Reaction Stoichiometry: Mole Method Calculations

Coefficients in balanced equations give the ratio by moles!!!

2 C4H10 + 13 O2 8 CO2 + 10 H2O

e.g. in the above reaction; 2 moles C4H10 react with 13 moles O2 to produce 8

moles CO2 and 10 moles H2O

Use these just like other conversion factors!

ProblemHow many moles of O2 are required to react with 0.500 moles of C4H10 according to the balanced equation?

0.500 mol C4H10 x = 3.25 mol O2 13 mol O2

2 mol C4H10

Ch. 4: Chemical Quantities and Aqueous Reactions

Example ProblemWhat 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 C4H10 = 58.12 g/mol

Step 1: “grams A moles A”

100 g C4H10 x = 1.721 mol C4H10

Step 2: “moles A moles B”

1.721 mol C4H10 x = 6.884 mol CO2

58.12 g C4H10

1 mol C4H10

2 mol C4H10

8 mol CO2

continued on next slide…

Example Problem, cont.Step 3: “moles B grams B”

6.884 mol CO2 x = 3.03 x 102 g CO2

Alternatively, the 3 steps can be combined into one unit analysis equation:

100 g C4H10 x x x =

3.03 x 102 g CO2

44.01 g CO2

1 mol CO2

1 mol C4H10

58.12 g C4H10

8 mol CO2

2 mol C4H10

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 ProblemIn 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 = 1.762 mol NH3

40.00 g O2 x = 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 = 1.762 mol NO

1 mol NH3

17.03 g NH3

1 mol O2

32.00 g O2

4 mol NO4 mol NH3

continued on next slide…

Example Problem, cont.Yield of NO based on O2:

1.250 mol O2 x = 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:

1.000 mol NO x = 30.01 g NO

4 mol NO5 mol O2

30.01 g NO1 mol 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 experimentallytheoretical yield -- amount of product predicted by balanced equationpercentage yield -- (actual yield/theoretical yield) x 100%

Example ProblemIn 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 ProblemThe 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 ProblemThe 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 Terminologysolution -- 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 concentrationsaturated 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 producte.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

ProblemWhat mass of NaCl is required to prepare 300 mL of 0.100 M

solution?1st--find moles of NaCl required:

300 mL x = 0.0300 mol NaCl

2nd--convert to grams of NaCl:

0.0300 mol NaCl x = 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.

0.100 mol NaCl

1000 mL solution

58.44 g NaCl

1 mole NaCl

Sample ProblemsCalculate 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 ProblemsCalculate 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 SolutionsConcentrated solution + H2O --> dilute solution

(moles solute)conc = (moles solute)dil

VcMc = VdMd

ProblemA 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!)

ProblemFor 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 = 0.100 mol CaCl2

2nd, find moles of AgNO3

(0.100 mol CaCl2) x = 0.200 mol AgNO3

0.400 mol CaCl21000 mL CaCl2 soln

2 mol AgNO3

1 mol CaCl2

Example Problem, cont.3rd, find volume of AgNO3 solution

(0.200 mole AgNO3) x = 800 mL AgNO3

solution

Part (b): mass of AgCl?

0.100 mol CaCl2 x = 0.200 mol AgCl

(0.200 mol AgCl) x = 28.6 g AgCl

WORK MORE PROBLEMS!!

1000 mL AgNO3 soln

0.250 mol AgNO3

2 mol AgCl

1 mol CaCl2

143 g AgCl

1 mol AgCl

Formation of a Solution

Formation of a solution of iodine molecules in ethyl alcohol. Ethyl alcohol is the solvent and iodine the solute.

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).

No Dissociation; I2 in H2O Dissociation; NaCl in H2O

Na+ and Cl- ions are stabilized when they are surrounded by H2O molecules.

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

A knowledge of these rules will allow you to predict a large number of precipitation reactions

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–, PO4

3– Li+, Na+, K+, NH4+

Sample QuestionsWrite the equations for the dissociation of the following

ionic compounds.a) KBr

b) Na2SO4

c) ZnCl2

Sample QuestionsWrite 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) + SO4

2–(aq)

c) ZnCl2

ZnCl2(s) --> Zn2+(aq) + 2 Cl–(aq)

Precipitation ReactionsWhen 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) + CO3

2–(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 CO3

2– are combined in aqueous solution, regardless of their sources.

Both atoms and charge must balance!

Strategy1. Dissociate the ionic compounds into their ions.2. Look at the reaction and see what the combinations of

ions could be. 3. Consult your memory or solubility charts to determine

the solubility of the product.4. Balance the equation(s).Example ProblemWill 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 PO4

3–(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

• 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

Summary of 3 Types of Balanced Chemical Equations, cont.

Sample QuestionsWrite 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 QuestionsWrite 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 + Base --> Salt + Watere.g. HNO3(aq) + NaOH(aq) --> NaNO3(aq) + H2O(l)

H2SO4(aq) + 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)

TitrationsTitration An unknown amount of one reactant is combined

exactly with a precisely measured volume of a standard solution of the other.

End-point When exactly stoichiometric amounts of two reactants have been combined.

Indicator Substance added to aid in detection of the endpoint (usually via a color change)

Example ProblemVinegar 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 x

= 0.007762 mol HC2H3O2

M = = 0.621 M acetic acid

0.504 mol NaOH

1000 mL NaOH soln

1 mole HC2H3O2

1 mole NaOH

0.007762 mol HC2H3O2

0.0125 L vinegar

Sample ProblemA 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 ProblemA 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

Reactant

Type

Reacts

with

Ion Exchange

Product

Decom-pose?

Gas

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)

Gas Evolution Reactions

Sample QuestionsWrite the molecular and net ionic equations for the

reactions below.

a) aluminum hydroxide + hydrochloric acid

b) calcium carbonate + hydrochloric acid

Sample QuestionsWrite 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 Terminologyredox 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.

OILRIGOILRIG

• O O Oxidation• II Is• LL Loss• RR Reduction• II Is• GG 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 States1. Oxidation state of a free element is zero.2. Oxidation state of a simple monoatomic ion is equal to

the charge on the ion.3. Sum of all the oxidation numbers must equal the

charge on the particle.4. In its compounds, H and Group 1A metals have an

oxidation number of +1. Group 2A metals have an oxidation number of +2.

5. In its compounds, fluorine has an oxidation state of -1.6. 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:

Ag2SClO3– ClO4

– Cr(NO3)3 H2O H2O2

Assigning Oxidation States1. Oxidation state of a free element is zero.2. Oxidation state of a simple monoatomic ion is equal to

the charge on the ion.3. Sum of all the oxidation numbers must equal the

charge on the particle.4. In its compounds, H and Group 1A metals have an

oxidation number of +1. Group 2A metals have an oxidation number of +2.

5. In its compounds, fluorine has an oxidation state of -1.6. 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:

Ag2SClO3– ClO4

– Cr(NO3)3 H2O H2O2

+1-2 +5 -2 +7 -2 +3 +5 -2+1 -2 +1 -1

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 ProblemIn 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 ProblemIn 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 Reactionse.g., N2(g) + O2(g) 2 NO(g)

4 Na(s) + O2(g) 2 Na2O(s)