Module 4: Chemical Equations and Reaction Stoichiometry Part 1

1
Module 4 Chemical Equations and Reaction
Stoichiometry Part 1

• By Alyssa Jean-Mary
• Source Modular Study Guide for First Semester
  Chemistry by Anthony J. Papaps and Marta E.
  Goicoechea-Pappas

2
Balancing Chemical Equations

• A chemical equation describes a chemical
  reaction
• Reactants ? Products
• The reactants are the substances that react, the
  starting substances.
• The products are the substances that are formed,
  the new substances.
• In addition to showing the reactants and
  products, the relative amount of substances
  involved in a chemical reaction are also shown.
• In a chemical equation, all the atoms that are
  present at the start (i.e. on the left side, the
  reactant side) must be present at the end (i.e.
  on the right side, the product side). This is
  because atoms cannot be created or destroyed
  they can only be transformed from one substance
  to another.
• Coefficients are used to balance a chemical
  equation. They are placed in front of each of the
  formulas. If the coefficient is one, it is not
  written, so any formula without a coefficient is
  assumed to have a coefficient of one.
• When balancing a chemical equation, never change
  the chemical formula (i.e. never change
  subscripts) - only add coefficients in front of
  the formula in order to balance the equation

3
Steps to Balancing Chemical Equations

• Steps to Balancing Chemical Equations
• Balance any polyatomic ions that are present on
  both sides of the equation as a group
• Balance any atom that is in only one formula on
  both sides of the equation
• If there is an odd number of atoms on one side of
  the equation and an even number of atoms on the
  other side, then the coefficients in front of the
  formulas containing these atoms will be the cross
  multiplication product of the subscripts (i.e. if
  one side has 3 and the other has 2, the
  coefficient for the formula with 3 will be 2 and
  the one for the formula with 2 will be 3).
• Balance any atom that is in more than one formula
  on either side of the equation
• If a fraction was used to balance any atom,
  multiply all the coefficients in the chemical
  equation by the lowest common denominator.
• A fraction is used when there is an odd number of
  atoms on one side of the equation and a diatomic
  molecule of that atom on the other side of the
  equation.
• After each atom has been balanced individually,
  double check the balancing since some of first
  atoms balanced could have become unbalanced by
  balancing later atoms

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Example 1 of Balancing Chemical Equations

• Balance __H2SO4 __ NH4OH ? __ (NH4)2SO4
  __H2O
• Polyatomic ions
• NH4 there is 1 NH4 on the left side and 2 NH4
  on the right side, so they need to be balanced by
  placing a 2 in front of the formula containing
  the NH4 on the left side __H2SO4 _2_ NH4OH ?
  __ (NH4)2SO4 __H2O
• SO4 there is 1 SO4 on the left side and 1 SO4
  on the right side, so they are balanced and thus
  no coefficients are added
• __H2SO4 _2_ NH4OH ? __ (NH4)2SO4 __H2O
• OH even though it is a polyatomic ion, it
  cannot be treated as a group since it is only
  present on one side of the equation
• Elements in only one compound
• O there is now 2 O on the left side (from the
  balancing of NH4) and 1 O on the right side, so
  it needs to be balanced by placing a 2 in front
  of the formula containing the O on the right
  side
• __H2SO4 _2_ NH4OH ? __ (NH4)2SO4 _2_H2O
• Elements in more than one compound
• H - there is now 4 H on the left side (2 from the
  balancing of NH4) and 4 H on the right side (from
  the balancing of O), so it is balanced, and thus
  no coefficients are added
• __H2SO4 _2_ NH4OH ? __ (NH4)2SO4 _2_H2O
• Eliminating fractions there are no coefficients
  that are fractions
• Double Check Balancing
• __H2SO4 _2_ NH4OH ? __ (NH4)2SO4 _2_H2O
• NH4 2 and 2 SO4 1 and 1 O 2 and 2 H
  4 and 4

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Example 2 of Balancing Chemical Equations

• Balance __ CaBr2 __ AgNO3 ? __ AgBr __
  Ca(NO3)2
• Polyatomic ions
• NO3 - there is 1 NO3 on the left side and 2 NO3
  on the right side, so they need to be balanced by
  placing a 2 in front of the formula containing
  the NO3 on the left side
• __ CaBr2 _2_ AgNO3 ? __ AgBr __ Ca(NO3)2
• Elements in only one compound
• Ag - there is now 2 Ag on the left side (from the
  balancing of NO3) and 1 Ag on the right side, so
  it needs to be balanced by placing a 2 in front
  of the formula containing the Ag on the right
  side
• _ CaBr2 _2_ AgNO3 ? _2_ AgBr __ Ca(NO3)2
• Br - there is 2 Br on the left side and now 2 Br
  on the right side (from the balancing of Ag), so
  it is balanced, and thus no coefficients are
  added
• __ CaBr2 _2_ AgNO3 ? _2_ AgBr __ Ca(NO3)2
• Ca - there is 1 Ca on the left side and 1 Ca on
  the right side, so it is balanced, and thus no
  coefficients are added
• __ CaBr2 _2_ AgNO3 ? _2_ AgBr __ Ca(NO3)2
• Elements in more than one compound none
• Eliminating fractions there are no coefficients
  that are fractions
• Double Check Balancing
• __ CaBr2 _2_ AgNO3 ? _2_ AgBr __ Ca(NO3)2
• NO3 2 and 2 Ag 2 and 2 Br 2 and 2 Ca
  1 and 1

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Example 3 of Balancing Chemical Equations

• Balance __ C6H14 __ O2 ? __ CO2 __ H2O
• Polyatomic ions none
• Elements in only one compound
• C - there is 6 C on the left side and 1 C on the
  right side, so it needs to be balanced by placing
  a 6 in front of the formula containing the C on
  the right side
• __ C6H14 __ O2 ? _6_ CO2 __ H2O
• H - there is 14 H on the left side and 2 H on the
  right side, so it needs to be balanced by placing
  a 7 in front of the formula containing the H on
  the right side
• __ C6H14 __ O2 ? _6_ CO2 _7_ H2O
• Elements in more than one compound
• O - there is 2 O on the left side and 19 O on the
  right side (from the balancing of C and H), so it
  needs to be balanced by placing (19/2) in front
  of the formula containing the O on the left side
  
• __ C6H14 _(19/2)_ O2 ? _6_ CO2 _7_ H2O
• Eliminating fractions Since there is a fraction,
  (19/2), all of the coefficients in the equation
  need to be multiplied by the lowest common
  denominator, which is 2
• _2_ C6H14 _19_ O2 ? _12_ CO2 _14_ H2O
• Double Check Balancing
• _2_ C6H14 _19_ O2 ? _12_ CO2 _14_ H2O
• C 12 and 12 H 28 and 28 O 38 and 38

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Example 4 of Balancing Chemical Equations

• Balance __ Fe2O3 __ C ? __ Fe3O4 __ CO
• Polyatomic ions none
• Elements in only one compound
• Fe - there is 2 Fe on the left side and 3 Fe on
  the right side, so, since there is a even number
  on one side and an odd number on the other side,
  the cross multiplication product of the
  subscripts is used to obtain the coefficients,
  so, it needs to be balanced by placing a 3 in
  front of the formula containing the Fe on the
  left side and a 2 in front of the formula
  containing the Fe on the right side
• _3_ Fe2O3 __ C ? _2_ Fe3O4 __ CO
• C - there is 1 C on the left side and 1 C on the
  right side, so it is balanced, and thus no
  coefficients are added
• _3_ Fe2O3 __ C ? _2_ Fe3O4 __ CO
• Elements in more than one compound
• O - there is 9 O on the left side (from the
  balancing of Fe) and 9 O on the right side (from
  the balancing of Fe), so it is balanced, and thus
  no coefficients are added
• _3_ Fe2O3 __ C ? _2_ Fe3O4 __ CO
• Eliminating fractions there are no coefficients
  that are fractions
• Double Check Balancing
• _3_ Fe2O3 __ C ? _2_ Fe3O4 __ CO
• Fe 6 and 6 C 1 and 1 O 9 and 9

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Reaction Stoichiometry

• Stoichiometry is the quantitative relationships
  among elements and compounds as they undergo
  chemical change.
• Before solving a stoichiometry problem, make sure
  that the equation is balanced

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An Example of a Stoichiometry Problem 1

• Given the following balanced equation and molar
  masses
• 2 C6H14 19 O2 ? 12 CO2 14 H2O
• MM 86.20g 16.00g 44.01g 18.02g
• How many moles of O2 are needed to completely
  react with 2.0 moles of C6H14?
• Answer Since the given and what it is looking
  for are both moles, just use the coefficients of
  the equation to convert
• 2.0mol C6H14 x (19mol O2/2mol C6H14) 19mol O2
• How many grams of H2O are formed if 2.0 moles of
  O2 react with an excess of C6H14?
• Answer Since the given is in moles and what it
  is looking for is in grams, first use the
  coefficients of the equation to convert to moles
  and then convert to grams using formula weight
• 2.0mol O2 x (14mol H2O/19mol O2) x (18.02g
  H2O/1mol H2O) 27g H2O

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An Example of a Stoichiometry Problem 2

• Given the following balanced equation and molar
  masses
• 2 C6H14 19 O2 ? 12 CO2 14 H2O
• MM 86.20g 16.00g 44.01g 18.02g
• How many moles of CO2 are formed if 3.6g of C6H14
  react with an excess of O2? How many grams?
• Answer 1 moles Since the given is in grams and
  what it is looking for is in moles, first convert
  the grams to moles using formula weight and then
  use the coefficients of the equation to convert
  to moles
• 3.6g C6H14 x (1mol C6H14/86.20g C6H14) x (12mol
  CO2/2mol C6H14) 0.25mol CO2
• Answer 2 grams Since moles were calculated in
  answer 1, they can be converted to grams using
  formula weight
• 0.25mol CO2 x (44.01g CO2/1mol CO2) 11g CO2
• How many moles of H2O (d 1.00 g/mL) are formed
  if 12mL of C6H14 (d 0.655 g/mL) react with an
  excess of O2? How many grams? How many mL?
• Answer 1 moles Since the given is in mL and
  what it is looking for is in moles, first convert
  the mL to grams using the density, then convert
  grams to moles using formula weight, and finally
  use the coefficients of the equation to convert
  to moles
• 12mL C6H14 x (0.655g C6H14/1mL C6H14) x (1mol
  C6H14/86.20g C6H14) x (14mol H2O/2mol C6H14)
  0.64mol H2O
• Answer 2 grams Since moles were calculated in
  answer 1, they can be converted to grams using
  formula weight
• 0.64mol H2O x (18.02g H2O/1mol H2O) 12g H2O
• Answer 3 mL Since grams were calculated in
  answer 2, they can be converted to mL using
  density
• 12g H2O x (1mL H2O/1.00g H2O) 12mL H2O

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An Example of a Stoichiometry Problem 3

• Given the following balanced equation and molar
  masses
• 2 C6H14 19 O2 ? 12 CO2 14 H2O
• MM 86.20g 16.00g 44.01g 18.02g
• How many molecules of CO2 are formed if 4.7g of
  C6H14 react with an excess of O2?
• Answer Since the given is in grams and what it
  is looking for is in molecules, first convert the
  grams to moles using formula weight, then use the
  coefficients of the equation to convert to moles,
  and finally convert to molecules using 6.02 x
  1023 molecules 1 mole
• 4.7g C6H14 x (1mol C6H14/86.20g C6H14) x (12mol
  CO2/2mol C6H14) x (6.02 x 1023 molecules CO2/1mol
  CO2) 2.0 x 1023moleclules CO2
• How many grams of C6H14 are needed to react with
  3.6g of O2 if the O2 is 97.1 pure?
• Answer Since the given is in grams of impure and
  what it is looking for is in grams, first convert
  the grams of impure to grams of pure using the
  percentage, then convert the grams to moles using
  formula weight, then use the coefficients of the
  equation to convert to moles, and finally convert
  to grams using formula weight
• 3.6g impure O2 x (97.1g pure O2/100g impure O2) x
  (1mol O2/16.00g O2) x (2mol C6H14/19mol O2) x
  (86.20g C6H14/1mol C6H14) 2.0g C6H14

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Limiting Reagent Concept

• Stoichiometry problems where the amounts of at
  least two reactants are given are referred to as
  limiting reagent (or excess) problems.
• The reactant that is used completely is the
  limiting reagent. After this reagent is
  completely used (i.e. the reaction has gone to
  completion), some of the other reagent will
  remain unreacted. This other reagent is the
  excess reagent.
• Limiting reagent problems involve doing two
  stoichiometry problems for each given reagent,
  calculate the amount that is produced of one of
  the products
• The reagent that predicts the smaller amount of
  that product is the limiting reagent. This is the
  amount of the product that is actually produced.
  The other reagent predicts the larger amount of
  the product. This is the excess reagent.

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Example 1 of the Limiting Reagent Concept

• Given the following balanced equation and molar
  masses
• 2 C6H14 19 O2 ? 12 CO2 14 H2O
• MM 86.20g 16.00g 44.01g 18.02g
• If 5.6g of C6H14 are mixed with 4.3g of O2,
• What is the limiting reagent?
• Answer In order to determine the limiting
  reagent, two stoichiometry problems need to be
  done (one with C6H14 and one with O2). The one
  that predicts the smaller amount of product (in
  this case, use H2O, but either H2O or CO2 could
  be used) is the limiting reagent.
• 5.6g C6H14 x (1mol C6H14/86.20g C6H14) x (14mol
  H2O/2mol C6H14) x (18.02g H2O/1mol H2O) 8.2g
  H2O
• 4.3g O2 x (1mol O2/16.00g O2) x (14mol H2O/19mol
  O2) x (18.02g H2O/1mol H2O) 3.6g H2O
• So, since O2 only produces 3.6g H2O, while C6H14
  produces 8.2g H2O, then O2 is the limiting
  reagent.
• What is the excess reagent?
• The excess reagent is the reagent that predicts
  the larger amount of product. In this case, C6H14
  is the excess reagent.
• At the end of the reaction, how much of the
  limiting reagent is left?
• 0g of O2 are left at the end of the reaction
  because all of the limiting reagent is used up in
  the reaction.
• How many grams of H2O are obtained?
• Since the limiting reagent is O2, the amount of
  H2O produced is the amount that it predicted
  would be produced 3.6g H2O.

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Example 2 of the Limiting Reagent Concept

• Given the following balanced equation and molar
  masses
• 2 C6H14 19 O2 ? 12 CO2 14 H2O
• MM 86.20g 16.00g 44.01g 18.02g
• If 5.6g of C6H14 are mixed with 4.3g of O2,
• How many grams of the excess reagent were used in
  the reaction?
• Answer Do a stoichiometry problem starting with
  the limiting reagent
• 4.3g O2 x (1mol O2/16.00g O2) x (2mol C6H14/19mol
  O2) x (86.20g C6H14/1mol C6H14) 2.4g C6H14
• At the end of the reaction, how many grams of the
  excess reagent were left?
• Answer Use the following equation excess
  reagent left initial amount of excess reagent
  amount of excess reagent used
• 5.6g 2.4g 3.2g C6H14