Download 4.2- Reaction Stoichiometry Reaction Stoichiometry

Principles of Chemistry:
A Molecular Approach,
1st Ed.
Nivaldo Tro
Chapter 4
Chemical quantities and Aqueous reactions
Chapt 4
Chemical reactions
4.2- Reaction Stoichiometry
Stoichiometry is the study of the numerical
relationship between chemical quantities (reactants
and products) in a chemical reaction
is based on the balanced chemical equation and
on the relationship between mass and moles
Such calculations are fundamental to most
quantitative work in chemistry
The coefficients in a balanced chemical equation
specify the relative amounts in moles of each of the
substances involved in the reaction
Chapt 4
Chemical reactions
Reaction Stoichiometry
Interpretation of the balanced chemical equation;
2 C8H18(l) + 25 O2(g)
16 CO2(g) + 18 H2O(g)
2 molecules C8H18 + 25 molecules O2
2 moles C8H18 + 25 moles O2
16 molecules CO2
+ 18 molecules of H2O
16 moles CO2 +18 moles H2O
Because moles can be converted to mass, you can
also give a mass interpretation of a chemical
equation
Chapt 4
Chemical reactions
1
Stoichiometry: Mole –to-Mole conversion
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
Chapt 4
Chemical reactions
Mass Relationships in Chemical
Equations(Mass –to – Mass conversion )
Estimate the mass of CO2 produced in 2006 by the
combustion of 3.5 × 1015 g petroleum(octane C8H18).
Solution: The equation for the reaction is:
2 C8H18(l) + 25 O2(g)
16 CO2(g) + 18 H2O(g)
g C8H18
mol C8H18
mol CO2
g CO2
Given: 3.4 × 1015 g C8H18
Find: g CO2
Chapt 4
Chemical reactions
Example cont’d
Molar mass:
Mole ratio :
1mol C8H18 = 114.22g
1 mol CO2 = 44.01 g
2 mol C8H18:16 mol CO2
Chapt 4
Chemical reactions
2
4.3 –Limiting reactant, Theoretical
and percent Yield
Limiting reactant or reagent(L.R)- The reactant
that makes the least amount of the product and is
completely consumed in the reaction
that limits the amount of the product in a chemical
reaction.
Excess Reactant- Any reactant that occurs in a quantity
greater than is required to completely react with L.R.
Theoretical Yield-The amount of product that can be made in
a chemical reaction based on the amount of L.R.
Actual Yield-The amount of product actually produced by a
reaction.
Chapt 4
Chemical reactions
Limiting reactant, Theoretical and
percent Yield
The percentage yield is the actual yield (experimentally
determined) expressed as a percentage of the theoretical yield
(calculated).
%Yield
actual yield
100%
theoretical yield
Chapt 4
Chemical reactions
Limiting reactant, Theoretical and
percent Yield: Making Pizzas
•If we had 4 crusts,15 oz. tomato sauce, and 10 cu cheese
how many pizzas can we make? We know that:
1 crust + 5 oz. tomato sauce + 2 cu cheese
1 pizza
tomato sauce- limiting reactant
theoretical yield (3 pizzas)
Chapt 4
Chemical reactions
3
Limiting and Excess Reactants in the
Combustion of Methane
If we have 5 molecules of CH4 and 8 molecules of O2,
how many molecules of CO2 can be produced? which
is the limiting reactant?
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
H H H H
C
C
H H H H
+
H H
C
H H
H H H H
C
C
H H H H
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
Chapt 4
?co2
Chemical reactions
Finding Limiting Reactant, Theoretical
Yield, and Percent Yield
How many grams of N2(g) can be made from 9.05 g
of NH3 reacting with 45.2 g of CuO?
Identify the limiting reactant and determine the
theoretical yield of N2(g).
2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are produced, what is the percent
yield?
Chapt 4
Chemical reactions
Solutions
Mostly chemical reaction occur in Solutions
Solutions are homogeneous mixtures
Solute + solvent
Solute: The component of the solution in small
quantity
Solvent: The component that is in large quantity
the composition can vary from one sample to
another
Chapt 4
Chemical reactions
4
Solution concentration and solution
stoichiometry
To describe solutions accurately, we must describe
how much of each component is present
Qualitative description
Dilute solutions have a small amount of solute
compared to solvent
Concentrated solutions have a large amount of
solute compared to solvent
Chapt 4
Chemical reactions
Solution Concentration: Molarity
Concentration - Amount of solute in a given
amount of solution.
Molarity (M)- is the amount of solute in moles
divided by volume of solution (Liters)
molar concentration (mol L-1)
Chapt 4
Chemical reactions
Preparing 1 L of a 1.00 M NaCl Solution
Chapt 4
Chemical reactions
5
Example:Calculating molarity
What is the molarity of a solution containing 3.4 g of
NH3 (MM 17.03) in 200.0 mL of solution?
g NH3
mol NH3
mL sol’n
L sol’n
Chapt 4
M
Chemical reactions
Using Molarity in Calculations
Molarity can be used as conversion factor between
the moles of solute and liters of solution.
If a sugar solution concentration is 2.0 M,
1 L solution : 2 moles sugar
Chapt 4
Chemical reactions
Example 4.6
a)
How many liters of 0.125 M NaOH contains
0.255 mol NaOH?
b)
Determine the mass of CaCl2 (MM = 110.98)
in 1.75 L of 1.50 M solution.
Chapt 4
Chemical reactions
6
Dilution
Stock solutions (concentrated)
To make solutions of lower concentrations from these
stock solutions, more solvent is added
The amount of solute doesn’t change, just the
volume of solution
moles solute in solution 1 = moles solute in solution 2
Dilution equation
M 1 ·V1 = M2 ·V2
The concentrations and volumes of the stock and
new solutions are inversely proportional.
Chapt 4
Chemical reactions
Preparing 3.00 L of 0.500 M CaCl2
from a 10.0 M stock
solution(dilution)
Chapt 4
Chemical reactions
Sample Problems
Diluting a solution
How would you prepare 200.0 mL of 0.25 M NaCl
solution from a 2.0 M solution?
M1V1 = M2V2
Chapt 4
Chemical reactions
7
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.
Chapt 4
Chemical reactions
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)
1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl
Chapt 4
Chemical reactions
Quantitative Analysis:Volumetric Analysis
Volumetric analysis is a method of analysis based
on titration
Titration is a procedure for determining the amount
of substance A by adding a carefully measured
volume of a solution with known concentration of B
until the reaction of A and B is just complete
End point- The point of completion of the reaction
indicated by the colour change of a chemical called
indicator
Chapt 4
Chemical reactions
8
Volumetric Analysis
Titration of an unknown amount of HCl with NaOH
HCl (aq)
End point : colorless to pink
Chapt 4
Chemical reactions
Volumetric Analysis
Consider the reaction of sulfuric acid, H2SO4, with sodium
hydroxide, NaOH:
H 2SO4 (aq) 2NaOH(aq)
2 H 2O ( l ) Na2SO4 (aq)
Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How
many milliliters of 0.250 M NaOH must be added to completely
react with the sulfuric acid?
Solution: Convert 0.0350L of H2SO4(35.0 mL) to moles of H2SO4
(using the molarity of the acid).
Then, convert to moles of NaOH (from the balanced chemical
equation)
Finally, convert to volume of NaOH solution (using the molarity
of NaOH
Chapt 4
Chemical reactions
Finally, convert to volume of NaOH solution (using
the molarity of NaOH
( 0.0350 L )
0.175 mole H 2SO 4 2 mol NaOH 1 L NaOHsoln.
1 L H 2SO 4 solution 1 mol H2SO4 0.250mol NaOH
0 . 0490 L NaOH solution
( 49.0 mL of NaOH solution)
Chapt 4
Chemical reactions
9
Practice problem
To neutralize 37.6 mL of Ba(OH)2 solution, 43.8 mL of 0.107 M HCl
is needed. What is the molarity of the base.
2 HCl + Ba(OH)2
Chapt 4
BaCl2 + 2 H2O
Chemical reactions
What happens when a solute dissolves?
There are attractive forces between the solute particles
holding them together
attractive forces between the solvent molecules
on mixing 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
Chapt 4
Chemical reactions
Electrolytes and Nonelectrolytes
Electrolytes- Materials that
dissolve in water to form
a solution that will conduct
electricity
Nonelectrolytes- Materials that
dissolve in water to form
a solution that will not
conduct electricity
Chapt 4
Chemical reactions
10
Electrolyte and non electrolyte
solution
Ionic compounds dissociate
into ions when they dissolve.
Chapt 4
Molecular compounds do not
dissociate when they dissolve.
Chemical reactions
Strong and Weak Electrolytes
Strong electrolytes are materials that dissolve completely as
ions and their solutions are good conductor of electricity
ionic compounds and strong acids
ionize virtually 100%
HCl(aq)
H+ (aq) + Cl-(aq)
Weak electrolytes are materials that dissolve mostly as
molecules (only partially ionize)
This situation is denoted by writing the equation with a
double arrow
HC2H3O2(aq)
H+ (aq) + C2H3O2-(aq)
Most soluble molecular compounds are either nonelectrolytes
or weak electrolytes
their solutions are poor conductor of electricity
Chapt 4
Chemical reactions
Practice Problem
Name and classify each of the following compounds
as strong electrolyte,weak electrolyte or nonelectrolyte. Write down the dissociation equation
where appropriate.
BaCl2
C6H6
HBr
HNO3
(NH4) 2CO3
Chapt 4
Chemical reactions
11
Solubility of Ionic Compounds
Solubility-The amount of a substance dissolved in a
given amount of solvent at a given temperature
defined as greater than 0.1g sloute/100g of the
solvent
If the solubility is low- precipitate will form
The degree of solubility depends on the
temperature
Even insoluble compounds dissolve, just not enough
to be meaningful
Solubility rules
Chapt 4
Chemical reactions
Chapt 4
Chemical reactions
Practice Problem
Determine if each of the following is soluble in water
KOH
AgBr
CaCl2
Pb(NO3) 2
PbSO4
Chapt 4
Chemical reactions
12
Types of the chemical reaction
Most of the reactions we will study fall into one of the
following categories:
Precipitation Reactions
Formation of an insoluble ionic substance
(precipitate)
Acid-Base Reactions
Reaction between an acid and base
Oxidation-Reduction Reactions
Involve transfer of electrons
Chapt 4
Chemical reactions
Precipitation reactions
Precipitation reactions- Reactions between aqueous
solutions of ionic compounds that produce an
insoluble ionic compound in water
the insoluble product is called a precipitate
For example, reaction between potassium iodide and
lead(II) nitrate solution
2 KI(aq) + Pb(NO3) 2(aq)
PbI2(s) + 2 KNO3(aq)
Chapt 4
Chemical reactions
Chapt 4
Chemical reactions
13
Process for Predicting the Products of
a Precipitation Reaction
1.
2.
3.
4.
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.
Chapt 4
Chemical reactions
Process for Predicting the Products of
a Precipitation Reaction
5.
6.
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
Balance the equation.
Chapt 4
Chemical reactions
Practice Problem
Predict the products and balance the equation.
KCl(aq) + AgNO3(aq)
Na2S(aq) + CaCl2(aq)
Write an equation for the reaction that takes place
when an aqueous solution of (NH4) 2SO4 is mixed with
an aqueous solution of Pb(C2H3O2) 2
(NH4) 2SO4(aq) + Pb(C2H3O2) 2(aq)
Chapt 4
Chemical reactions
14
Quantitative analysis: Gravimetric
analysis
Gravimetric Analysis- is a type of quantitative
analysis in which the amount of a species in a
material is determined by converting the species into
a product that can be isolated and weighed.
Precipitation reactions are often used in
gravimetric analysis
Chapt 4
Chemical reactions
Gravimetric analysis
Consider the problem of determining the amount of
lead in a sample of drinking water.
Adding sodium sulfate (Na2SO4) to the sample will
precipitate lead(II) sulfate.
Na2 SO4 (aq) Pb2 (aq)
2 Na (aq) PbSO4 ( s)
The PbSO4 can then be filtered, dried, and
weighed
Chapt 4
Chemical reactions
Gravimetric analysis
Suppose a 1.00 L sample of polluted water was analyzed for lead(II)
ion, Pb2+ , by adding an excess of sodium sulfate to it. The mass of
lead(II) sulfate that precipitated was 229.8 mg. What is the mass of
lead in a liter of the water? Express the answer as mg of lead per liter
of solution.
Na 2SO 4 ( aq ) Pb 2 ( aq )
2 Na ( aq ) PbSO 4 ( s )
First we must obtain the mass percentage of lead in lead(II) sulfate, by
dividing the molar mass of lead by the molar mass of PbSO4, then
multiplying by 100.
% Pb
207.2 g/mol
303.3 g/mol
100
68 .32 %
Use the % of Pb to calculate the mass of lead in 229.8 mg of
lead(II) sulfate
Chapt 4
Chemical reactions
15
4.7 Representing aqueous reactions
Molecular equation: in which reactants and products are
written as molecules, even though they may actually exist in
solution as ions
2 KOH(aq) + Mg(NO3) 2(aq)
2 KNO3(aq) + Mg(OH)2(s)
(strong)
(strong)
(strong)
(insoluble)
Complete ionic equation: Equation that describe the actual
dissolved species in solution(ions)
Aqueous strong electrolytes are written as ions
Insoluble substances, weak electrolytes, and nonelectrolytes
are written in molecule form
2 K+ (aq) + 2 OH-(aq) + Mg2+(aq) + 2 NO3-(aq)
Chapt 4
2 K+ (aq) + 2 NO3-(aq) + Mg(OH)2(s)
Chemical reactions
Net Ionic equation
Spectator ions- Ions that appear on both side of
the equation and they can be cancelled
e.g. K+ and NO3 2 K+ (aq) + 2 OH-(aq) + Mg2+(aq) + 2 NO3-(aq)
Mg(OH)2(s)
2 K+ (aq) + 2 NO3-(aq) +
Net Ionic equation- An equation showing only the
species that actually change(react) during the
reaction
Eliminating spectator ions
2 OH-(aq) + Mg2+(aq)
Chapt 4
Mg(OH)2(s)
Chemical reactions
Examples
1.
Write a balanced equation for the reaction that takes
place when an aqueous solution of NiCl2 is mixed
with an aqueous solution of Na3PO4.
Molecular equation (to be balanced)
NiCl
2
Na 3 PO 4
Ni 3 ( PO 4 ) 2
NaCl
Ionic equation
Net Ionic equation
Chapt 4
Chemical reactions
16
4.8- Acid–Base Reactions
neutralization reactions
Acid-Base reactions also called neutralization
reactions because the acid and base neutralize each
other’s properties
Arrhenius Acids ionize in water to form H+ ions.
More precisely, the H+ from the acid molecule is
donated to a H2O molecule to form hydronium
ion, H3 O+ .
Most chemists use H+ and H3O+
interchangeably.
H ( aq ) H 2 O ( l )
Chapt 4
H 3 O ( aq )
Chemical reactions
Polyprotic Acids
Polyprotic Acids-contain more than one ionizable
proton and release them equentially. e.g. sulfuric
acid(H2SO4)or phosphoric acid
Dissociation of H3 PO4 (phosphoric acid)
All three dissociations are incomplete
H3PO4 (aq) + H2O (l)
H2PO4- (aq) + H3O+ (aq)
H2PO4 (aq) + H2O (l)
HPO42- (aq) + H3O+ (aq)
HPO42- (aq) + H2O (l)
PO43- (aq) + H3O+ (aq)
Chapt 4
Acids are found in lemons,
limes, and vinegar.
Vitamin C and aspirin are
also acids
Chemical reactions
Many common household products,
such as bleach and ammonia,
are bases.
Chapt 4
Chemical reactions
17
Acid-Base reactions
Arrhenius Bases- dissociate in water to form
OH ions.
NaOH (s )
H 2O
Na ( aq ) OH (aq )
Acid-Base neutralization reactions- H+ ion from
acid combine with OH- from the base to form water
Net ionic equation: H+ (aq) + OH (aq)
Chapt 4
H2O(l)
Chemical reactions
Common Acids and bases
Chapt 4
Chemical reactions
Bronsted- Lowery Concept
The Brønsted-Lowry acid is defined as the species
(molecule or ion) that donates a proton(H+ ) to another species
in a proton-transfer reaction
HNO 3 ( aq )
acid
H 2O ( l )
NO 3 ( aq )
H 3 O ( aq )
base
base
acid
A Brønsted-Lowry base is defined as the species (molecule
or ion) that accepts the proton (H+ ) in a proton-transfer
reaction
NH 3 (aq ) H 2O( l )
base
NH 4 (aq ) OH (aq )
acid
Chapt 4
Chemical reactions
18
Summary of Acid-Base concepts
In summary, the Arrhenius concept and the
Brønsted-Lowry concept are essentially the same in
aqueous solution.
The Arrhenius Concept:
acid: proton (H+ ) donor
base: hydroxide ion (OH-) donor
The Brønsted-Lowry Concept:
acid: proton (H+ ) donor
base: proton (H+ ) acceptor
Chapt 4
Chemical reactions
Practice Example
1.
Write the molecular, ionic, and net-ionic equation for
the reaction of aqueous nitric acid with aqueous
calcium hydroxide
Chapt 4
Chemical reactions
Acid-Base Neutralization reaction
I. Neutralization reaction of the strong acid, HCl(aq) and a strong
base, KOH(aq).
Molecular equation.
HCl ( aq ) KOH ( aq )
KCl ( aq ) H 2 O ( l )
Net ionic equation: H+ (aq) + OH (aq)
H2O(l)
II. Neutralization Reaction of Weak acid and strong base
In a reaction involving HCN(aq), a weak acid, and KOH(aq), a
strong base, the product is KCN, a strong electrolyte.
The net ionic equation
HCN ( aq )
OH ( aq )
Chapt 4
CN ( aq )
H 2O ( l )
Chemical reactions
19
Gas-Evolution reactions
Gas-Evolution reactions -Aqueous reactions that
form a gas when two solutions are mixed
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)
Chapt 4
Chemical reactions
Chapt 4
Chemical reactions
Gas-Evolution reactions
Carbonates react with acids to form CO2, carbon dioxide gas
Sulfites react with acids to form SO2, sulfur dioxide gas
Sulfides react with acids to form H2S, hydrogen sulfide gas
Chapt 4
Chemical reactions
20
Example 4.14
Write a molecular equation for the reaction that
occurs 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 and possible
products
Na2CO3(aq) + HNO3(aq) H2CO3(l) + NaNO3(aq)
Check to see if either product decomposes
Na2CO3(aq) + 2 HNO3(aq)
Chapt 4
2 NaNO3(aq) + CO2(g) + H2O(l)
Chemical reactions
Oxidation-Reduction
reactions(Redox)
Oxidation–reduction reactions in which
electrons are transferred from one reactant to
another
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)
2 Fe2O3(s)
Oxidation is defined as the loss of electrons
OIL (oxidation is loss of electrons )
Reduction is defined as the gain of electrons.
RIG (Reduction Is Gain of electrons)
Ger
Leo
Chapt 4
Chemical reactions
Oxidation-Reduction Reaction
Oxidation and reduction always occur simultaneously.
The # of e- lost by the substance being oxidized = the # of
e - gained by the substance being reduced
Atoms that lose electrons are being oxidized
Atoms that gain electrons are being reduced
2 Na(s) + Cl2(g)
2 Na+Cl–(s)
Na
Na+ + 1 e–
oxidation
Cl2 + 2 e–
2 Cl– reduction
Na is oxidized and Cl2 is reduced
Chapt 4
Chemical reactions
21
Combustion as Redox
2 H2 ( g) + O2 ( g)
2 H 2 O(g)
Chapt 4
Chemical reactions
Oxidation numbers or state
Oxidation number is a simple way of keeping track of
electrons in a reaction
The oxidation number (or oxidation state) of an atom in a
substance is the actual charge of the atom if it exists as a
monatomic ion
oxidation states are not ion charges!
Oxidation states are imaginary charges assigned
based on a set of rules.
Ion charges are real, measurable charges
Rules for assigning oxidation states(page 148)
Chapt 4
Chemical reactions
Rules for assigning oxidation states
Rule
Applies to
Statement
1
Elements
The oxidation number of an atom in an element is
zero.
2
Monatomic ions
The oxidation number of an atom in a monatomic ion
equals the charge of the ion.
3
Oxygen
The oxidation number of oxygen is –2 in most of its
compounds. (An exception is O in H2O2 and other
peroxides, where the oxidation number is –1.)
4
Hydrogen
The oxidation number of hydrogen is +1 in
most of its compounds except in metal
hydrides.
5
Halogens
Fluorine is –1 in all of its compounds. The other
halogens are –1 unless the other element is
another halogen or oxygen.
6
Compounds
and ions
The sum of the oxidation numbers of the atoms
in a compound is zero. The sum in a polyatomic
ion equals
the charge on the ion.
Chapt 4
Chemical reactions
22
Sample problems
Oxidation no. of Cr in Cr2O7 2- (dichromate ions) ?
2(Cr) + 7(-2) = -2
2Cr = -2 -7(-2)
2Cr = -2 +14 = +12
and
Cr = +6
Oxidation no. of Cl in ClO4- (perchlorate ion)
Cl + 4(-2) = -1
Cl = -1-4 (-2) = -1 + 8 = + 7
Oxidation no. of N in NH4+ ion?
N + 4(+1) = +1
N = +1 - 4(+1) = -3
1.
Assign Oxidation numbers to all atoms in the following:
CO2 , K2Cr2O7 ,PCl5, HNO2
Chapt 4
Chemical reactions
Identifying Redox Reactions
Oxidation occurs when an atom’s oxidation state
increases during a reaction
Reduction occurs when an atom’s oxidation state
decreases during a reaction.
CH4 + 2 O2
–4 +1
0
CO2 + 2 H2O
+4 –2
+1 –2
oxidation
reduction
Chapt 4
Chemical reactions
Redox reaction
An oxidizing agent is a species that oxidizes
another species; it is itself reduced.
A reducing agent is a species that reduces another
species; it is itself oxidized.
Loss of 2 e- oxidation
reducing agent
Fe(s ) Cu 2 (aq )
Fe 2 (aq ) Cu(s )
oxidizing agent
Gain of 2 e- reduction
Chapt 4
Chemical reactions
23
Redox reaction
Activity Series - ranks the elements (neutral atom)
in order of their reducing ability of cation (Mn+ ) in
aqueous solution
1. Any element higher in the activity series will react with
the ion of any element(in sol’n) lower in the activity series
2. Position of Hydrogen- indicates which metal will react with aq.
acid H+ (aq) to produce H2 gas
3. Most reactive metals are at the top of the activity series (strong
reducing agent)
4. Least reactive metal - bottom of the activity series ( weak
reducing agent)
Chapt 4
Chemical reactions
Activity Series
Zn0 (s) + 2Ag+ (aq)
2Ag(s) + Zn2+(aq)
Chapt 4
Chemical reactions
Practice Problem
Assign oxidation states, determine the elements
oxidized and reduced, and determine the oxidizing
agent and reducing agent in the following reaction.
Sn4+ + Ca
F2 + S
Sn2+ + Ca2+
SF4
Chapt 4
Chemical reactions
24
Balancing Simple OxidationReduction Reactions
Oxidation half – one-half of the reaction in which oxidation occurs
Reduction half - one-half of the reaction in which reduction occurs
a balanced equation must have a charge balance as well as a
mass balance.
2
E.g.
Zn ( s )
Ag ( aq )
Zn
( aq )
Ag ( s )
The number of electrons lost in the oxidation half-reaction must
equal the number gained in the reduction half-reaction,
Ag ( aq )
Zn ( s )
Zn
2
e
( aq )
Chapt 4
Ag ( s )
2e
reduction half-reaction
oxidation half-reaction
Chemical reactions
Adding the two half-reactions together, the electrons cancel,
Zn(s) Zn2 (aq) 2e
2Ag (aq) 2e
2Ag(s)
Zn(s) 2Ag (aq)
oxidation half-reaction
reduction half-reaction
Zn 2 (aq) 2Ag(s)
the balanced oxidation-reduction reaction
Chapt 4
Chemical reactions
Redox reaction:Combustion
Reactions
Some Common Oxidation-Reduction Reactions
Most of the oxidation-reduction reactions fall into one of the
following simple categories:
Combustion Reactions
Combination Reactions
Decomposition Reactions
Displacement Reactions
Combustion reactions-Reactions in which O2(g) is a reactant
Combustion reactions release lots of energy
2 C8H18(g) + 25 O2(g)
Chapt 4
16 CO2(g) + 18 H2O(g)
Chemical reactions
25
Combination Reactions
A combination reaction is a reaction in which two substances
combine to form a third substance.
Combination reaction of sodium and chlorine
Decomposition Reaction: A decomposition reaction is a
reaction in which a single compound reacts to give two or more
substances.
Decomposition reaction of mercury(II) oxide
Displacement Reactions
A displacement reaction (also called a single- replacement
reaction) is a reaction in which an element reacts with a
compound, displacing an element from it.
– Displacement reaction of zinc and hydrochloric acid.
Chapt 4
Chemical reactions
Combustion Products
1.
3.
Write the equation for the complete combustion of
CH3OH(l).
Write the skeleton equation
CH3OH(l) + O2(g) CO2(g) + H2O(g)
Balance the equation
Chapt 4
Chemical reactions
Key Skills
Relating quantities in a chemical equation.
Determining limiting reactant, calculating theoretical and % yield
Calculating and using molarity as a conversion factor
Determining solution dilutions; using solution stoichiometry to find
volume and amounts
Predicting the solubility of a compound
Writing equations for precipitation reaction; molecular, ionic and net
ionic equations.
Writing equations for Acid-base reactions, Gas-evolution reactions
Assigning oxidation states and Identifying Redox reactions
Writing equations for combustion reactions
Chapt 4
Chemical reactions
26
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