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Stoichiometry

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					Stoichiometry

   Chapter 5
            Stoichiometry
• Quantitative relationships between
  reactants and products
• The balanced chemical equation gives us
  the relationships in moles
• Consider: N2 + 3H2  2NH3
• Three mol-mol conversion factors or mole
  ratios can be derived from balanced
  equation
            Stoichiometry
             N2 + 3H2  2NH3

• One mol of N2 produces 2 mol of NH3


• One mol of N2 reacts with 3 mol of H2
            Stoichiometry
• 3 mol of H2 produces 2 mol of NH3



• Inverse can also be used as a conversion
  factor
                  Example
• How many moles of NH3 can be produced from
  33.6 g of N2?
• Convert grams of N2 to mol of N2, then convert
  mol of N2 to mol of NH3
    Stoichiometry - Procedure
1. Write down what is given and what is
   requested in the problem
2. a. if a mass is given, use the molar mass
   to convert mass to moles of what is given
   b). if a number of molecules is given, use
   Avogadro’s number to convert to moles
   of what is given
      Stoichiometry procedure
3. Using the correct mole ratio from the balanced
   equation, convert moles of what is given to
   moles of what is requested
4. a. If a mass is required, convert moles of what
   is requested to mass of what is requested
   b. If a number of molecules is requested, use
   Avogadro’s number to convert to numbers of
   molecules of what is requested
5. Procedure is summarized on next slide
Stoichiometry
            Limiting reactant
• If specific amounts of each reactant are mixed,
  the reactant that produces the least amount of
  product is called the limiting reactant
• Think of hot dogs and buns
  – hot dogs are sold in packs of ten
  – hot dog buns are sold in packs of eight
  – how many hot dog-hog dog bun combinations can
    you make with one pack of hot dogs and one pack of
    hot dog buns?
Limiting reactant considerations
• When reactants are mixed in exactly the
  mass ratio determined from the balanced
  equation, the mixture is said to be a
  stoichiometric mixture
• Example:
  4.0 g H2 + 32.0 g O2  36.0 g H2O
• Other mass ratios require calculations to
  determine the limiting reactant
   Limiting reactant procedure
• Convert amount of each reactant to the
  number of moles of product using mole
  ratios
• The limiting reactant is the one that
  produces the smallest amount of product
    Limiting reactant example
• 2 CH3OH(l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (g)
• Mix 40.0 g of methanol with 40.0 g of O2,
  what is the mass of CO2 produced?
• Methanol, CH3OH, has a molar mass of
  40.0 g
• O2 has a molar mass of 32.00 g
Process
               Percent yield
• The actual yield is the amount of products
  obtained when the reaction is run
• The theoretical yield is the calculated amount of
  product that would be obtained if all of the
  limiting reactant was converted to a given
  product
• The percent yield is the actual yield in grams or
  moles divided by the theoretical yield in grams
  or moles times 100%
       Incomplete conversion
• In some cases, a reverse reaction occurs
  whereby reactants are reformed from products
• This limits the percent of reactants that are
  converted to products
• Such reactions are known as reversible
  reactions
• The reaction between N2 and H2 to produce NH3
  is a reaction which is reversible. This has
  severe consequences for the commercial
  production of ammonia
            Example calculation
• For the conversion of N2 and H2 to NH3
   – 4.70 g H2 react with N2
   – 12.5 g of NH3 is formed




• The theoretical yield is 26.5 g of NH3 in this reaction
  47.5% (commercially the yield is only 28%)
      Working with Solutions
  Solution Concentration: Molarity
• The amount of solute dissolved in a given
  solvent reported as moles of solute per
  liter of solution.
• Calculation of molarity is done by
  calculating the moles of solute present and
  dividing by the total volume of the solution
  in liters.
  Preparing Solutions of Known
         Concentration
Begin with pure solute

Example: Make a 1.5 M solution of NaOH
Molecular weight of NaOH: Na = 24; O = 16; H = 1 Total
= 40
1.5 *40 = 60 g NaOH / liter final volume.
Dissolve 60 g NaOH in 400 mL distilled water and dilute
to l liter final volume. Mix well.
• Begin with concentrated solution (dilution
  calculation - MV = MV)

• Example: Make 100 mL 0.05 M NaOH from a
  1.5 M solution.

  0.05 * 100 = 1.5 * ?

  (0.05 * 100 )/1.5 = 3.33 mL of 1.5 M NaOH
  diluted to 100 mL with distilled water
  Stoichiometry of Reactions in
        Aqueous Solution

1. Write the balanced equation
2. Calculate moles from masses or moles
from molarity
3. Use a stoichiometric factor
4. Calculate mass from moles or molarity
from moles and volume.
              Titrations
Essence of any technique of quantitative
chemical analysis - determination of the
quantity of a given constituent in a mixture
-If you know the balanced equation for the
reaction and the exact quantity of one of
the reactants, then you can calculate the
quantity of any other substance consumed
or produced in the reaction.
                 Titration
• Titration - procedure that allows for the
  determination of an unknown by adding
  carefully measured quantities of one
  solution into another solution when the
  exact concentration or amount of one of
  the solutions is known.
• Indicator - dye that changes color when
  the solution used for analysis if complete
• Buret - a measuring cylinder that most
  commonly has a volume of 50.0 mL and
  is calibrated in 0.1 mL divisions
• Equivalence point - the point at which the
  reaction is complete.
 Pre Lab Analysis of a Commercial
             Bleach
• Use titration to determine the molarity of
  the NaOCl as well as the percent by mass
  of the NaOCl in a sample of bleach using
  sodium thiosulfate as the titrant and iodine
  as the indicator.
                              Reactions
•   1. Acidified iodide ion is added to hypochlorite ion solution, and the iodide
    is oxidized to iodine.

    2 H+(aq) + ClO- (aq) + 2 I- (aq)  Cl- (aq) + I2 (aq) + H2O ( l)

•   2. Iodine is only slightly soluble in water. In an aqueous solution of iodide
    ion, iodine dissolves very readily. The triiodide ion forms in this situation.
    Triiodide is a combination of a neutral I2 molecule with an I- ion. The
    triiodide ion is yellow in dilute solution, and dark red-brown when
    concentrated.

    I2 (aq) + I-(aq)  I3 -(aq)

•   3. The triiodide is then titrated with a standard solution of thiosulfate ions,
    which reduces the iodine back to iodide ions:

    I3 -(aq) + 2 S2O32- (aq)  3 I-(aq) + S4O62- (aq)

				
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