College Prep' Chemistry Chapter 9 p' 1

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College Prep. Chemistry Chapter 9 p. 1

• Stoichiometry the calculation of quantities in
  chemical reactions.
• Chemists use balanced chemical equations as a
  basis to calculate how much reactant is needed or
  product is formed in a reaction. Like a recipe.
• Interpreting Chemical Equations
• N2(g) 3H2 (g) ? 2NH3 (g)
• A balanced chemical equation can be interpreted
  in terms of atoms, molecules, or

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College Prep. Chemistry Ch. 9 p. 2

• Moles.
• Coefficients represent moles or molecules.
• 1 mole of N2 reacts with 3 moles of H2 to produce
  2 moles of NH3
• 1 molecule of N2 reacts with 3 molecules of H2 to
  produce 2 molecules of NH3
• Mass is conserved on both sides of the equation.
  Law of Conservation of Mass

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College Prep. Chemistry Ch. 9 p. 3

• Mole Ratio- a conversion factor derived from the
  coefficients of a balanced chemical equation
  interpreted in terms of moles.
• Relates moles of reactants to moles of products.
• N2 3H2 ? 2NH3
• Ex. How many moles of NH3 are formed from .6
  moles of N2?
• ?moles NH3 .6 moles N2 x 2 moles NH3/1mole N2
  1.2 moles NH3
• Section Review p. 301 1-4 Sample Problem A p.
  305, Practice p. 306, 1,2

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College Prep. Chemistry Ch.9 p. 4

• Mass-Mass Calculation- substances are measured in
  grams, not moles.
• Using the Molar mass, you can convert grams to
  moles.
• N2 3H2 ? 2NH3
• How many grams of NH3 are formed from 5.4 g of H2
  ?
• ? g NH3 5.4 g H2 x mole H2/2gH2 x 2 moles
  NH3/3 moles H2 x 17 g NH3/mole NH3 31 g NH3
• Mole-Mass problem skips one step. Sample Problem
  B p. 306 ?gC6H12O6 3 mole H2O x 1mole
  C6H12O6/6 mole H2O x 180 g C6H12O6/1mole C6H12O6
  90.1 g C6H12O6 Sample Problem C, D, E p.
  307-310 Practice p. 308, 309, 311 Section Review
  p. 311

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College Prep. Chemistry Ch. 9 p. 5

• Limiting and Excess Reagents- when making a
  recipe or doing a chemical reaction, the reagent
  that you run out of first is called the limiting
  reagent.
• Limiting reagent- reagent which determines the
  amount of product that can be formed in a
  reaction.
• Excess Reagent the reactant that is not
  completely used up in a reaction.
• 2Cu S ? Cu2S
• How many grams of Cu2S can be formed in the
  reaction of 80 g of Cu and 25 g of S?

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College Prep. Chemistry Ch. 9 p. 6

• ?gCu2S 80 g Cu x mole Cu/63.5g Cu x 1 mole
  Cu2S/2 moles Cu .63 moles Cu2S
• ?g Cu2S 25 g S x mole S/32 g S x 1 mole
  Cu2S/1mole S .78 moles Cu2S
• 80 g of Cu is the Limiting Reagent, since it
  produces the least moles of product.
• ?g Cu2S .63 moles Cu2S x 159 g Cu2S/mole Cu2S
  100 g Cu2S
• Sample Problem F, G p. 313-314
• Practice p. 313, 315

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College Prep. Chemistry Ch. 9 p. 7

• Percent Yield ratio of the actual yield to the
  theoretical yield expressed as a percentage.
• Theoretical Yield maximum amount of products
  that could be formed from a given amount of
  reactants.
• Actual Yield amount of product that forms when
  the reaction is carried out in the laboratory.
• Percent Yield measures the efficiency of a
  reaction.

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College Prep. Chemistry Ch. 9 p. 8

• Yield Actual Yield/ Theoretical Yield x 100
• Yield should not be larger than 100.
• Most reactions have yields smaller than 100,
  some reactions dont go to completion.
• Ex. Calculate the yield of the following
  reaction when 24.8 g of CaCO3 reacts to produce
  13.1 g CaO.
• CaCO3 ? CaO CO2

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College Prep. Chemistry Ch. 9 p. 9

• ? g CaO 24.8 g CaCO3 x 1 mole CaCO3/100 g CaCO3
  x 1 mole CaO/1mole CaCO3 x 56 g CaO/mole CaO
  13.9 g CaO
• yield 13.1 g/13.9 g x100 94.2
• Sample Problem H p. 317, Practice p. 318 (1,2),
  Section Review p. 318 (1-4)
• P. 320-322 1,2, 6, 10, 14, 15, 18-21, 25, 28,
  31, 33