Stoichiometry

1
Stoichiometry

• Chapter 5

2
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

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Stoichiometry

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

4
Stoichiometry

• 3 mol of H2 produces 2 mol of NH3
• Inverse can also be used as a conversion factor

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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 givenb. if a
   number of molecules is given, use Avogadros
   number to convert to moles of what is given

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Stoichiometry procedure

1. Using the correct mole ratio from the balanced
   equation, convert moles of what is given to moles
   of what is requested
2. a. If a mass is required, convert moles of what
   is requested to mass of what is requestedb. If a
   number of molecules is requested, use Avogadros
   number to convert to numbers of molecules of what
   is requested
3. Procedure is summarized on next slide

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Stoichiometry
8
Example

• How many moles of NH3 can be produced from 33.6 g
  of N2?

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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?

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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
• Example4.0 g H2 32.0 g O2 ? 36.0 g H2O
• Other mass ratios require calculations to
  determine the limiting reactant

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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

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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 32.0 g
• O2 has a molar mass of 32.00 g

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Process
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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

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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

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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

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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.

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Preparing Solutions of Known Concentration

• Begin with pure solute
• Example Make a 1.5 M solution of
  NaOHMolecular weight of NaOH Na 24 O 16
  H 1 Total 401.5 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.

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• 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

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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.

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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.

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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.

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• 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.

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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.

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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)