Chapter 4 Chemical Quantities and Aqueous Reactions - PowerPoint PPT Presentation

1 / 81
About This Presentation
Title:

Chapter 4 Chemical Quantities and Aqueous Reactions

Description:

Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 4 Chemical Quantities and Aqueous Reactions Roy Kennedy Massachusetts Bay Community College – PowerPoint PPT presentation

Number of Views:220
Avg rating:3.0/5.0
Slides: 82
Provided by: RoyK184
Category:

less

Transcript and Presenter's Notes

Title: Chapter 4 Chemical Quantities and Aqueous Reactions


1
Chapter 4Chemical Quantities and Aqueous
Reactions
Chemistry A Molecular Approach, 1st Ed.Nivaldo
Tro
Roy Kennedy Massachusetts Bay Community
College Wellesley Hills, MA
2008, Prentice Hall
2
Reaction Stoichiometry
  • the numerical relationships between chemical
    amounts in a reaction is called stoichiometry
  • the coefficients in a balanced chemical equation
    specify the relative amounts in moles of each of
    the substances involved in the reaction

2 C8H18(l) 25 O2(g) ? 16 CO2(g) 18 H2O(g) 2
molecules of C8H18 react with 25 molecules of
O2 to form 16 molecules of CO2 and 18 molecules
of H2O 2 moles of C8H18 react with 25 moles of
O2 to form 16 moles of CO2 and 18 moles of H2O 2
mol C8H18 25 mol O2 16 mol CO2 18 mol H2O
3
Predicting Amounts from Stoichiometry
  • the amounts of any other substance in a chemical
    reaction can be determined from the amount of
    just one substance
  • 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 gasoline
  • assuming that gasoline is octane, C8H18, the
    equation for the reaction is
  • 2 C8H18(l) 25 O2(g) ? 16 CO2(g) 18 H2O(g)
  • the equation for the reaction gives the mole
    relationship between amount of C8H18 and CO2, but
    we need to know the mass relationship, so the
    Concept Plan will be

5
Practice
  • According to the following equation, how many
    milliliters of water are made in the combustion
    of 9.0 g of glucose?
  • C6H12O6(s) 6 O2(g) 6 CO2(g) 6 H2O(l)
  • convert 9.0 g of glucose into moles (MM 180)
  • convert moles of glucose into moles of water
  • convert moles of water into grams (MM 18.02)
  • convert grams of water into mL
  • How? what is the relationship between mass and
    volume?

density of water 1.00 g/mL
6
Limiting Reactant
  • for reactions with multiple reactants, it is
    likely that one of the reactants will be
    completely used before the others
  • when this reactant is used up, the reaction stops
    and no more product is made
  • 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

7
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

8
Limiting and Excess Reactants in the Combustion
of Methane
  • CH4(g) 2 O2(g) CO2(g) 2 H2O(g)
  • Our balanced equation for the combustion of
    methane implies that every 1 molecule of CH4
    reacts with 2 molecules of O2

9
Limiting and Excess Reactants in the Combustion
of Methane
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?

10
Example 4.4Finding Limiting Reactant,
Theoretical Yield, and Percent Yield
11
  • Example
  • 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.

12
Practice 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)
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
Describing Solutions
  • since solutions are mixtures, the composition can
    vary from one sample to another
  • pure substances have constant composition
  • salt water samples from different seas or lakes
    have different amounts of salt
  • so to describe solutions accurately, we must
    describe how much of each component is present
  • we saw that with pure substances, we can describe
    them with a single name because all samples
    identical

15
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

16
Solution ConcentrationMolarity
  • moles of solute per 1 liter of solution
  • used because it describes how many molecules of
    solute in each liter of solution

17
Preparing 1 L of a 1.00 M NaCl Solution
18
Example 4.5 Find the molarity of a solution
that has 25.5 g KBr dissolved in 1.75 L of
solution
19
Using Molarity in Calculations
  • molarity shows the relationship between the moles
    of solute and liters of solution
  • If a sugar solution concentration is 2.0 M, then
    1 liter of solution contains 2.0 moles of sugar
  • 2 liters 4.0 moles sugar
  • 0.5 liters 1.0 mole sugar
  • 1 L solution 2 moles sugar

20
Example 4.6 How many liters of 0.125 M NaOH
contains 0.255 mol NaOH?
21
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
  • the concentrations and volumes of the stock and
    new solutions are inversely proportional
  • M1V1 M2V2

22
Example 4.7 To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
23
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

24
Example 4.8 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)
25
What Happens When a Solute Dissolves?
  • there are attractive forces between the solute
    particles holding them together
  • there are also attractive forces between the
    solvent molecules
  • when we mix the solute with the solvent, there
    are attractive forces between the solute
    particles and the solvent molecules
  • if the attractions between solute and solvent are
    strong enough, the solute will dissolve

26
Table Salt Dissolving in Water
Each ion is attracted to the surrounding water
molecules and pulled off and away from the crystal
When it enters the solution, the ion is
surrounded by water molecules, insulating it from
other ions
The result is a solution with free moving charged
particles able to conduct electricity
27
Electrolytes and Nonelectrolytes
  • materials that dissolve in water to form a
    solution that will conduct electricity are called
    electrolytes
  • materials that dissolve in water to form a
    solution that will not conduct electricity are
    called nonelectrolytes

28
Molecular View of Electrolytes and
Nonelectrolytes
  • in order to conduct electricity, a material must
    have charged particles that are able to flow
  • electrolyte solutions all contain ions dissolved
    in the water
  • ionic compounds are electrolytes because they all
    dissociate into their ions when they dissolve
  • nonelectrolyte solutions contain whole molecules
    dissolved in the water
  • generally, molecular compounds do not ionize when
    they dissolve in water
  • the notable exception being molecular acids

29
Salt vs. Sugar Dissolved in Water
30
Acids
  • acids are molecular compounds that ionize when
    they dissolve in water
  • the molecules are pulled apart by their
    attraction for the water
  • when acids ionize, they form H cations and
    anions
  • the percentage of molecules that ionize varies
    from one acid to another
  • acids that ionize virtually 100 are called
    strong acids
  • HCl(aq) ? H(aq) Cl-(aq)
  • acids that only ionize a small percentage are
    called weak acids
  • HF(aq) ? H(aq) F-(aq)

31
Strong and Weak Electrolytes
  • strong electrolytes are materials that dissolve
    completely as ions
  • ionic compounds and strong acids
  • their solutions conduct electricity well
  • weak electrolytes are materials that dissolve
    mostly as molecules, but partially as ions
  • weak acids
  • their solutions conduct electricity, but not well
  • when compounds containing a polyatomic ion
    dissolve, the polyatomic ion stays together
  • Na2SO4(aq) ? 2 Na(aq) SO42-(aq)
  • HC2H3O2(aq) ? H(aq) C2H3O2-(aq)

32
Classes of Dissolved Materials
33
Solubility of Ionic Compounds
  • some ionic compounds, like NaCl, dissolve very
    well in water at room temperature
  • other ionic compounds, like AgCl, dissolve hardly
    at all in water at room temperature
  • compounds that dissolve in a solvent are said to
    be soluble, while those that do not are said to
    be insoluble
  • NaCl is soluble in water, AgCl is insoluble in
    water
  • the degree of solubility depends on the
    temperature
  • even insoluble compounds dissolve, just not
    enough to be meaningful

34
When Will a Salt Dissolve?
  • Predicting whether a compound will dissolve in
    water is not easy
  • The best way to do it is to do some experiments
    to test whether a compound will dissolve in
    water, then develop some rules based on those
    experimental results
  • we call this method the empirical method

35
Solubility RulesCompounds that Are Generally
Soluble in Water
Compounds Containing the Following Ions are Generally Soluble Exceptions (when combined with ions on the left the compound is insoluble)
Li, Na, K, NH4 none
NO3, C2H3O2 none
Cl, Br, I Ag, Hg22, Pb2
SO42 Ag, Ca2, Sr2, Ba2, Pb2
36
Solubility RulesCompounds that Are Generally
Insoluble
Compounds Containing the Following Ions are Generally Insoluble Exceptions (when combined with ions on the left the compound is soluble or slightly soluble)
OH Li, Na, K, NH4, Ca2, Sr2, Ba2
S2 Li, Na, K, NH4, Ca2, Sr2, Ba2
CO32, PO43 Li, Na, K, NH4
37
Precipitation Reactions
  • reactions between aqueous solutions of ionic
    compounds that produce an ionic compound that is
    insoluble in water are called precipitation
    reactions and the insoluble product is called a
    precipitate

38
2 KI(aq) Pb(NO3)2(aq) ? PbI2(s) 2 KNO3(aq)
39
No Precipitate Formation No Reaction
KI(aq) NaCl(aq) ? KCl(aq) NaI(aq) all ions
still present, ? no reaction
40
Process for Predicting the Products ofa
Precipitation Reaction
  • Determine what ions each aqueous reactant has
  • Determine formulas of possible products
  • Exchange ions
  • () ion from one reactant with (-) ion from other
  • Balance charges of combined ions to get formula
    of each product
  • Determine Solubility of Each Product in Water
  • Use the solubility rules
  • If product is insoluble or slightly soluble, it
    will precipitate
  • If neither product will precipitate, write no
    reaction after the arrow

41
Process for Predicting the Products ofa
Precipitation Reaction
  1. If either product is insoluble, write the
    formulas for the products after the arrow
    writing (s) after the product that is insoluble
    and will precipitate, and (aq) after products
    that are soluble and will not precipitate
  2. Balance the equation

42
Example 4.10 Write the equation for the
precipitation reaction between an aqueous
solution of potassium carbonate and an aqueous
solution of nickel(II) chloride
  • Write the formulas of the reactants
  • K2CO3(aq) NiCl2(aq) ?
  • Determine the possible products
  • Determine the ions present
  • (K CO32-) (Ni2 Cl-) ?
  • Exchange the Ions
  • (K CO32-) (Ni2 Cl-) ? (K Cl-) (Ni2
    CO32-)
  • Write the formulas of the products
  • cross charges and reduce
  • K2CO3(aq) NiCl2(aq) ? KCl NiCO3

43
Example 4.10 Write the equation for the
precipitation reaction between an aqueous
solution of potassium carbonate and an aqueous
solution of nickel(II) chloride
  • Determine the solubility of each product
  • KCl is soluble
  • NiCO3 is insoluble
  • If both products soluble, write no reaction
  • does not apply since NiCO3 is insoluble

44
Example 4.10 Write the equation for the
precipitation reaction between an aqueous
solution of potassium carbonate and an aqueous
solution of nickel(II) chloride
  • Write (aq) next to soluble products and (s) next
    to insoluble products
  • K2CO3(aq) NiCl2(aq) ? KCl(aq) NiCO3(s)
  • Balance the Equation
  • K2CO3(aq) NiCl2(aq) ? 2 KCl(aq) NiCO3(s)

45
Ionic Equations
  • equations which describe the chemicals put into
    the water and the product molecules are called
    molecular equations
  • 2 KOH(aq) Mg(NO3)2(aq) 2 KNO3(aq)
    Mg(OH)2(s)
  • equations which describe the actual dissolved
    species are called complete ionic equations
  • aqueous strong electrolytes are written as ions
  • soluble salts, strong acids, strong bases
  • insoluble substances, weak electrolytes, and
    nonelectrolytes written in molecule form
  • solids, liquids, and gases are not dissolved,
    therefore molecule form
  • 2K1(aq) 2OH-1(aq) Mg2(aq) 2NO3-1(aq)
    2K1(aq) 2NO3-1(aq) Mg(OH)2(s)

46
Ionic Equations
  • ions that are both reactants and products are
    called spectator ions
  • 2K1(aq) 2OH-1(aq) Mg2(aq) 2NO3-1(aq)
    2K1(aq) 2NO3-1(aq) Mg(OH)2(s)
  • an ionic equation in which the spectator ions
    are
  • removed is called a net ionic equation
  • 2OH-1(aq) Mg2(aq) Mg(OH)2(s)

47
Acid-Base Reactions
  • also called neutralization reactions because the
    acid and base neutralize each others properties
  • 2 HNO3(aq) Ca(OH)2(aq) ? Ca(NO3)2(aq) 2
    H2O(l)
  • the net ionic equation for an acid-base reaction
    is
  • H(aq) OH?(aq) ? H2O(l)
  • as long as the salt that forms is soluble in water

48
Acids and Bases in Solution
  • acids ionize in water to form H ions
  • more precisely, the H from the acid molecule is
    donated to a water molecule to form hydronium
    ion, H3O
  • most chemists use H and H3O interchangeably
  • bases dissociate in water to form OH? ions
  • bases, like NH3, that do not contain OH? ions,
    produce OH? by pulling H off water molecules
  • in the reaction of an acid with a base, the H
    from the acid combines with the OH? from the base
    to make water
  • the cation from the base combines with the anion
    from the acid to make the salt
  • acid base ???salt water

49
Common Acids
50
Common Bases
51
HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
52
Example - Write the molecular, ionic, and
net-ionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide
  • Write the formulas of the reactants
  • HNO3(aq) Ca(OH)2(aq) ?
  • Determine the possible products
  • Determine the ions present when each reactant
    dissociates
  • (H NO3-) (Ca2 OH-) ?
  • Exchange the ions, H1 combines with OH-1 to make
    H2O(l)
  • (H NO3-) (Ca2 OH-) ? (Ca2 NO3-)
    H2O(l)
  • Write the formula of the salt
  • cross the charges
  • (H NO3-) (Ca2 OH-) ? Ca(NO3)2 H2O(l)

53
Example - Write the molecular, ionic, and
net-ionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide
  • Determine the solubility of the salt
  • Ca(NO3)2 is soluble
  • Write an (s) after the insoluble products and a
    (aq) after the soluble products
  • HNO3(aq) Ca(OH)2(aq) ? Ca(NO3)2(aq) H2O(l)
  • Balance the equation
  • 2 HNO3(aq) Ca(OH)2(aq) ? Ca(NO3)2(aq) 2
    H2O(l)

54
Example - Write the molecular, ionic, and
net-ionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide
  • Dissociate all aqueous strong electrolytes to get
    complete ionic equation
  • not H2O
  • 2 H(aq) 2 NO3-(aq) Ca2(aq) 2 OH-(aq) ?
    Ca2(aq) 2 NO3-(aq) H2O(l)
  • Eliminate spectator ions to get net-ionic
    equation
  • 2 H1(aq) 2 OH-1(aq) ? 2 H2O(l)
  • H1(aq) OH-1(aq) ? H2O(l)

55
Titration
  • often in the lab, a solutions concentration is
    determined by reacting it with another material
    and using stoichiometry this process is called
    titration
  • in the titration, the unknown solution is added
    to a known amount of another reactant until the
    reaction is just completed, at this point, called
    the endpoint, the reactants are in their
    stoichiometric ratio
  • the unknown solution is added slowly from an
    instrument called a burette
  • a long glass tube with precise volume markings
    that allows small additions of solution

56
Acid-Base Titrations
  • the difficulty is determining when there has been
    just enough titrant added to complete the
    reaction
  • the titrant is the solution in the burette
  • in acid-base titrations, because both the
    reactant and product solutions are colorless, a
    chemical is added that changes color when the
    solution undergoes large changes in
    acidity/alkalinity
  • the chemical is called an indicator
  • at the endpoint of an acid-base titration, the
    number of moles of H equals the number of moles
    of OH?
  • aka the equivalence point

57
Titration
58
Titration
The base solution is the titrant in the burette.
As the base is added to the acid, the H reacts
with the OH to form water. But there is still
excess acid present so the color does not change.
At the titrations endpoint, just enough base has
been added to neutralize all the acid. At this
point the indicator changes color.
59
Example 4.14The 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?
60
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)

61
NaHCO3(aq) HCl(aq) ? NaCl(aq) CO2(g) H2O(l)
62
Compounds that UndergoGas Evolving Reactions
Reactant Type Reacting With Ion Exchange Product Decom-pose? Gas Formed Example
metalnS, metal HS acid H2S no H2S K2S(aq) 2HCl(aq) ? 2KCl(aq) H2S(g)
metalnCO3, metal HCO3 acid H2CO3 yes CO2 K2CO3(aq) 2HCl(aq) ? 2KCl(aq) CO2(g) H2O(l)
metalnSO3 metal HSO3 acid H2SO3 yes SO2 K2SO3(aq) 2HCl(aq) ? 2KCl(aq) SO2(g) H2O(l)
(NH4)nanion base NH4OH yes NH3 KOH(aq) NH4Cl(aq) ? KCl(aq) NH3(g) H2O(l)
63
Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of
nitric acid, a gas evolves
  • Write the formulas of the reactants
  • Na2CO3(aq) HNO3(aq) ?
  • Determine the possible products
  • Determine the ions present when each reactant
    dissociates
  • (Na1 CO3-2) (H1 NO3-1) ?
  • Exchange the anions
  • (Na1 CO3-2) (H1 NO3-1) ? (Na1 NO3-1)
    (H1 CO3-2)
  • Write the formula of compounds
  • cross the charges
  • Na2CO3(aq) HNO3(aq) ? NaNO3 H2CO3

64
Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of
nitric acid, a gas evolves
  • Check to see either product H2S - No
  • Check to see if either product decomposes Yes
  • H2CO3 decomposes into CO2(g) H2O(l)
  • Na2CO3(aq) HNO3(aq) ? NaNO3 CO2(g) H2O(l)

65
Example 4.15 - When an aqueous solution of sodium
carbonate is added to an aqueous solution of
nitric acid, a gas evolves
  • Determine the solubility of other product
  • NaNO3 is soluble
  • Write an (s) after the insoluble products and a
    (aq) after the soluble products
  • Na2CO3(aq) 2 HNO3(aq) ? 2 NaNO3(aq) CO2(g)
    H2O(l)
  • Balance the equation
  • Na2CO3(aq) 2 HNO3(aq) ? 2 NaNO3 CO2(g)
    H2O(l)

66
Other Patterns in Reactions
  • the precipitation, acid-base, and gas evolving
    reactions all involved exchanging the ions in the
    solution
  • other kinds of reactions involve transferring
    electrons from one atom to another these are
    called oxidation-reduction reactions
  • also known as redox reactions
  • many involve the reaction of a substance with
    O2(g)
  • 4 Fe(s) 3 O2(g) ? 2 Fe2O3(s)

67
Combustion as Redox2 H2(g) O2(g) ? 2 H2O(g)
68
Redox without Combustion2 Na(s) Cl2(g) ? 2
NaCl(s)
2 Na ? 2 Na 2 e?
Cl2 2 e? ? 2 Cl?
69
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 Na2O (s)
  • 2 Na(s) Cl2(g) ? 2 NaCl(s)

70
Oxidation and Reduction
  • in order to convert a free element into an ion,
    the atoms must gain or lose electrons
  • of course, if one atom loses electrons, another
    must accept them
  • reactions where electrons are transferred from
    one atom to another are redox reactions
  • atoms that lose electrons are being oxidized,
    atoms that gain electrons are being reduced

2 Na(s) Cl2(g) ? 2 NaCl(s) Na ? Na 1 e
oxidation Cl2 2 e ? 2 Cl reduction
71
Electron Bookkeeping
  • 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

72
Rules for Assigning Oxidation States
  • rules are in order of priority
  • free elements have an oxidation state 0
  • Na 0 and Cl2 0 in 2 Na(s) Cl2(g)
  • monatomic ions have an oxidation state equal to
    their charge
  • Na 1 and Cl -1 in NaCl
  • (a) the sum of the oxidation states of all the
    atoms in a compound is 0
  • Na 1 and Cl -1 in NaCl, (1) (-1) 0

73
Rules for Assigning Oxidation States
  • (b) the sum of the oxidation states of all the
    atoms in a polyatomic ion equals the charge on
    the ion
  • N 5 and O -2 in NO3, (5) 3(-2) -1
  • (a) Group I metals have an oxidation state of 1
    in all their compounds
  • Na 1 in NaCl
  • (b) Group II metals have an oxidation state of
    2 in all their compounds
  • Mg 2 in MgCl2

74
Rules for Assigning Oxidation States
  • in their compounds, nonmetals have oxidation
    states according to the table below
  • nonmetals higher on the table take priority

Nonmetal Oxidation State Example
F -1 CF4
H 1 CH4
O -2 CO2
Group 7A -1 CCl4
Group 6A -2 CS2
Group 5A -3 NH3
75
Practice Assign an Oxidation State to Each
Element in the following
  • Br2
  • K
  • LiF
  • CO2
  • SO42-
  • Na2O2

76
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
77
OxidationReduction
  • oxidation and reduction must occur simultaneously
  • if an atom loses electrons another atom must take
    them
  • the reactant that reduces an element in another
    reactant is called the reducing agent
  • the reducing agent contains the element that is
    oxidized
  • the reactant that oxidizes an element in another
    reactant is called the oxidizing agent
  • the oxidizing agent contains the element that is
    reduced

2 Na(s) Cl2(g) ? 2 NaCl(s) Na is oxidized, Cl
is reduced Na is the reducing agent, Cl2 is the
oxidizing agent
78
Identify the Oxidizing and Reducing Agents in
Each of the Following
  • 3 H2S 2 NO3 2 H 3 S 2 NO 4 H2O
  • MnO2 4 HBr MnBr2 Br2 2 H2O

79
Combustion Reactions
  • Reactions in which O2(g) is a reactant are called
    combustion reactions
  • Combustion reactions release lots of energy
  • Combustion reactions are a subclass of
    oxidation-reduction reactions

2 C8H18(g) 25 O2(g) ? 16 CO2(g) 18 H2O(g)
80
Combustion Products
  • to predict the products of a combustion reaction,
    combine each element in the other reactant with
    oxygen

Reactant Combustion Product
contains C CO2(g)
contains H H2O(g)
contains S SO2(g)
contains N NO(g) or NO2(g)
contains metal M2On(s)
81
Practice Complete the Reactions
  • combustion of C3H7OH(l)
  • combustion of CH3NH2(g)
Write a Comment
User Comments (0)
About PowerShow.com