Chemical Equations and Reaction Stoichiometry

1
Chemical Equations and Reaction Stoichiometry

• Chapter 3

2
Chemical Reactions
3
Physical State Representation

• (g) _________
• (l) _________
• (s) _________
• (aq) _______________
• These representations will commonly be observed
  in chemical equations.

4
Chemical Equations

• There is no loss in the quantity of matter from a
  chemical reaction. This is the _____ __
  __________ __ _____.
• What is the total mass on both sides of the
  chemical reaction shown previously
• A balanced chemical equation must have the _____
  of each kind of atom on both sides of the
  equation.
• Smallest possible whole-number coefficients.

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Balancing Chemical Equations

• __Al(s) __Br2(l) ___Al2Br6(s) (covalent)
• __C3H8(g) __O2(g) __CO2(g) __H2O(g)
• __C4H10(l) __O2(g) __CO2(g) __H2O(g)
• __P4(s) __Cl2(g) __PCl3(s)
• Demo On demo. CD
• Be sure to balance equations before proceeding
  with calculations.

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Calculations Based on Balanced Chemical Equations

• The ______ in a balanced equation represent the
  relative ratio of moles or molecules (formula
  units).
• Calculations utilize a series of _________ to
  obtain the desired answer.
• Make sure that the ________ cancel.

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Calculations Based on Balanced Chemical Equations

• Conversion process
• Convert grams to moles (using molar mass)
• Convert moles to moles (from balanced equation)
• This can also be molecules. Why
• Convert moles to grams (using molar mass)
• This is the general steps that will be followed
  to solve equations.

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Solving Problems from Chemical Equations

• CH4(g) 2O2(g) CO2(g) 2H2O(g)
• How many CH4 molecules required to react
  completely with 82 molecules of O2
• __C4H10(l) __O2(g) __CO2(g) __H2O(g)
• How many O2 molecules required to react
  completely with 48 molecules of C4H10

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Solving Problems from Chemical Equations

• __Fe2O3(s) __CO(g) __Fe(s) __CO2(g)
• How many moles of Fe2O3 are required to react
  completely with 4.9 moles of CO
• What mass of CO2 can be produced from 0.540 moles
  of Fe2O3 with excess CO
• __P4(s) __Cl2(g) __PCl3(s)
• How many moles of PCl3 can be produced from 1.48
  moles of Cl2(g) Assume the P4 is in excess.
• How many grams of PCl3 can be produced from 42.3
  grams of P4

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More Problems

• __NH3 __O2 __NO __H2O
• How many grams of NO can be produced from 17.80
  grams of O2 NH3 is in excess.
• How many molecules of NH3 are required to produce
  7.31 10-10 grams of H2O

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Limiting Reactant

• In a typical reaction there is one substance
  that is generally used up first. This reactant
  is termed as the limiting reactant since is
  limits how much product will be formed. The
  reactants will be in excess.
• DEMO Setting a table for six people using 10
  plates 8 spoon and 5 forks.
• What is limiting and what is in excess This is
  the same with a reaction.

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Limiting Reactant

• DEMO CD (5-4) NH3 formation
• Method for calculating the limiting reactant
• Calculate the amount of product that would be
  formed from each reactant.
• The one that produces the least amount of product
  is the limiting reactant/reagent.

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Limiting Reactant

• What is the maximum amount of sulfur dioxide that
  can be produced from 95.6 grams of carbon
  disulfide and 110 grams of oxygen
• CS2 O2 CO2 SO2
• What is the maximum amount of HCl that can be
  produced from 28.2 grams of NaCl 23.8 grams of
  H2O and 32.1 grams of SiO2
• NaCl H2O SiO2 HCl Na2SiO3

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Limiting Reactant

• Determine the amount of Cu3P formed if 225 grams
  of copper is reacted with 65.0 grams of P4. What
  is the limiting reactant
• ___Cu ___P4 ____Cu3P

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Determining Percent Yields

• In all the previous calculations it is assumed
  that the yield is 100. In the real world
  however the actual yield is always below 100.
  Why
• Many reactions do not go to completion.
• Two or more reaction pathways occur
  simultaneously.
• Some of the product is lost upon isolation.

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Determining Percent Yields

• Percent yield
• A 10.0-gram sample of ethanol C2H5OH was
  reacted with excess CH3COOH to produce 14.8 grams
  of CH3COOC2H5. What is the percent yield
• CH3COOH C2H5OH CH3COOC2H5 H2O

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Determining Percent Yields

• Aspirin can be synthesized as shown below. For
  the reaction 12.4 grams of C4H6O3 is reacted with
  an excess of C7H6O3. The amount of aspirin
  C9H8O4 acquired from the reaction is 26.28
  grams. What is the percent yield
• 2C7H6O3(s) C4H6O3(l) 2C9H8O4(s) H2O(l)

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Concentrations of Solutions

• In many chemical reactions the reactants are
  that mixed are present as solutions.
• Solution consists of a substance (solute)
  dissolved in another substance (solvent).
• DEMO CuSO4(aq) and Ca(NO3)2(aq)
• Mix Ca(NO3)2 and Na2(CO3)
• Ca(CO3) is the solid. Write the chemical
  equation.
• Solutions are _________ mixtures. Mixing occurs
  at the molecular level.
• Aqueous solutions solvent in ______.

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Concentration of Solutions

• Concentration expresses the amount of solute
  dissolved in a give amount of solvent or
  solution.
• Percent by mass of solute
• What mass of NaCl is required to prepare 75.0
  grams of a 5.00 solution of NaCl
• DEMO Prepare this solution.

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Concentrations of Solutions

• A 7.50 solution of C6H12O6 contains 50.0 grams
  of C6H12O6. What is the mass of the solution
• Calculate the mass of NaOH in 150.0 mL of an
  8.00 solution of NaOH. The density of the
  solution is 1.09 g/mL.

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Concentrations of Solutions

• Molarity is defined as the number of moles of
  solute per liter of solution.
• DEMO 1.0 and 0.1 M solution of CuSO4 How many
  grams of CuSO4 required to make 1 liter of these
  solutions
• Prepare 50.0 mL of a 0.100 M solution of aqueous
  K2Cr2O7.
• What is the molarity of a solution that contains
  4.91 grams of HNO3 in 500.0 milliliters of
  solution

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Concentration of Solutions

• Determine the mass of BaSO4 required to produce
  250.0 milliliters of a 0.125 M solution of barium
  sulfate.
• Optional The density of a concentrated HCl
  solution is 1.185 g/mL and it is 36.31 by mass.
  What is the molarity

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Solution by Dilution

• When a specified amount of concentrated solution
  is diluted the molarity decreases and the volume
  increases. The number of moles however remains
  constant.
• Concentrated solution
• Number of moles M1 V1
• Diluted solution
• Same number of moles M2 V2
• Therefore M1 V1 M2 V2

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Solutions by Dilution

• M1 V1 M2 V2 This relationship is used to
  calculate the amount of concentrated solution
  V1 needed to prepare a diluted solution of given
  volume V2 and molarity M2.
• How would you prepare 25.0 mL of a 8.00 10-3 M
  solution of K2Cr2O7 from 0.100 M K2Cr2O7
• How many mL of 8.0 M sulfuric acid is required to
  produce 50.0 mL of a 0.150 M solution of sulfuric
  acid

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Solutions in Chemical Reactions

• In many chemical reaction it is beneficial or
  even necessary to add reactants in solution. If
  molarity is known the amount of solute (or
  reactant) to be added to a reaction can be
  determined.
• What volume of 0.500 M BaCl2 is required to react
  completely with 4.32 grams of Na2SO4

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Solutions in Chemical Reactions

• What volume of 0.200 M NaOH will react with 50.0
  mL of 0.200 M aluminum nitrate What will be the
  mass of the aluminum hydroxide precipitate