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Chapter 4: Chemical Quantities and Aqueous Reactions

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Title: Chapter 4: Chemical Quantities and Aqueous Reactions


1
Chapter 4 Chemical Quantities and Aqueous
Reactions
2
Reaction Stoichiometry
  • the numerical relationships between chemical
    amounts in a reaction is called stoichiometry

2 C8H18(l) 25 O2(g) ? 16 CO2(g) 18 H2O(g) 2
mol C8H18 25 mol O2 16 mol CO2 18 mol H2O
3
Predicting Amounts from Stoichiometry
  • How much CO2 can be made from 22.0 moles of C8H18
    in the combustion of C8H18?
  • 2 C8H18(l) 25 O2(g) ? 16 CO2(g) 18 H2O(g)
  • 2 moles C8H18 16 moles CO2

4
Example Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g C8H18
  • 2 C8H18(l) 25 O2(g) ? 16 CO2(g) 18 H2O(g)

5
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6
Limiting Reactant
7
Limiting Reactant
  • the reactant that limits the amount of product is
    called the limiting reactant
  • sometimes called the limiting reagent
  • the limiting reactant gets completely consumed
  • reactants not completely consumed are called
    excess reactants
  • the amount of product that can be made from the
    limiting reactant is called the theoretical yield

8
Things Dont Always Go as Planned!
9
Things Dont Always Go as Planned!
  • many things can happen during the course of an
    experiment that cause the loss of product
  • the amount of product that is made in a reaction
    is called the actual yield
  • generally less than the theoretical yield, never
    more!
  • the efficiency of product recovery is generally
    given as the percent yield

10
  • Example 1 CH4(g) 2 O2(g) CO2(g) 2 H2O(g)
  • If we have 5 molecules of CH4 and 8 molecules of
    O2, which is the limiting reactant?

11
  • Example 2 When 28.6 kg of C are allowed to react
    with 88.2 kg of TiO2 in the reaction below, 42.8
    kg of Ti are obtained. Find the Limiting
    Reactant, Theoretical Yield, and Percent Yield.

Molar Mass TiO2 79.87 g/mol Molar Mass Ti
47.87 g/mol Molar Mass C 12.01 g/mol
Limiting Reactant TiO2 Theoretical Yield 52.9
kg Percent Yield 80.9
12
Example 3. How many grams of N2(g) can be made
from 9.05 g of NH3 reacting with 45.2 g of CuO?2
NH3(g) 3 CuO(s) ? N2(g) 3 Cu(s) 3 H2O(l)
1 mol NH3 17.03g, 1 mol CuO 79.55g, 1 mol N2
28.02 g
13
Solutions
  • when table salt is mixed with water, it seems to
    disappear, or become a liquid the mixture is
    homogeneous
  • the salt is still there, as you can tell from the
    taste, or simply boiling away the water
  • homogeneous mixtures are called solutions
  • the component of the solution that changes state
    is called the solute
  • the component that keeps its state is called the
    solvent
  • if both components start in the same state, the
    major component is the solvent

14
Solution Concentration
  • qualitatively, solutions are often described as
    dilute or concentrated
  • dilute solutions have a small amount of solute
    compared to solvent
  • concentrated solutions have a large amount of
    solute compared to solvent
  • quantitatively, the relative amount of solute in
    the solution is called the concentration

15
Solution ConcentrationMolarity
  • moles of solute per 1 liter of solution
  • used because it describes how many molecules of
    solute in each liter of solution

16
Preparing 1 L of a 1.00 M NaCl Solution
17
Example 4.5 Find the molarity of a solution
that has 25.5 g KBr dissolved in 1.75 L of
solution
KBr 119.00 g/mol
18
Dilution
  • often, solutions are stored as concentrated stock
    solutions
  • to make solutions of lower concentrations from
    these stock solutions, more solvent is added
  • the amount of solute doesnt change, just the
    volume of solution
  • moles solute in solution 1 moles solute in
    solution 2
  • M1V1 M2V2

19
  • Example To what volume should you dilute 0.200 L
    of 15.0 M NaOH to make 3.00 M NaOH?

20
Solution Stoichiometry
  • since molarity relates the moles of solute to the
    liters of solution, it can be used to convert
    between amount of reactants and/or products in a
    chemical reaction

21
  • Example What volume of 0.150 M KCl is required
    to completely react with 0.150 L of 0.175 M
    Pb(NO3)2 in the reaction
  • 2 KCl(aq) Pb(NO3)2(aq) ? PbCl2(s) 2 KNO3(aq)

22
Titration
23
Titration
24
  • Example The titration of 10.00 mL of HCl
    solution of unknown concentration requires 12.54
    mL of 0.100 M NaOH solution to reach the end
    point. What is the concentration of the unknown
    HCl solution?

25
Types of Chemical Reactions
  • Precipitation Reactions A process in which an
    insoluble solid precipitate drops out of the
    solution.

26
  • Most precipitation reactions occur when the
    anions and cations of two ionic compounds change
    partners. (double replacement)
  • Pb(NO3)2(aq) 2 KI(aq) ? 2 KNO3(aq) PbI2(s)

27
Electrolytes in Solution
  • Electrolytes Dissolve in water to produce ionic
    solutions.
  • Nonelectrolytes Do not form ions when they
    dissolve in water.

a) NaCl soln conducts electricity, completes
circuit (charged particles)
b) C6H12O6 does not
28
Electrolytes in Solution
  • Dissociation
  • The process by which a compound splits up to
    form ions in the solution.

29
Electrolytes in Solution
  • Strong Electrolyte Total dissociation when
    dissolved in water.
  • Weak Electrolyte Partial dissociation when
    dissolved in water.

30
Solubility Rules Precipitation
  • Allow you to predict whether a reactant or a
    product is a precipitate.
  • Soluble compounds are those which dissolve to
    more than 0.01 M.
  • There are three basic classes of salts

31
Solubility RulesCompounds that Are Generally
Soluble in Water
32
Solubility RulesCompounds that Are Generally
Insoluble
33
Example 1 Solubility Rules Precipitation
  • Predict the solubility of
  • (a) CdCO3 (b) MgO (c) Na2S
  • (d) PbSO4 (e) (NH4)3PO4
  • Write the balanced reaction and predict whether a
    precipitate will form for
  • (a) NiCl2(aq) (NH4)2S(aq) ?
  • (b) Na2CrO4(aq) Pb(NO3)2(aq) ?
  • (c) AgClO4(aq) CaBr2(aq) ?

34
Solubility Rules Precipitation
  • a) insol. d) insol.
  • b) insol. e) sol.
  • c) sol.
  • a) NiS(s) NH4Cl
  • b) NaNO3 PbCrO4(s)
  • c) AgBr(s) Ca(ClO4)2

35
Net Ionic Equation for Precipitation Reactions
  • Subscripts aq- aqueous, fully dissociation in
    water.
  • s- solid, Does not dissolve in water.
  • l- liquid, pure liquid that doesnt dissociate in
    water or water itself.
  • Always balance the chemical equation first!!

36
Net Ionic Equations for Precipitation Reactions
  • Write net ionic equation for the following
    reaction
  • 2 AgNO3(aq) Na2CrO4(aq) ? Ag2CrO4(s) 2
    NaNO3(aq)
  • 1. Is it balanced? If not do it!
  • 2. Separate all except solids into ions
    (complete ionic equation)
  • 3. Cancel out spectator ions on both sides
  • 4. Rewrite (net ionic equation)
  • Complete Ionic Equation
  • 2 Ag 2 NO3- 2Na CrO42- ? Ag2CrO4(s) 2
    Na 2 NO3-
  • Net Ionic Equation
  • 2 Ag CrO42- ? Ag2CrO4(s)

37
  • CdCl2(aq) Na2CO3 (aq) CdCO3 (s)
    NaCl

38
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39
  • Na2CrO4(aq) Pb(NO3)2(aq) ? PbCrO4 NaNO3

40
Acid-base Neutralization Reactions
  • Neutralization Reaction produces salt water.
  • HA(aq) MOH(aq) ? H2O(l) MA(aq)
  • HA acid MOH Base

41
Neutralization Reactions
  • General Equation Strong Acid Strong Base
  • HCl(aq) NaOH(aq) ? H2O(l) NaCl(aq)
  • Complete Ionic Equation
  • H Cl- Na OH- ? H2O(l) Na Cl-
  • Net Ionic Equation
  • H OH- ? H2O(l)
  • or
  • H3O OH- ? 2 H2O(l)

42
Neutralization Reactions
  • General Equation Weak Acid Strong Base
  • HF(aq) NaOH(aq) ? H2O(l) NaF(aq)
  • Complete Ionic Equation
  • HF Na OH- ? H2O(l) Na F-
  • Net Ionic Equation
  • HF OH- ? H2O(l) F-

43
Neutralization Reactions
  • General Equation Strong Acid Strong Base
  • HBr(aq) Ba(OH)2(aq) ? H2O(l) BaBr2(aq)
  • Balanced General Equation
  • 2 HBr(aq) Ba(OH)2(aq) ? 2 H2O(l) BaBr2(aq)
  • Complete Ionic Equation
  • 2 H 2 Br- Ba2 2 OH- ? 2 H2O(l) Ba2 2
    Br-
  • Net Ionic Equation
  • 2H 2 OH- ? 2 H2O(l) or H OH- ? H2O(l) or
  • H3O OH- ? H2O(l)

44
Example Neutralization Reactions
  • Write ionic and net ionic equations for the
    following
  • (a) NaOH(aq) CH3CO2H(aq)
  • (b) HCl(aq) NH3(aq) ?

NaOH strong base will dissociate well CH3CO2H
weak acid doesnt dissociate well HCl is a strong
acid and therefore a strong electrolyte NH3 is a
weak base and is a weak electrolyte
45
Neutralization Reactions
  • Balanced General Equation
  • NaOH(aq) CH3CO2H(aq) ? H2O(l) CH3CO2Na(aq)
  • Complete Ionic Equation
  • Na OH- CH3CO2H ? H2O(l) CH3CO2- Na
  • Net Ionic Equation
  • OH- CH3CO2H ? H2O l) CH3CO2-

46
Neutralization Reactions
  • Balanced General Equation proton transfer
  • HCl(aq) NH3(aq) ? NH4(aq) Cl-(aq)
  • Complete Ionic Equation
  • H Cl- NH3 ? NH4 Cl-
  • Net Ionic Equation
  • H NH3 ? NH4 or
  • H3O NH3 ? NH4 H2O(l)

47
Gas Evolving Reactions
  • Some reactions form a gas directly from the ion
    exchange
  • K2S(aq) H2SO4(aq) ? K2SO4(aq) H2S(g)
  • Other reactions form a gas by the decomposition
    of one of the ion exchange products into a gas
    and water
  • K2SO3(aq) H2SO4(aq) ? K2SO4(aq) H2SO3(aq)
  • H2SO3 ? H2O(l) SO2(g)

48
Combustion as Redox2 H2(g) O2(g) ? 2 H2O(g)
49
Redox without Combustion2 Na(s) Cl2(g) ? 2
NaCl(s)
2 Na ? 2 Na 2 e?
Cl2 2 e? ? 2 Cl?
50
Reactions of Metals with Nonmetals
  • consider the following reactions
  • 4 Na(s) O2(g) ? 2 Na2O(s)
  • 2 Na(s) Cl2(g) ? 2 NaCl(s)
  • the reaction involves a metal reacting with a
    nonmetal
  • in addition, both reactions involve the
    conversion of free elements into ions
  • 4 Na(s) O2(g) ? 2 Na2O2 (s)
  • 2 Na(s) Cl2(g) ? 2 NaCl(s)

51
Oxidation state
  • for reactions that are not metal nonmetal, or
    do not involve O2, we need a method for
    determining how the electrons are transferred
  • chemists assign a number to each element in a
    reaction called an oxidation state that allows
    them to determine the electron flow in the
    reaction
  • even though they look like them, oxidation states
    are not ion charges!
  • oxidation states are imaginary charges assigned
    based on a set of rules
  • ion charges are real, measurable charges

52
OxidationReduction Reactions
  • Assigning Oxidation Numbers All atoms have an
    oxidation number regardless of whether it
    carries an ionic charge.
  • rules are in order of priority!!
  • 1. An atom in its elemental state has an
    oxidation number of zero.

Elemental state as indicated by single elements
with no charge. Exception diatomics H2 N2 O2
F2 Cl2 Br2 and I2
53
OxidationReduction Reactions
  • 1. Elemental state 0
  • 2. Group 1A ion 1
  • 3. Group 2 A ion 2
  • 4. H ion 1 unless it is bounded with a metal.
  • 5. O ion -2
  • 6. Group 7A ion -1
  • 7. Group 6A ion -2

54
OxidationReduction Reactions
  • The sum of the oxidation numbers must be zero
    for a neutral compound and must be equal to the
    net charge for a polyatomic ion.
  • A. H2SO4 neutral atom, no net charge
  • SO42- sulfate polyatomic ion

55
OxidationReduction Reactions
  • B. ClO4 , net charge of -1

56
Example 7 OxidationReduction Reactions
  • Assign oxidation numbers to each atom in the
    following substances
  • A. CdS G. V2O3
  • B. AlH3 H. HNO3
  • C. Na2Cr2O7 I. FeSO4
  • D. SnCl4 J. Fe2O3
  • E. MnO4 K. H2PO4-
  • F. VOCl3

57
  • A. CdS Cd2 S2-
  • B. AlH3 Al3 H-
  • C. Na2Cr2O7 Na, Cr6, O2-
  • D. SnCl4 Sn4, Cl1-
  • E. MnO4 Mn7, O2-
  • F. VOCl3 V5, O2-, Cl1-
  • G. V2O3 V3, O2-
  • H. HNO3 H1, N5, O2-
  • I. FeSO4 Fe2, S6, O2-
  • J. Fe2O3 Fe3, O2-
  • K. H2PO4- H1, P5, O2-

58
  • LEO the lion says GER.
  • LEO Loss of Electrons is Oxidation (reducing
    agent)
  • GER Gain of Electrons is Reduction (oxidation
    agent)

59
Oxidation and ReductionAnother Definition
  • oxidation occurs when an atoms oxidation state
    increases during a reaction
  • reduction occurs when an atoms oxidation state
    decreases during a reaction

CH4 2 O2 ? CO2 2 H2O -4 1
0 4 2 1 -2
60
OxidationReduction
  • oxidation and reduction must occur simultaneously
  • the reactant that reduces an element in another
    reactant is called the reducing agent
  • the reactant that oxidizes an element in another
    reactant is called the oxidizing agent

2 Na(s) Cl2(g) ? 2 NaCl(s) Na is oxidized, Cl
is reduced Na is the reducing agent, Cl2 is the
oxidizing agent
61
Example OxidationReduction Reactions
  • For each of the following, identify which species
    is the reducing agent and which is the oxidizing
    agent.
  • Ca(s) 2 H(aq) ? ? Ca2(aq) H2(g)
  • 2 Fe2(aq) Cl2(aq) ? 2 Fe3(aq) 2 Cl(aq)
  • C) SnO2(s) 2 C(s) ? Sn(s) 2 CO(g)

62
OxidationReduction Reactions
  • A) Ca0(s) 2 H(aq) ? ? Ca2(aq) H20(g)
  • Ca0 ? Ca2 oxidized, reducing agent
  • 2 H ? H20 reduced, oxidizing agent
  • B) 2 Fe2(aq) Cl2(aq) ? 2 Fe3(aq) 2
    Cl(aq)
  • Fe2 ? Fe3 oxidized, reducing agent
  • Cl20 ? 2 Cl- reduced, oxidizing agent
  • C) SnO2(s) 2 C(s) ? Sn(s) 2 CO(g)
  • Sn4 ? Sn0 reduced, oxidizing agent
  • 2 C0 ? 2 C2 oxidized, reducing agent

63
Balancing Redox Reactions
  • Half-Reaction Method Allows you to focus on the
    transfer of electrons. This is important when
    considering batteries and other aspects of
    electrochemistry.
  • The key to this method is to realize that the
    overall reaction can be broken into two parts, or
    half-reactions. (oxidation half and reduction
    half)

64
Balancing Redox Reactions
  • Balance for an acidic solution
  • MnO4(aq) Br(aq) ? Mn2(aq) Br2(aq)
  • 1. Determine oxidation and reduction
    half-reactions
  • Oxidation half-reaction Br(aq) ? Br20(aq)
  • Reduction half-reaction MnO4(aq) ? Mn2(aq)
  • 2. Balance for atoms other than H and O
  • Oxidation 2 Br(aq) ? Br2(aq)
  • Reduction MnO4(aq) ? Mn2(aq)

65
Balancing Redox Reactions
  • 3. Balance for oxygen by adding H2O to the side
    with less oxygen
  • Oxidation 2 Br(aq) ? Br2(aq)
  • Reduction MnO4(aq) ? Mn2(aq) 4 H2O(l)
  • 4. Balance for hydrogen by adding H to the side
    with less hydrogens
  • Oxidation 2 Br(aq) ? ? Br2(aq)
  • Reduction MnO4(aq) 8 H(aq) ? Mn2(aq) 4
    H2O(l)

66
Balancing Redox Reactions
  • 5. Balance for charge by adding electrons (e)
  • Oxidation 2 Br(aq) ? Br2(aq) 2 e
  • Reduction MnO4(aq) 8 H(aq) 5 e ?
    Mn2(aq) 4 H2O(l)
  • 6. Balance for numbers of electrons by
    multiplying
  • Oxidation 52 Br(aq) ? Br2(aq) 2 e
  • Reduction 2MnO4(aq) 8 H(aq) 5 e ?
    Mn2(aq) 4 H2O(l)

67
Balancing Redox Reactions
  • 7. Combine and cancel to form one equation
  • Oxidation 10 Br(aq) ? 5 Br2(aq) 10 e
  • Reduction 2 MnO4(aq) 16 H(aq) 10 e ?2
    Mn2(aq) 8 H2O(l)
  • 2 MnO4(aq) 10 Br(aq) 16 H(aq) ?2 Mn2(aq)
    5 Br2(aq) 8 H2O(l)

68
Example Balancing Redox Reactions
  • Balance the following in an acidic soln
  • NO3(aq) Cu(s) ? NO(g) Cu2 (aq)

69
Balancing Redox Reactions
  • 2 (4H NO3- 3 e ? NO 2 H2O)
  • 3 (Cu0 ? Cu2 2 e)
  • 8 H 2 NO3- 3 Cu0 ? 2 NO 4 H2O 3 Cu2
  • We will not be balancing in basic solutions!!
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