Chapter 8 Quantities in Chemical Reactions

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Chapter 8Quantities inChemical Reactions

• Tro, 2nd ed.

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STOICHIOMETRY SOME REVIEW

• The molar mass of an element is its atomic mass
  in grams.
• It contains 6.0221 x 1023 atoms (Avogadros
  number) of the element.
• The molar mass of an element or compound is the
  sum of the atomic masses of all its atoms.

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Avogadros Number of Particles
6.0221 x 1023 Particles
1 MOLE
Molar Mass
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How many moles of NaCl are present in 292.215
grams of NaCl? The molar mass of NaCl 58.443 g.
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STOICHIOMETRY

• Consider the combustion of acetylene
• 2 C2H2(g) 5 O2(g) ? 4 CO2(g)
  2 H2O(l)
• 2 molecules 5 molecules 4 molecules
  2 molecules
• 200 500 400
  200
• 12x1023 31x1023 24x1023 12x1023
• 2 moles 5 moles 4 moles
  2 moles
• Consider each of the formulas as molecules or
  moles
• Coefficients represent number of moles
• Recipe just like a cooking recipe
• 2 cups of acetylene requires 5 cups of oxygen,
  etc.
• When we use the coefficients to relate the
  chemicals in an equation, we are doing
  STOICHIOMETRY

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STOICHIOMETRY

• 2 C2H2(g) 5 O2(g) ? 4 CO2(g) 2 H2O(l)
• EXAMPLE If 6 moles of C2H2 are burned,
• then 15 moles of O2 are needed,
• 12 moles of CO2 are made and
• 6 moles of H2O are made
• How did we know? We used ratios.
• Practice with 0.102 moles of C2H2
• 0.102 mol 5 mol O2/2 mol C2H2 0.255 mol O2
• 0.102 mol 4 mol CO2/2 mol C2H2 0.204 mol CO2
• 0.102 mol 2 mol H2O/2 mol C2H2 0.102 mol H2O

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Introduction to StoichiometryThe Mole-Ratio
Method

• Stoichiometry The area of chemistry that deals
  with the quantitative relationships between
  reactants and products.
• Mole Ratio a ratio between the moles of any two
  substances involved in a chemical reaction.
• The coefficients used in mole ratio expressions
  are derived from the coefficients used in the
  balanced equation.

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The Mole-Ratio Method
N2 3H2 ? 2NH3
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The Mole-Ratio Method
N2 3H2 ? 2NH3
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The Mole-Ratio Method

• The mole ratio is used to convert the number of
  moles of one substance to the corresponding
  number of moles of another substance in a
  stoichiometry problem.
• The mole ratio is used in the solution of every
  type of stoichiometry problem.

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The Mole Ratio Method for Stoichiometry

• 1. Have a balanced chemical equation.
• 2. Convert the quantity of starting substance to
  moles (if it is not already moles)
• 3. Convert the moles of starting substance to
  moles of desired substance.
• 4. Convert the moles of desired substance to the
  units specified in the problem.

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In the following reaction how many moles of PbCl2
are formed if 5.000 moles of NaCl
react? 2NaCl(aq) Pb(NO3)2(aq) ? PbCl2(s)
2NaNO3(aq)
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• Calculate the number of moles of phosphoric acid
  (H3PO4) formed by the reaction of 10 moles of
  sulfuric acid (H2SO4).
• Ca5(PO4)3F 5H2SO4 ? 3H3PO4 HF 5CaSO4

1 mol
5 mol
3 mol
1 mol
5 mol
Step 2 Moles starting substance 10.0 mol H2SO4
Step 3 The conversion needed is
moles H2SO4 ? moles H3PO4
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Mole-Mass Calculations

• The object of this type of problem is to
  calculate the mass of one substance that reacts
  with or is produced from a given number of moles
  of another substance in a chemical reaction.
• If the mass of the starting substance is given,
  we need to convert it to moles. (Step 2)
• We use the mole ratio to convert moles of
  starting substance to moles of desired substance.
  (Step 3)
• We can then change moles of desired substance to
  mass of desired substance if called for by the
  problem. (Step 4)

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• Calculate the number of moles of H2SO4 necessary
  to yield 784 g of H3PO4.
• Ca5(PO4)3F 5H2SO4 ? 3H3PO4 HF 5CaSO4

Method 1 Step by Step Calculations
Step 1 Balance reaction given, with 784 grams of
H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
Step 3 Convert moles of H3PO4 to moles of H2SO4
by the mole-ratio method.
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• Calculate the number of moles of H2SO4 necessary
  to yield 784 g of H3PO4
• Ca5(PO4)3F 5H2SO4 ? 3H3PO4 HF 5CaSO4

Method 2 Continuous Calculation
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Mass-Mass Calculations

• Solving mass-mass stoichiometry problems requires
  all four steps of the mole-ratio method.
• You list the four steps here

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Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2. N2 3 H2 ? 2 NH3
Method 1 Step by Step Calculations

Step 2 Convert 112 g of H2 to moles. grams ?
moles
Step 3 Calculate the moles of NH3 by the mole
ratio method.
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Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2. N2 3 H2 ? 2 NH3

Step 4 Convert moles NH3 to grams NH3. moles ?
grams
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Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2. N2 3 H2 ? 2 NH3
Method 2 Continuous Calculation

grams H2 ? moles H2 ? moles NH3 ? grams NH3
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GROUP PRACTICE

• First write the balanced chemical reaction
  ammonium phosphate reacts with calcium hydroxide
  to produce calcium phosphate, ammonia and water.
  Then answer
• a. What weight of calcium hydroxide is need to
  produce 155 g of calcium phosphate?
• b. How many molecules of ammonia gas are
  released?
• c. What volume of water will be produced,
  assuming normal conditions?
• 2 (NH4)3PO4(aq) 3 Ca(OH)2(aq) ? Ca3(PO4)2(aq)
  6 NH3(g) 6 H2O(l)

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• a. 155 g Ca3(PO4)2 1mol/310 g 0.500 mol
  Ca3(PO4)2
• 0.500 mol Ca3(PO4)2 3Ca(OH)2/1 Ca3(PO4)2 1.50
  mol Ca(OH)2
• 1.50 mol Ca(OH)2 74 g/mol 111 g
• b. 0.5 mol Ca3(PO4)2 6 H2O/1 Ca3(PO4)2 3.0 mol
  H2O
• 3.0 mol H2O 18 g/mol 1 mL/1 g 54 mL water
• c. 0.5 mol Ca3(PO4)2 6 NH3/1 Ca3(PO4)2 3.0
  mol NH3
• 3.0 mol NH3 6.0221 x 1023 molecules/mol 1.8 x
  1024 molecules

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Limiting-Reactant and Yield Calculations

• The limiting reactant is one of the reactants in
  a chemical reaction.
• It is called the limiting reactant because the
  amount of it present is insufficient to react
  with the amounts of other reactants that are
  present.
• The limiting reactant limits the amount of
  product that can be formed.

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H2 Cl2 ? 2HCl
4 molecules Cl2 can form 8 molecules HCl
Cl2 is the limiting reactant
3 molecules of H2 remainH2 is in excess
7 molecules H2 can form 14 molecules HCl
9.3
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Steps Used to Determine the Limiting Reactant

1. Calculate the amount of product (moles or grams,
   as needed) formed from each reactant.
2. Determine which reactant is limiting. The
   reactant that gives the least amount of product
   is the limiting reactant the other reactant is
   in excess.
3. Calculate the amount of the other reactant
   required to react with the limiting reactant,
   then subtract this amount from the starting
   quantity of the reactant. This gives the amount
   of the that substance that remains unreacted.

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How many moles of HCl can be produced by reacting
4.0 mol H2 and 3.5 mol Cl2? Which compound is
the limiting reactant?
H2 Cl2 ? 2 HCl

• Step 1 Calculate the moles of HCl that can form
  from each reactant.

Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it
produces less HCl than H2.
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You Practice

• Work on Skillbuilders 8.4 8.5 with a partner.
  Turn in one piece of paper with both your names
  on it showing all work.

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How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain
unreacted?
MgBr2(aq) 2 AgNO3 (aq) ? 2 AgBr(s)
Mg(NO3)2(aq)

• Step 1 Calculate the grams of AgBr that can form
  from each reactant.

The conversion needed is g reactant ? mol
reactant ? mol AgBr ? g AgBr
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How many grams of the excess reactant (AgNO3)
remain unreacted?

• Step 3 Calculate the grams of unreacted AgNO3.
  First calculate the number of grams of AgNO3 that
  will react with 50 g of MgBr2.

The conversion needed is g MgBr2 ? mol MgBr2 ?
mol AgNO3 ? g AgNO3
The amount of AgNO3 that remains is
100.0 g AgNO3 -
92.3 g AgNO3
7.7 g AgNO3
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Theoretical Yield

• The quantities of products calculated from
  equations represent the maximum yield (100) of
  product according to the reaction represented by
  the equation.
• Many reactions fail to give a 100 yield of
  product.
• This occurs because of side reactions and the
  fact that many reactions are reversible.

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Theoretical Yield Percent Yield

• The theoretical yield of a reaction is the
  calculated amount of product that can be obtained
  from a given amount of reactant.
• The actual yield is the amount of product finally
  obtained from a given amount of reactant.
• The percent yield of a reaction is the ratio of
  the actual yield to the theoretical yield
  multiplied by 100.
• Actual yield 100 Percent Yield
• Theor yield

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Silver bromide was prepared by reacting 200.0 g
of magnesium bromide and an adequate amount of
silver nitrate. Calculate the percent yield if
375.0 g of silver bromide was obtained from the
reaction
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can be formed.

The conversion needed is g MgBr2 ? mol MgBr2 ?
mol AgBr ? g AgBr
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Silver bromide was prepared by reacting 200.0 g
of magnesium bromide and an adequate amount of
silver nitrate. Calculate the percent yield if
375.0 g of silver bromide was obtained from the
reaction
MgBr2(aq) 2AgNO3 (aq) ? 2AgBr(s) Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
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You Practice

• Work on Skillbuilder 8.6 and problem 60 on page
  257.