Download Chapter 1: Fundamental Concepts

Ch. 4: Chemical Quantities and Aqueous Reactions
Reaction Stoichiometry: Mole Method Calculations
Coefficients in balanced equations give the ratio by moles!!!
2 C4H10 + 13 O2
e.g. in the above reaction;
8 CO2 + 10 H2O
2 moles C4H10 react with 13 moles O2 to produce 8 moles CO2
and 10 moles H2O
Use these just like other conversion factors!
Problem
How many moles of O2 are required to react with 0.500 moles
of C4H10 according to the balanced equation?
0.500 mol C4H10 x
13 mol O2
2 mol C4H10
= 3.25 mol O2
Example Problem
What mass of CO2 could be produced from the combustion of
100 g (3 sig fig) of butane (C4H10)?
Always convert given quantities (e.g. grams) to moles!!!
grams A
moles A
moles B
grams B
will need formula masses to convert between grams and moles:
CO2 = 44.01 g/mol
Step 1: “grams A
100 g C4H10 x
Step 2: “moles A
C4H10 = 58.12 g/mol
moles A”
1 mol C4H10
58.12 g C4H10
= 1.721 mol C4H10
moles B”
1.721 mol C4H10 x
8 mol CO2
= 6.884 mol CO2
2 mol C4H10
continued on next slide…
Example Problem, cont.
Step 3: “moles B
grams B”
6.884 mol CO2 x
44.01 g CO2
1 mol CO2
= 3.03 x 102 g CO2
Alternatively, the 3 steps can be combined into one unit analysis
equation:
100 g C4H10 x
1 mol C4H10
58.12 g C4H10
x
8 mol CO2
2 mol C4H10
3.03 x 102 g CO2
x
44.01 g CO2
1 mol CO2
=
Limiting Reactant Calculations
• In practice, reagents are often combined in a ratio that is
different from that in the balanced chemical equation. One
of the reactants will be completely consumed and some of
the other will remain unreacted.
• The limiting reactant is the one that is completely
consumed. It determines the maximum amount (yield) of
the products.
• Whenever quantities of both reactants are given, the
limiting reactant must be determined!!!
Example Problem
In a commercial process, nitric oxide (NO) is produced:
4 NH3 + 5 O2
4 NO + 6 H2O
What mass (in grams) of NO can be made from the reaction of
30.00 g NH3 and 40.00 g O2?
1st, find moles of reactants:
30.00 g NH3 x
40.00 g O2 x
1 mol NH3
17.03 g NH3
1 mol O2
32.00 g O2
= 1.762 mol NH3
= 1.250 mol O2
2nd, calculate amount of product based on each reactant,
separately:
Yield of NO based on NH3:
1.762 mol NH3 x
4 mol NO
4 mol NH3
= 1.762 mol NO
continued on next slide…
Example Problem, cont.
Yield of NO based on O2:
1.250 mol O2 x
4 mol NO
5 mol O2
= 1.000 mol NO
Therefore, O2 is the limiting reactant! (excess of NH3 exists)
A maximum of 1.000 moles of NO can be produced.
3rd, calculate yield of product based on limiting reactant:
30.01 g NO
1.000 mol NO x 1 mol NO = 30.01 g NO
Theoretical and Percentage Yield
• In actual experiments, the amount of a product that is
actually obtained is always somewhat less than that
predicted by the stoichiometry of the balanced chemical
equations.
• This is due to competing reactions and/or mechanical
losses in isolation of the product.
actual yield -- amount of product obtained experimentally
theoretical yield -- amount of product predicted by balanced
equation
percentage yield -- (actual yield/theoretical yield) x 100%
Example Problem
In the previous experiment for the production of NO from 30.00
g NH3 and 40.00 g O2, the chemist obtained 25.50 g NO.
What is the percentage yield of this reaction?
4 NH3 + 5 O2
4 NO + 6 H2O
Theoretical yield = 30.01 g NO (based on limiting reactant)
Actual yield = 25.50 g (given in problem)
% yield = (25.50 g NO/30.01 g NO) x 100% = 85.0%
(Note: the % yield can never be more than 100%!)
Sample Problem
The simplest borane known as “diborane-6,” B2H6, can be prepared
from boron trichloride and lithium hydride according to the following
balanced equation.
2 BCl3 + 6 LiH
B2H6 + 6 LiCl
In one experiment, 20.0 g of BCl3 was allowed to react with 5.00 g of
LiH. After isolation and purification of the reaction product, 2.05 g of
pure B2H6 was actually obtained. Calculate the percentage yield of
this reaction.
Formula masses (g/mol): BCl3 = 117.2
LiH = 7.95
B2H6 = 27.7
LiCl = 42.4
Sample Problem
The simplest borane known as “diborane-6,” B2H6, can be prepared
from boron trichloride and lithium hydride according to the following
balanced equation.
2 BCl3 + 6 LiH
B2H6 + 6 LiCl
In one experiment, 20.0 g of BCl3 was allowed to react with 5.00 g of
LiH. After isolation and purification of the reaction product, 2.05 g of
pure B2H6 was actually obtained. Calculate the percentage yield of
this reaction.
Formula masses (g/mol): BCl3 = 117.2
LiH = 7.95
B2H6 = 27.7
LiCl = 42.4
Answer: 86.9%
Reactions in Aqueous Solution
Solution Terminology
solution -- homogeneous (uniform) mixture consisting of:
solvent - the bulk medium, e.g. H2O,
solute(s) - the dissolved substance(s), e.g. NaCl
concentration -- measure of relative solute/solvent ratio
--concentrated vs. dilute
standard solution -- accurately known concentration
saturated solution -- contains maximum amount of solute
solubility -- concentration of a saturated solution, e.g.:
solubility of NaCl is about 36 g NaCl/100 g H2O (“soluble”)
solubility of CuS is about 10-5 g CuS/100 g H2O (“insoluble”)
precipitate -- an “insoluble” reaction product
e.g. a precipitation reaction where the precipitate is AgCl:
NaCl(aq) + AgNO3(aq) --> AgCl(s) + NaNO3(aq)
Solution Concentration
• Molar Concentration
Molarity (M) = moles solute/liter of solution
Units: moles/L or moles/1000 mL (just a conversion factor!)
e.g. a “0.10 M” NaCl solution contains 0.10 mole NaCl per liter of solution
Problem
What mass of NaCl is required to prepare 300 mL of 0.100 M solution?
1st--find moles of NaCl required:
300 mL x
0.100 mol NaCl
= 0.0300 mol NaCl
1000 mL solution
2nd--convert to grams of NaCl:
0.0300 mol NaCl x
58.44 g NaCl
1 mole NaCl
= 1.75 g NaCl
Prepare this solution by weighing 1.75 g NaCl, dissolving in some H2O
(about 250 mL), and then diluting to the 300 mL mark.
Sample Problems
Calculate the molarity of a solution made by dissolving 23.4 g
of sodium sulfate in enough water to form 125 mL of
solution.
The label of a stock bottle of aqueous ammonia indicates that
the solution is 28% ammonia by mass and has a density of
0.898 g/mL. Calculate the molarity of the solution.
Sample Problems
Calculate the molarity of a solution made by dissolving 23.4 g
of sodium sulfate in enough water to form 125 mL of
solution.
Answer: 1.32 M Na2SO4
The label of a stock bottle of aqueous ammonia indicates that
the solution is 28% ammonia by mass and has a density of
0.898 g/mL. Calculate the molarity of the solution.
Answer: 15 M NH3
Dilution of Concentrated Solutions
Concentrated solution + H2O --> dilute solution
(moles solute)conc = (moles solute)dil
VcMc = VdMd
Problem
A 5.00 M NaCl “stock” solution is available. How would you
prepare 300 mL of a 0.100 M NaCl “standard” solution?
Vc x (5.00 M) = (300 mL) x (0.100 M)
Vc = (300 mL) x (0.100 M)/(5.00 M) = 6.00 mL
Measure out 6.00 mL of the 5.00 M “stock” solution, then
add H2O to a total volume of 300 mL.
Stoichiometry Problems -- Reactions in Solution
Start with a Balanced Equation and Use the Mole Method
(Molarity is just a conversion factor!)
Problem
For the following reaction,
2 AgNO3(aq) + CaCl2(aq) --> 2 AgCl(s) + Ca(NO3)2(aq)
a) What volume of 0.250 M AgNO3 is required to react completely with 250 mL
of 0.400 M CaCl2?
b) What mass of AgCl should be produced?
Part (a): volume of AgNO3?
1st, find moles of CaCl2
(250 mL CaCl2 soln) x
0.400 mol CaCl2
= 0.100 mol CaCl2
1000 mL CaCl2 soln
2nd, find moles of AgNO3
(0.100 mol CaCl2) x
2 mol AgNO3
1 mol CaCl2
= 0.200 mol AgNO3
Example Problem, cont.
3rd, find volume of AgNO3 solution
(0.200 mole AgNO3) x
1000 mL AgNO3 soln
0.250 mol AgNO3
= 800 mL AgNO3 solution
Part (b): mass of AgCl?
0.100 mol CaCl2 x
(0.200 mol AgCl) x
2 mol AgCl
1 mol CaCl2
143 g AgCl
1 mol AgCl
= 0.200 mol AgCl
= 28.6 g AgCl
WORK MORE PROBLEMS!!
Formation of a Solution
No Dissociation; I2 in H2O
Formation of a solution of iodine
molecules in ethyl alcohol. Ethyl
alcohol is the solvent and iodine
the solute.
Dissociation; NaCl in H2O
Na+ and Cl- ions are stabilized
when they are surrounded by
H2O molecules.
Water is probably the most common solvent. When a substance is
dissolved into water it is said to be aqueous. (from the Latin, aqua, water).
Electrolytes
Dissociation Reactions of Salts (in aqueous solution)
Electrolytes are solutes that produce ions in solution via
dissociation (their solutions can conduct electricity), e.g.
–
–
NaCl(s) --> Na+(aq) + Cl–(aq)
(NH4)2SO4(s) --> 2 NH4+(aq) + SO42–(aq)
These are “strong” electrolytes -100% ionized (e.g. “soluble” salts,
strong acids)
Some substances are “weak”
electrolytes -- partially ionized (< 100%)
(e.g. weak acids)
or, “non-electrolytes” -not ionized at all (e.g. sugar, iodine)
Solubility Rules for Ionic Compounds in Water I
• Soluble Compounds
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+
Solubility Rules for Ionic Compounds in Water II
• Insoluble Compounds
Compounds Containing
the Following Ions are
Generally Insoluble
OH–
S2–
CO32–, PO43–
Exceptions
(when combined with ions on the left
the compound is soluble or slightly
soluble)
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
A knowledge of these rules will allow you to predict a large
number of precipitation reactions
Sample Questions
Write the equations for the dissociation of the following ionic
compounds.
a) KBr
b) Na2SO4
c) ZnCl2
Sample Questions
Write the equations for the dissociation of the following ionic
compounds.
a) KBr
KBr(s) --> K+(aq) + Br–(aq)
b) Na2SO4
Na2SO4(s) --> 2 Na+(aq) + SO42–(aq)
c) ZnCl2
ZnCl2(s) --> Zn2+(aq) + 2 Cl–(aq)
Precipitation Reactions
When two dissociated ionic substances combine, an insoluble
precipitate may form. e.g. the reaction between lead(II)
nitrate and potassium iodide, below, forms insoluble lead(II)
iodide.
+
Ionic Reactions in Aqueous Solution
• Equations for Ionic Reactions
Metathesis Reaction (also called “double displacement”)
Ions from two different reactants simply trade partners, e.g.:
Na2CO3(aq) + Ba(NO3)2(aq) --> BaCO3(s) + 2 NaNO3(aq)
This was written as a molecular equation in which all reactants
and products are shown as complete, neutral chemical formulas.
It could also have been written as a complete ionic equation in
which all soluble ionic compounds are split up into their ions, e.g.;
2 Na+(aq) + CO32–(aq) + Ba2+(aq) + 2 NO3–(aq) -->
BaCO3(s) + 2 Na+(aq) + 2 NO3–(aq)
Equations for Ionic Rxns, cont.
2 Na+(aq) + CO32–(aq) + Ba2+(aq) + 2 NO3–(aq) -->
BaCO3(s) + 2 Na+(aq) + 2 NO3–(aq)
Here, the Na+ and NO3– ions are called “spectator ions” because they
appear unchanged on both sides of the equation.
The spectator ions do not participate in the chemically important part
of the reaction -- the precipitation of BaCO3.
The essential chemical process can be written without the spectator
ions in the
net ionic equation, e.g.:
Ba2+(aq) + CO32–(aq) --> BaCO3(s)
The net ionic equation shows that, in general, a precipitate of BaCO3
will form whenever the ions Ba2+ and CO32– are combined in
aqueous solution, regardless of their sources.
Both atoms and charge must balance!
Strategy
1.
2.
3.
4.
Dissociate the ionic compounds into their ions.
Look at the reaction and see what the combinations of
ions could be.
Consult your memory or solubility charts to determine the
solubility of the product.
Balance the equation(s).
Example Problem
Will a precipitate form when aqueous solutions of CaCl2 and
K3PO4 are mixed? Write balanced molecular, ionic, and
net ionic equations for the reaction.
Step 1: Write the possible ions: Ca2+, Cl–, K+, PO43–
Step 2: Look at the combinations; KCl, Ca3(PO4)2
Step 3: You know KCl is soluble, Ca3(PO4)2 is not, so a
precipitate will form.
Step 4: Balance the equation:
3 CaCl2 + 2 K3PO4 --> Ca3(PO4)2 + 6 KCl
Example Problem, cont.
Molecular equation is:
3 CaCl2(aq) + 2 K3PO4(aq) --> Ca3(PO4)2(s) + 6 KCl(aq)
The ionic equation:
3 Ca2+(aq) + 6 Cl–(aq) + 6 K+(aq) + 2 PO43–(aq) -->
Ca3(PO4)2(s) + 6 K+(aq) + 6 Cl–(aq)
Ions that do not participate in the reaction are called spectator
ions.
Net ionic equation is:
3 Ca2+(aq) + 2 PO43–(aq) --> Ca3(PO4)2(s)
Summary of 3 Types of Balanced Chemical Equations
• Molecular Equation
– Shows all compounds with complete, neutral molecular
formulas
– Useful in planning experiments and stoichiometry calculations
• Ionic Equation (complete)
– All strong electrolytes are shown in their dissociated, ionic
forms
– Insoluble substances and weak electrolytes are shown in their
molecular form
– “spectator ions” are included
– Useful for showing all details of what is happening in the
reaction
Summary of 3 Types of Balanced Chemical Equations, cont.
• Net Ionic Equation
– “spectator ions” are omitted
– Only the essential chemical process is shown, i.e. formation of
a:
• Solid precipitate
• Gaseous product, or
• Weak electrolyte (e.g. water)
– Useful for generalizing the reaction -- same important product
can often be formed from different sets of reactants
• When will a precipitate form?
– KNOW THE SOLUBILITY RULES -- TABLE 4.1
Sample Questions
Write balanced molecular equations for the following
reactions.
a) AgNO3 + NaCl -->
b) BaCl2 + Na2SO4 -->
c) Pb(NO3)2 + NaC2H3O2 -->
d) NaOH + nickel(II) nitrate -->
e) AgNO3 + Na2S -->
Sample Questions
Write balanced molecular equations for the following
reactions.
a) AgNO3 + NaCl -->
AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)
b) BaCl2 + Na2SO4 -->
BaCl2(aq) + Na2SO4(aq) --> BaSO4(s) + 2 NaCl(aq)
c) Pb(NO3)2 + NaC2H3O2 -->
Pb(NO3)2(aq) + 2 NaC2H3O2(aq) --> Pb(C2H3O2)2(aq) + 2 NaNO3(aq)
d) NaOH + nickel(II) nitrate -->
2 NaOH(aq) + Ni(NO3)2(aq) --> Ni(OH)2(s) + 2 NaNO3(aq)
e) AgNO3 + Na2S -->
2 AgNO3(aq) + Na2S(aq) --> 2 NaNO3(aq) + Ag2S(s)
Arrhenius Acid-base Concept
Acid = H+ supplier
e.g. HNO3, HCl, H2SO4, etc
In water, HNO3 forms H+(aq) + NO3–(aq)
Base = OH– supplier
e.g. NaOH, Mg(OH)2, etc.
In water, NaOH forms Na+(aq) + OH–(aq)
• Acid-Base Neutralization Reactions
Acid +
e.g. HNO3(aq) +
H2SO4(aq) +
Base -->
Salt +
Water
NaOH(aq) --> NaNO3(aq) + H2O(l)
2 KOH(aq) --> K2SO4(aq) + 2 H2O(l)
Sample Problem
• Write a balanced chemical equation for the complete
neutralization reaction of barium hydroxide with sulfuric
acid. Circle the species in your equation that is the “salt.”
Sample Problem
• Write a balanced chemical equation for the complete
neutralization reaction of barium hydroxide with sulfuric
acid. Circle the species in your equation that is the “salt.”
Ba(OH)2(s) + H2SO4(aq)  2 H2O(l) + BaSO4(s)
Titrations
Titration
End-point
Indicator
An unknown amount of one reactant is combined
exactly with a precisely measured volume of a
standard solution of the other.
When exactly stoichiometric amounts of two
reactants have been combined.
Substance added to aid in detection of the endpoint
(usually via a color change)
Example Problem
Vinegar is an aqueous solution of acetic acid, HC2H3O2. A 12.5
mL sample of vinegar was titrated with a 0.504 M solution of
NaOH. The titration required 15.40 mL of the base solution
in order to reach the endpoint. What is the molar
concentration of HC2H3O2 in vinegar?
NaOH(aq) + HC2H3O2(aq) --> NaC2H3O2(aq) + H2O(l)
(15.40 mL NaOH) x
0.504 mol NaOH
1000 mL NaOH soln
x
1 mole HC2H3O2
1 mole NaOH
= 0.007762 mol HC2H3O2
M=
0.007762 mol HC2H3O2
0.0125 L vinegar
= 0.621 M acetic acid
Sample Problem
A new fluorocarbon material recently prepared at TCU is
believed to be a diprotic acid (call it H2A for short). A 0.860
g sample of this unknown acid was titrated (reaction below)
with a 0.1025 M solution of NaOH. In the titration, 38.50 mL
of the NaOH solution was required to reach the endpoint.
Calcuate the molecular mass of the new acid.
H2A + 2 NaOH --> Na2A + 2 H2O
Sample Problem
A new fluorocarbon material recently prepared at TCU is
believed to be a diprotic acid (call it H2A for short). A 0.860
g sample of this unknown acid was titrated (reaction below)
with a 0.1025 M solution of NaOH. In the titration, 38.50 mL
of the NaOH solution was required to reach the endpoint.
Calcuate the molecular mass of the new acid.
H2A + 2 NaOH --> Na2A + 2 H2O
Answer: 436 g/mol
Gas Evolution Reactions
Reactant
Type
Reacts
with
Ion
Exchange
Product
Decom
Gas
-pose? Formed
Example
MnS,
MHS
acid
H2S
no
H2S
K2S(aq) + 2HCl(aq) 
2KCl(aq) + H2S(g)
MnCO3,
MHCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2HCl(aq) 
2KCl(aq) + CO2(g) + H2O(l)
MnSO3
MHSO3
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)
Sample Questions
Write the molecular and net ionic equations for the reactions
below.
a) aluminum hydroxide + hydrochloric acid
b) calcium carbonate + hydrochloric acid
Sample Questions
Write the molecular and net ionic equations for the reactions
below.
a) aluminum hydroxide + hydrochloric acid
Al(OH)3(s) + 3 HCl(g) --> AlCl3(aq) + 3 H2O(l)
b) calcium carbonate + hydrochloric acid
CaCO3(s) + 2 HCl(g) --> CaCl2(aq) + CO2(g) + H2O(l)
Oxidation - Reduction Reactions
Redox Terminology
redox reaction -- electron transfer process
e.g.,
2 Na + Cl2 --> 2 NaCl
Related terms:
oxidizing agent = the substance that gets reduced (Cl2)
reducing agent = the substance that gets oxidized (Na)
Oxidation and reduction always occur together so that there is no
net loss or gain of electrons overall.
OILRIG
•
•
•
•
•
•
O
I
L
R
I
G
Oxidation
Is
Loss
Reduction
Is
Gain
Oxidation Numbers (Oxidation States)
Oxidation number
a “charge” that is assigned to an atom to
aid in balancing redox reactions
The oxidation state is the atomic number (no. of positive
charges) minus the number of electrons that it is said to
have.
Oxidation: the oxidation state increases.
Reduction: the oxidation state decreases.
Rules for assigning oxidation numbers -- see page 164
Learn the rules by working examples!
Assigning Oxidation States
1.
2.
3.
4.
5.
6.
Oxidation state of a free element is zero.
Oxidation state of a simple monoatomic ion is equal to the
charge on the ion.
Sum of all the oxidation numbers must equal the charge
on the particle.
In its compounds, H and Group 1A metals have an
oxidation number of +1. Group 2A metals have an
oxidation number of +2.
In its compounds, fluorine has an oxidation state of -1.
In its compounds, oxygen has an oxidation state of -2.
If there is a conflict between two rules, apply the rule with the
lower number and ignore the conflicting rule.
e.g. assign all oxidation numbers in:
Ag2S
ClO3– ClO4– Cr(NO3)3
H 2O
H2O2
Assigning Oxidation States
1.
2.
Oxidation state of a free element is zero.
Oxidation state of a simple monoatomic ion is equal to the
charge on the ion.
Sum of all the oxidation numbers must equal the charge
on the particle.
In its compounds, H and Group 1A metals have an
oxidation number of +1. Group 2A metals have an
oxidation number of +2.
In its compounds, fluorine has an oxidation state of -1.
In its compounds, oxygen has an oxidation state of -2.
3.
4.
5.
6.
If there is a conflict between two rules, apply the rule with the
lower number and ignore the conflicting rule.
e.g. assign all oxidation numbers in:
+1
-2
Ag2S
+5 -2
+7 -2
+3
+5 -2
ClO3– ClO4– Cr(NO3)3
+1 -2
+1 -1
H 2O
H2O2
Sample Problems
--Note that fractional values of oxidation numbers are allowed.
More examples:
NH3
PCl5
FeS
Na2O
Mg2TiO4
V(C2H3O2)3
HPO42–
K2Cr2O7
Sample Problems
--Note that fractional values of oxidation numbers are allowed.
More examples:
NH3
N -3, H +1
PCl5
P +5, Cl -1
FeS
Fe +2, S -2
Na2O
Na +1, O -2
Mg2TiO4
Mg +2, Ti +4, O -2
V(C2H3O2)3 V +3, C 0, H +1, O -2
HPO42–
H +1, P +5, O -2
K2Cr2O7
K +1, Cr +6, O -2
Redox Reactions
• The transfer of electrons between species, meaning that
one species is oxidized and one is reduced. The two
processes will always occur together.
• When has a redox reaction occurred?
– If there is a change in the oxidation state of any element in the
reaction, a redox reaction has happened.
– Remember that if something is oxidized, something else must
be reduced! (and vice-versa)
More examples:
2 Ca(s) + O2(g) --> 2 CaO(s)
2 Al(s) + 3 Cl2(g) --> 2 AlCl3(s)
Example Problems
• When H2O2 (hydrogen peroxide) is used as an antiseptic it
oxidizes bacteria. What product might it form, O2 or H2O?
• Answer: In the starting material O is in the -1 oxidation
state. If it oxidizes bacteria, it must get reduced, which
means its oxidation number goes down. O2 has an oxidation
number of 0, while the O in H2O has an oxidation number of
-2. Therefore H2O is the product of reduction of hydrogen
peroxide.
In the following reactions, state which compounds are oxidized
and which are reduced.
• 2 Al(s) + 3 I2(s) --> 2 AlI3(s)
• 2 PbS(s) + 3 O2(g) --> 2 PbO(s) + 2 SO2(g)
• 2 KCl(aq) + MnO2(s) + 2 H2SO4(aq) --> K2SO4(aq) + MnSO4(aq) +
Cl2(g) + 2 H2O(l)
• Answer: Oxidized: Al, S, Cl, and Reduced: I, O, Mn.
Sample Problem
In the following redox reaction,
14 HNO3(aq) + 4 K2C2O4(aq) + K2Cr2O7(aq) -->
8 CO2(g) + 2 Cr(NO3)2(aq) + 10 KNO3(aq) + 7 H2O(l)
A) what is the substance oxidized?
B) what is the oxidizing agent?
C) what are the oxidation numbers of each atom in Cr(NO3)2?
Sample Problem
In the following redox reaction,
14 HNO3(aq) + 4 K2C2O4(aq) + K2Cr2O7(aq) -->
8 CO2(g) + 2 Cr(NO3)2(aq) + 10 KNO3(aq) + 7 H2O(l)
A) what is the substance oxidized?
B) what is the oxidizing agent?
C) what are the oxidation numbers of each atom in Cr(NO3)2?
Answer:
a) K2C2O4 is oxidized
b) The oxidizing agent is K2Cr2O7
c) Cr is +2, N is +5, O is -2
Combustion Reactions
¤ Combustion of Hydrocarbons (produces CO2 and H2O)
e.g., 2 C8H18(l) + 25 O2(g) --> 16 CO2(g) + 18 H2O(l)
2 C2H5OH(l) + 6 O2(g) --> 4 CO2(g) + 6 H2O(l)
¤ Other Combustion Reactions
e.g., N2(g) + O2(g)  2 NO(g)
4 Na(s) + O2(g)  2 Na2O(s)