
nNa2CO3= 200 mL x (0.110mol/L)
            = 22.0 mmol
Therefore Na2CO3(aq) is the limiting reagent
                               Example 2:
   100.0 g of iron (III) chloride and 50.00g of
    hydrogen sulfide react. What is the limiting

       2FeCl3 + 3H2S          Fe2S3 + 6HCl
   Iron(III) chloride
    100g/ 162.204 g/mol
    =0.6165 mol
   Hydrogen Sulfide
    50.00g/ 34.081 g/mol
    =1.467 mol
    1.467 mol/ 3
   Iron (III) chloride is the limiting reagent
                              Example 3:
   In an experiment, 26.8 g of iron(III) chloride
    in solution is combined with 21.5 g of sodium
    hydroxide in solution. Which reactant is in
    excess, and by how much? What mass of
    each product will be obtained?
FeCl3(aq) + 3NaOH(aq)               Fe(OH)3(s) + 3NaCl(aq)
26.8 g     21.5g                   m            m
162.20g/mol 40.00 g/mol       106.88 g/mol 58.44 g/mol

nFeCl3 = 26.8 g x (1mol/ 162.20g0
        =0.165 mol

nNaOH = 21.5 g x (1 mol / 40.00g)
      = 0.538 mol
nNaOH=0.165 mol x (3/1)

nNaOH= 0.538 mol - 0.496 mol
       = 0.042 mol

mNaOH = 0.042 mol x (40.00g/1 mol)
        = 1.7g
mFe(OH)3= 0.165 molFeCl3 x (1 mol Fe(OH)3/1
 molFeCl3) x (106.88 gFe(OH)3/ 1 mol Fe(OH)3)
         = 17.7 g Fe(OH)3

m NaCl = 0.165 molFeCl3 x (3 mol NaCl/ 1 mol FeCl3)
 x (58.44 g NaCl / 1 mol NaCl)
      = 29.0 NaCl

Therefore sodium hydroxide is in excess by 1.7
  g, the mass of iron (III) hydroxide produced is
  17.7 g and the mass of sodium chloride
  produced is 29.0g.
                        Theoretical yields
                            Actual yields
   The theoretical yield of a chemical reaction is
    the amount of product formed if all of the
    limiting reagent reacts.
       Calculated using stoichiometry.
   The actual yield is the actual quantity of
    products formed after a chemical reaction.
   Usually the theoretical yield will be greater
    than the actual yield that is produced.
            Reasons for Discrepancy
   The actual yield of a chemical reaction is
    usually less than the theoretical yield for
    these reasons
       Purity of chemicals being used
       Errors in measurements
       Experimental factors that may have lead to loss of
               % Error Calculations:
   A reasonable quantity of reasonable excess
    reagent is 10%
                                  Example 4:
     You decide to test the method of
      stoichiometry using the reaction of 2.00 g of
      copper(II) sulfate in solution with an excess of
      sodium hydroxide in solution. What would be
      a reasonable mass of sodium hydroxide to
CuSO4(aq) + 2NaOH(aq)           Cu(OH)2(aq) + Na2SO4(aq)

2.00 g           m
159.62 g/mol      40.00 g/mol
nCuSO4 = 2.00g x (1 mol/159.62 g)
         =0.0125 mol
nNaOH =0.0125 mol x (2/1)
        =0.0251 mol
mNaOH = 0.0251 mol x (40g/1mol)
         = 1.00 g
Now add 10% to this amount to determine the
  reasonable mass of sodium hydroxide to use.
1.00 g + 0.10g
                For More Information:

   Nelson Chemistry Text
       Pages 320-327
   The Key- Chemistry 20
   D2L Web Lessons

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