BOOK PROBLEMS C 3
1. You should know the name & formula for all these ions, and draw Lewis structures for them!
a.
ClO- , hypochlorite ion. The active ingredient in Chlorox®.. Oxygen in cpds is –2 except in O2
(0) and peroxides (H2O2, -1)
Cl + -2 = -1 Cl = +1
b.
ClO2- , chlorite ion. Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
Cl + 2(-2) = -1 Cl = +3
c.
ClO3- , chlorate ion. Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
Cl + 3(-2) = -1 Cl = +5
d.
ClO4- , perchlorate ion. Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
Cl + 4(-2) = -1 Cl = +7
3. You should be able to name and give formulae for all the nitrogen oxides! These are all
nonmetal-nonmetal cpds, so you count the atoms in Greek. You should also be able to draw Lewis
structures for them.
a. N2O : Dinitrogen oxide, an anaesthetic gas, aka nitrous oxide. Oxygen in cpds is –2 except in O2
(0) and peroxides (H2O2, -1)
2(N) + -2 = 0 2N = +2, N = 1
b. NO: Nitrogen Oxide. Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
N + -2 = 0 N = 2
c. NO2: Nitrogen dioxide. This is NOT nitrous oxide!!!!
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
N + 2(-2) = 0 N = 4
d. NO3: Nitrogen trioxide
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
N + 3(-2) = 0 N = 6
e. N2O3: Dinitrogen trioxide
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
2(N) + 3(-2) = 0: 2N = 6, N = 3
f. N2O4: Dinitrogen tetroxide
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
2(N) + 4(-2) = 0: 2N = 8, N = 4
g. N2O5: Dinitrogen pentoxide
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
2(N) + 5(-2) = 0 2N = 10, N = 5
You should ALSO know that N2 is a DIATOMIC ELMENT with oxidation # 0. What are the other
diatomic elements?
7. You should be able to name and give formulae for these ions. You should also be able to draw
Lewis structures for them.
Formal charge is compared w/ oxidation # in Example 3.3 on p R3.8 in Munowitz. Read it over!
a. NO3-, Nitrate ion.
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
N + 3(-2) = -1 N = 5
To draw the Lewis structure, REM:
Decide which atoms are bonded. Decide which is the central atom.
For an NO3- ion, N is central.
1. Count all valence electrons.
There are a total of 24 valence electrons in NO3- (5 from the N, and 6 each from the 3 oxygens,, & 1
from the –1 charge).
2. Connect each atom by 1 bond; that is: Place two electrons in each bond.
3. Complete the octets of the peripheral atoms attached to the central atom by adding electrons in
pairs.
4. Place any remaining electrons on the central atom in pairs.
(already has 24, no remaining electrons in this example)
5. If the central atom does not have an octet, SHARE! Form double bonds. If necessary form triple
bonds.
6. If you can make more than 1 structure just by moving pairs of e- , then do it. The nitrate ion has 3
resonance contributors:
Remember: The above representation DOES NOT SHOW (though it may look like it) resonance
contributors flipping from one to another. Resonance structures are an indication that simple Lewis
structures don’t adequately model reality. Reality for the nitrate ion is NOT 2 single and 1 double
bond, but 3 identical bonds which have shorter bond length than a single bond, but not as short as a
double bond, and higher bond energy than a single bond, but not has high as a double bond.
GEOMETRY: N has 3 things around it. The geometry is trigonal planar, with 120o bond .
To compute formal charges, remember that elemental N has 5 valence e- (it’s Group V) and
elemental O has 6 valence e- (it’s Group VI). Formal charge compares the valences of N & O in the
ion with the valences in the atoms. Formal charges give you an idea of what parts of the ion are +
and what parts are -.
Atom
N
-O
=O
Valence
e(VE)
5
6
6
nonbonding e(NBE)
0
6 (3 lone pairs)
4 (2 lone pairs)
bonding e(BE)
8
2
4
Formal Charge
(VE – NBE –1/2BE)
+1
-1
0
0
+1 -1
-1
b. Formal charge is compared w/ oxidation # in Example 3.3 on p R3.8 in Munowitz. Read it over!
a. CO32-, carbonate ion.
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
C + 3(-2) = -2 C = +4
To draw the Lewis structure, REM:
Decide which atoms are bonded. Decide which is the central atom.
For a CO32- ion, C is central.
1. Count all valence electrons.
There are a total of 24 valence electrons in CO32- (4 from the C, and 6 each from the 3 oxygens,, & 2
from the –2 charge).
2. Connect each atom by 1 bond; that is: Place two electrons in each bond.
3. Complete the octets of the peripheral atoms attached to the central atom by adding electrons in
pairs.
4. Place any remaining electrons on the central atom in pairs.
(already has 24, no remaining electrons in this example)
5. If the central atom does not have an octet, SHARE! Form double bonds. If necessary form triple
bonds.
6. If you can make more than 1 structure just by moving pairs of e- , then do it. The carbonate ion has 3
resonance contributors:
-2
C
-2
-2
C
C
Remember: The above representation DOES NOT SHOW (though it may look like it) resonance
contributors flipping from one to another. Resonance structures are an indication that simple Lewis
structures don’t adequately model reality. Reality for the carbonate ion is NOT 2 single and 1 double
bond, but 3 identical bonds which have shorter bond length than a single bond, but not as short as a
double bond, and higher bond energy than a single bond, but not has high as a double bond.
GEOMETRY: The C has 3 things around it. The geometry is trigonal planar, with 120o bond .
To compute formal charges, remember that elemental C has 4 valence e- (it’s Group IV) and
elemental O has 6 valence e- (it’s Group VI). Formal charge compares the valences of C & O in the
ion with the valences in the atoms. Formal charges give you an idea of what parts of the ion are +
and what parts are -.
-2
C
Atom
Valence
e(VE)
4
6
6
C
-O
=O
-2
0
C
+1 -1
-1
nonbonding e(NBE)
0
6 (3 lone pairs)
4 (2 lone pairs)
bonding e(BE)
8
2
4
Formal Charge
(VE – NBE –1/2BE)
0
-1
0
c. SO2, Sulfur dioxide.
Oxygen in cpds is –2 except in O2 (0) and peroxides (H2O2, -1)
S + 2(-2) = 0; S = +4
To draw the Lewis structure, REM:
Decide which atoms are bonded. Decide which is the central atom.
For a SO2 molecule, S is central.
1. Count all valence electrons.
There are a total of 18 valence electrons in SO2 (6 from the S, and 6 each from the 2 oxygens)
2. Connect each atom by 1 bond; that is: Place two electrons in each bond.
3. Complete the octets of the peripheral atoms attached to the central atom by adding electrons in
pairs.
4. Place any remaining electrons on the central atom in pairs.
(There are 2 e- left over, a lone pair )
5. If the central atom does not have an octet, SHARE! Form double bonds. If necessary form triple
bonds.
6. If you can make more than 1 structure just by moving pairs of e- , then do it. Sulfur dioxide has 2
resonance contributors:
..
S
..
CS
Remember: The above representation DOES NOT SHOW (though it may look like it) resonance
contributors flipping from one to another. Resonance structures are an indication that simple Lewis
structures don’t adequately model reality. Reality for sulfur dioxide is NOT 1 single and 1 double
bond, but 2 identical bonds which have shorter bond length than a single bond, but not as short as a
double bond, and higher bond energy than a single bond, but not has high as a double bond.
GEOMETRY: The S has 3 things around it (2 O’s and a lone pair (-:) of e-). The geometry is trigonal
planar, with 120o bond  (or bent, if you pretend the lone pair isn’t there)
To compute formal charges, remember that elemental S has 6 valence e- (it’s Group VI) and
elemental O has 6 valence e- (it’s Group VI too). Formal charge compares the valences of S& O in the
ion with the valences in the atoms. Formal charges give you an idea of what parts of the molecule are
+ and what parts are -.
..0
S
-1
+1
Atom
0
Valence
e(VE)
6
6
6
S
-O
=O
nonbonding e(NBE)
2
6 (3 lone pairs)
4 (2 lone pairs)
bonding e(BE)
8
2
4
Formal Charge
(VE – NBE –1/2BE)
0
-1
0
..
S
Th This is another plausible Lewis structure/ resonance contributor, the validity of which
is supported by experimental data. S is a 3rd row element, big enough to cope with an expanded octet (10
valence e-) .
Atom
Valence
nonbonding e- bonding eFormal Charge
e
(NBE)
(BE)
(VE – NBE –1/2BE)
(VE)
S
6
2
8
0
=O
6
4 (2 lone pairs)
4
0
=O
6
4 (2 lone pairs)
4
0
8. You should be able to name and give the correct formula for the thiocyanate ion. You should also
be able to draw Lewis structures for it.
Formal charge is compared w/ oxidation # in Example 3.3 on p R3.8 in Munowitz. Read it over!
a. NCSTo draw the Lewis structure, REM:
Decide which atoms are bonded. Decide which is the central atom.
For an NCS- ion, C is central.
7. Count all valence electrons.
There are a total of 16 valence electrons in NCS- (5 from the N, 4 from C, 6 form S, & 1 from the –1
charge).
8. Connect each atom by 1 bond; that is: Place two electrons in each bond.
9. Complete the octets of the peripheral atoms attached to the central atom by adding electrons in
pairs.
10. Place any remaining electrons on the central atom in pairs.
(already has 16, no remaining electrons in this example)
11. If the central atom does not have an octet, SHARE! Form double bonds. If necessary form triple
bonds.
12. If you can make more than 1 structure just by moving pairs of e- , then do it. The nitrate ion has 3
resonance contributors:
..
..
N=C=S
..
..
Atom
N
C
S
_
Valence
e(VE)
5
4
6
This is a linear molecule. Why? The central C has 2 things bonded to it.
nonbonding e(NBE)
2
0
4 (2 lone pairs)
bonding e(BE)
4
8
4
Formal Charge
(VE – NBE –1/2BE)
-1
0
0
You get formal charges for the other two structures the same way:
_
..
..
C=S=N
..
..
Atom
C
S
N
..
..
S=N=C
..
..
Atom
S
N
C
Valence
e(VE)
4
6
5
nonbonding e(NBE)
4
0
4
bonding e(BE)
4
8
4
Formal Charge
(VE – NBE –1/2BE)
-2
+2
-1
_
Valence
e(VE)
6
5
4
nonbonding e(NBE)
4
0
4
bonding e(BE)
4
8
4
Formal Charge
(VE – NBE –1/2BE)
0
+1
-2
The most plausible structure for thiocyanate is N=C=S- , in which the formal charges are smallest
and N, which is the most electronegative atom in the group, has the – charge density.
9. . Balancing chemical equations is an exercise which operates on one of the fundamental laws of the
Universe: “Matter and energy cannot be created or destroyed, only interconverted.” Also know as the
First Law of Accounting – “The Books Must Balance”, and TANSTAAFL – “There Ain’t No Such
Thing As A Free Lunch”.
REM: An ACID, according to the BrØnsted-Lowry definition, DONATES a PROTON (H+).
e.g. HCl + H2O  H3O+ + Cl- . HCl is donating an H+ to water.
A BASE, according to the BrØnsted-Lowry definition, ACCEPTS a PROTON (H+)
e.g. HCl + H2O  H3O+ + Cl- . H2O is accepting an H+ from hydrochloric acid.
Another definition of acids and bases is more generally applicable:
An ACID, according to the LEWIS definition, ACCEPTS A PAIR OF ELECTRONS.
..
e.g. H :Cl: + : ö –H  H :
..
H
H
ö +-- H
+ Cl- .
H+ is accepting a pair of electrons from water.
A BASE, according to the Lewis definition, DONATES a PAIR OF ELECTRONS.
In the above example, H2O is donating a pair of electrons to H+.
YOU SHOULD BE ABLE TO NAME ALL OF THE COMPOUNDS LISTED IN THE EXAMPLES!
a)
acid
base
salt
HNO3 + NaOH  HOH (H2O) + NaNO3 (Molecular equation)
Nitric acid + sodium hydroxide  water + sodium nitrate.
Nitric acid is a strong acid. It dissociates completely into ions in water:
HNO3(l) + nH2O(l)  H3O+(aq) + NO3-(aq)
Sodium hydroxide is a strong base. It dissociates completely into ions in water:
NaOH(s) + nH2O(l)  Na+(aq) + OH-(aq)
Water stays as water in water.
Sodium nitrate is soluble in water, It dissociates completely into ions in water:
NaNO3(s) + nH2O(l)  Na+(aq) + NO3-(aq)
NOTE: YOU CAN FIND OUT IF SOMETHING IS SOLUBLE IN WATER BY CHECKING IN
APPENDIX C-17 IN MUNOWITZ.
The subscript (aq) indicates a solvated, in this case hydrated, ion; that is, an ion completely
surrounded by a cage of water molecules.
The above being the case, it is better to write the molecular equation shown above as an ionic
equation:
H3O+(aq) + NO3-(aq) + Na+(aq) + OH-(aq)  2H2O(l) + Na+(aq) + NO3-(aq) (ionic equation)
Notice that Na+(aq) + NO3-(aq) appear on both sides of the equation, so they cancel. They are
spectator ions.
H3O+(aq) + NO3-(aq) + Na+(aq) + OH-(aq)  2H2O(l) + Na+(aq) + NO3-(aq)
So, finally, you get the NET IONIC EQUATION:
H3O+(aq) + OH-(aq)  2H2O(l)
Acid
base
b. acid
base
salt
CH3COOH + KOH 
H2O + CH3COO-K+ (Molecular equation)
Acetic acid + potassium hydroxide  water + potassium acetate.
Acetic acid is a weak acid. It does not dissociate completely into ions in water:
Potassium hydroxide is a strong base. It dissociates completely into ions in water:
KOH(s) + nH2O(l)  K+(aq) + OH-(aq)
Water stays as water in water.
Potassium acetate is soluble in water, It dissociates completely into ions in water:
CH3COO-K+ (s) + nH2O(l)  K+(aq) + CH3COO-(aq)
(acetate ion)
NOTE: YOU CAN FIND OUT IF SOMETHING IS SOLUBLE IN WATER BY CHECKING IN
APPENDIX C-17 IN MUNOWITZ.
The subscript (aq) indicates a solvated, in this case hydrated, ion; that is, an ion completely
surrounded by a cage of water molecules.
The above being the case, it is better to write the molecular equation shown above as an ionic
equation:
CH3COOH(aq) + K+(aq) + OH-(aq)  H2O(l) + K+(aq) + CH3COO-(aq) (ionic equation)
Notice that K+(aq) appears on both sides of the equation, so they cancel. It is a spectator ion.
CH3COOH(aq) + K+(aq) + OH-(aq)  H2O(l) + K+(aq) + CH3COO-(aq)
So, finally, you get the NET IONIC EQUATION:
CH3COOH(aq) + OH-(aq)  H2O(l) + CH3COO-(aq)
Acid
base
salt
c. H3O+(aq) + OH-(aq)  2H2O(l)
Acid
base
A generic acid/base reaction. H3O+ donates a proton (H+) to OH-, a base. The base (:OH-) donates
a pair of electrons to the H+.
d. H2SO4(l) + NaOH(s)  H2O + NaHSO4(s)
Acid
base
salt
Sulfuric acid + sodium hydroxide  water + sodium hydrogen sulfate (sodium bisulfate).
Sulfuric acid is a strong acid. It dissociates completely into ions in water:
H2SO4(l) + nH2O(l)  H3O+(aq) + HSO4- aq)
Sodium hydroxide is a strong base. It dissociates completely into ions in water:
NaOH(s) + nH2O(l)  Na+(aq) + OH-(aq)
Water stays as water in water.
Sodium hydrogen sulfate is soluble in water, It dissociates completely into ions in water:
NaHSO4(s) + nH2O(l)  Na+(aq) + HSO4- (aq)
NOTE: YOU CAN FIND OUT IF SOMETHING IS SOLUBLE IN WATER BY CHECKING IN
APPENDIX C-17 IN MUNOWITZ.
The subscript (aq) indicates a solvated, in this case hydrated, ion; that is, an ion completely
surrounded by a cage of water molecules.
The above being the case, it is better to write the molecular equation shown above as an ionic
equation:
H3O+(aq) + HSO4-(aq) + Na+(aq) + OH-(aq)  2H2O(l) + Na+(aq) + HSO4- (aq) (ionic equation)
Notice that Na+(aq) + HSO4- (aq) appear on both sides of the equation, so they cancel. They are
spectator ions.
H3O+(aq) + HSO4- (aq) + Na+(aq) + OH-(aq)  2H2O(l) + Na+(aq) + HSO4- (aq)
So, finally, you get the NET IONIC EQUATION:
H3O+(aq) + OH-(aq)  2H2O(l)
Acid
base
10. Balancing chemical equations is an exercise which operates on one of the fundamental laws of the
Universe: “Matter and energy cannot be created or destroyed, only interconverted.” Also know as the First
Law of Accounting – “The Books Must Balance”, and TANSTAAFL – “There Ain’t No Such Thing As A
Free Lunch”.
Acids donate protons (Bronsted-Lowry) and accept a pair of electrons (Lewis).
Once again, you should be able to NAME everything in this problem!
Acid
base
salt
a. 2HCOOH + Ca(OH)2  Ca(HCO2)2 + 2H2O
formic acid + calcium hydroxide  calcium formate + water.
Formic acid is a weak acid, calcium hydroxide is a strong base, water is water and calcium formate is
soluble:
2HCOOH(aq) + Ca2+(aq) + 2OH- (aq)  Ca2+(aq) + 2HCOO-(aq) + 2H2O(l)
Calcium ion is a apectator ion.
2HCOOH(aq) + 2OH- (aq)  2HCOO-(aq) + 2H2O(l)
acid
base
salt
b. This problem is incorrect. Ammonium hydroxide (NH4OH) is not a good representation of the
weak base ammonia (:NH3). Ammonia has a pair of e- to donate. It should read:
acid base
salt
HCl + NH3  H2O + NH4Cl
Hydrochloric acid + ammonia  water + ammonium chloride.
Hydrochloric acid is a strong acid. Ammonia is a weak base. Water is water. Ammonium chloride
is soluble in water.
H3O+ (aq) + Cl-(aq) + :NH3(aq)  NH4+(aq) + Cl-(aq) + H2O(l)
Chloride ion is a spectator ion.
H3O+ (aq) + :NH3(aq)  NH4+(aq) + H2O(l)
c H2C2O4 +
2OH- 
C2O4= + 2H2O
Oxalic acid + hydroxide ion  oxalate ion + water
Acid + any strong base (KOH, NaOH, etc)  salt and water. The base cation (K+, Na+ or
whatever) is a spectator ion and cancels.
10 d. H3O+ + CH3NH2  H2O + CH3NH3+
hydronium ion + methyl amine  water + methyl ammonium ion.
Any strong acid (HCl, HNO3 etc) + methyl amine ( a weak base. The N has a lone pair of e- to
donate. Hydronium ion has an H+ (proton) which can accept them. The acid anion (Cl-, NO3- or
whatever) is a spectator ion and cancels.
11. You should be able to name all the ions in the following problems!
a. Fe3+ (Iron (III) ion, ferric ion) can accept a pair of e-. It is a Lewis ACID (But not a Bronsted acid).
:CN:- (Cyanide ion) can doate a pair of e- or accept a proton. It is both a Lewis and a Bronsted
BASE.
b. :NH3 (Ammonia, a weak base) candonate a pair of electrons, or accept a proton. It is a Lewis and
B-L BASE.
BF3, (Boron trifluoride) can accept a pair of e-. It is a Lewis ACID (But not a Bronsted acid).
c. H3O+ (hydronium ion is both a B-L and a Lewis ACID.
:OH- (Hydroxide ion) is both a B-L and a Lewis BASE.
d. Al3+ (Aluminum ion (NOT aluminum (III) ion) is a Lewis acid, but not a B-L acid.
:OH- (Hydroxide ion) is both a B-L and a Lewis BASE.
e. CH3COO:- (acetate ion) is both a B-L and a Lewis BASE.
CH3COOH (acetic acid) is both a B-L and a Lewis ACID.
12. You should be able to name everything in the following problems!
a. CH3COO- + H2O  CH3COOH + OHAcetate ion + water  acetic acid + hydroxide ion.
Base
acid
acid
base
b. NH3 + H2O  NH4+ + OHammonia + water  ammonium ion + hydroxide ion
base
acid
acid
base
c.
NH3
+ HCl 
NH4Cl
ammonia + hydrochloric acid  ammonium chloride
base
+ acid
salt (in water, ammonium chloride would dissociate into
NH4+ (ammonium ion, an acid) and chloride ion (Cl-, a
weak base)
d. CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
Calcium carbonate (marble) is insoluble in water, hydrochloric acid is a strong acid, calcium chloride
is soluble, and H2CO3, carbonic acid, breaks down to carbon dioxide and water.
CaCO3(s) + 2H3O+ + 2 Cl-  Ca2+(aq) + 2Cl-(aq) + H2CO3  H2O(l) + CO2(g)
Chloride ion is a spectator ion, and cancels:
CaCO3(s) + 2H3O+  Ca2+(aq) + H2CO3  H2O(l) + CO2(g) NET IONIC EQN.
14. REM: Oxidation is the loss of electrons. Something that gets oxidized gets a more positive
oxidation #. A reducing agent gets oxidized. Oxygen has an oxidation number of 0 (as O 2), and –2
in a compound (except for peroxides). Hydrogen has an oxidation # of 0 (as H2) and +1 when
combined with nonmetals. The Hydride ion (H:-) is found in combination with metals and has an
oxidation # of –1. You should be able to name all the chemical species in this problem.
a. 2IO3- + 5Cu + 12H3O+  I2 + 5Cu2+ + 18H2O
2 iodate ions + 5 copper metal + 12 hydronium ions  iodine (one of the diatomic elements) + 5
copper(II) (cupric) ions + 18 waters.
IO3
+5 +3(-2) = -1
Cu  Cu2+
0
+2
I2
0
Iodine goes from +5 to 0. It gets reduced. IO3- is the oxidizing agent.
Copper goes from 0 to +2. It gets oxidized. Cu is the reducing agent.
b. Cd + NiO2 + 2H2O  Cd(OH)2 + Ni(OH)2
cadmium metal + nickel(IV)oxide + 2 waters  cadmium hydroxide + nickel (II) hydroxide.
This is what goes on in a NiCad battery.
Cd 
Cd(OH)2
0
+2 + 2(-2) + 2(+1) = 0
agent.
(charge on hydroxide ion is –1)
Cadmium goes from 0 to +2. It gets oxidized. It is the reducing
NiO2
 Ni(OH)2
+4 + 2(-2) = 0
+2 + 2(-2) + 2(+1) = 0
oxidizing agent.
Nickel goes from +4 to +2 It gets reduced. It is the
14c. Pb + PbO2 + 4H3O+ + 2SO4=  2PbSO4 + 6H2O
lead metal + lead (IV) oxide + 4 hydronium ions + 2 sulfate ions  2lead(II)sulfate + 6 waters.
This is what goes on in your car battery.
Pb 
0
PbSO4
+2 + 6 + 4(-2) = 0
(charge on sulfate ion is –2)
Lead goes from 0 to +2. It gets oxidized. It is the reducing agent.
PbO2
 2PbSO4 (charge on sulfate ion is –2)
+4 + 2(-2) = 0
+2 + 6 + 4(-2) = 0
Lead goes from +4 to +2 It gets reduced. It is the oxidizing
agent.
d. 6I- + 2Al3+  3I2 + 2Al
6 iodide ions + 2 aluminum ions  3 iodines + 2 aluminum metal.
I- 
-1
Al3+
+3
I2
0
Iodine goes from -1 to 0. It gets oxidized. It is the reducing agent.
 Al
0
Aluminum goes from +3 to 0 It gets reduced. It is the oxidizing agent.
16. . Balancing chemical equations is an exercise which operates on one of the fundamental laws of the
Universe: “Matter and energy cannot be created or destroyed, only interconverted.” Also know as the First
Law of Accounting – “The Books Must Balance”, and TANSTAAFL – “There Ain’t No Such Thing As A Free
Lunch”. These are all Oxidation/reduction reactions (of which more later!) and they are all balanced using the
same protocol (tho’ you can do them by trial & error more easily at this point. In fact, I recommend that
approach here, even though I have done it the more general way.)
a. __Br- + __F2  __Br2 + __F- Bromide ion + fluoride  bromine + fluoride ion.
1. Break into half reactions:
Br-  Br2
F2  F2. Balance everything but O & H:
2Br-  Br2
F2  2FNo O & H to worry about.
Bromine & Fluorine are diatomic elements.
3. Balance charge using e--:
2Br-  Br2 + 2e2e- + F2  2F-
4. Multiply each eq’n so e- balance: DONE
2Br-  Br2 + 2e2e- + F2  2F- 2 moles of e- are transferred. Bromide ion is oxidized. Fluorine is
reduced.
5. Add ‘em up:
2Br- + F2  Br2 + 2F- (e- cancel)
b. __Cr + __Cu2+  __Cr3+ + __Cu
Let’s see:
(Chromium metal + cupric (copper (II) ion  chromium (III) ion + copper metal)
1. Break into half reactions:
Cr  Cr3+
Cu2+  Cu
2.
3.
4.
5.
Balance everything but O & H: DONE
Balance O with H2O: DONE
Balance H with H+: DONE
Balance charge using e--:
Cr  Cr3+ + 3e2e- + Cu2+  Cu
6. Multiply each eq’n so e- balance:
2Cr  2Cr3+ + 6e6e- + 3Cu2+  3Cu 6 moles of e- are transferred. Chromium is oxidized, Copper (II) is reduced.
7.
Add ‘em up:
2Cr + 3Cu2+  2Cr3+ + 3Cu
c. __Fe3+ + __I-  __Fe2+ + __I2
Let’s see:
(Ferric (Iron (III) ion + iodide  ferrous (Iron (II) ion + iodine)
1. Break into half reactions:
Fe3+  Fe2+
I-  I2
2. Balance everything but O & H:
Fe3+  Fe2+
2 I-  I2
3. Balance O with H2O: DONE
4. Balance H with H+: DONE
5. Balance charge using e--:
1e- + Fe3+  Fe2+
2I-  I2 + 2e6. Multiply each eq’n so e- balance:
2e- + 2Fe3+  2Fe2+
2I-  I2 + 2e- 2 moles of e- are transferred. Iodide is oxidized, ferric ion is reduced.
7.
Add ‘em up:
2Fe3+ + 2I-  2Fe2+ + I2
20. Free radicals have unpaired electrons and are highly reactive.
a. Draw the Lewis dot structure for CH3 and you will find it has an unpaired e-:
CH3. + CH3.  CH3-CH3
2 methyl radicals make ethane.
b. Draw the Lewis dot structure for CH3O and you will find it has an unpaired e-:
CH3O. + HBr  CH3OH + Br. . Bromine is a free radical.
a methoxy radical + hydrogen bromide make methanol + a bromine radical.
21. You should be able to name all the ions.
a. MgSO4(s) + nH2O(l)  Mg2+(aq) + SO4=(aq) . Magnesium sulfate + water  magnesium
ion + sulfate ion.
b. HCl(g) + nH2O(l)  H3O+(aq) + Cl-(aq). Hydrochloric acid + water  hydronium ion +
chloride ion.
c. Ca(OH)2(s) + nH2O(l)  Ca2+(aq) + 2OH-(aq) . Calcium hydroxide + water  calcium ion
+ 2 hydroxide ions
d. CH3COONa(s) + nH2O(l)  Na+(aq) + CH3COO-(aq) . Sodium acetate + water  sodium
ion + acetate ion.
e. This is incorrect. Ammonium hydroxide is not a good representation for ammonia.
It should be:
NH3(g) + nH2O(l)  NH4+ (aq) + OH- (aq) . Ammonia + water  ammonium ion +
hydroxide ion. The double arrow indicates that ammonia is a weak base, and does not
react with water 100%. Most of the ammonia will remain NH3(aq).
f. CaCO3(s) + nH2O(l)  Ca2+(aq) + CO3=(aq) . Calcium carbonate + water   calcium
ion + carbonate ion. The double arrow indicates that calcium carbonate is not very soluble
in water.
22. You should be able to name everything in this problem!
a. CH3COONa(aq) + H2O(l)  NaOH(aq) + CH3COOHaq) sodium acetate + water  sodium hydroxide + acetic acid.
Sodium acetate + sodium hydroxide are soluble. Acetic acid is a weak acid.
CH3COO-(aq)- + Na+ + H2O  Na+(aq) + OH-(aq) + CH3COOH IONIC EQN sodium ion is a spectator.
CH3COO-(aq)- + H2O  + OH-(aq) + CH3COOH NET IONIC EQN
b. Na2SO4(aq) + Pb(NO3)2(aq)  PbSO4(s) + 2NaNO3(aq)
sodium sulfate + lead(II)nitrate  lead(II)sulfate + 2sodium nitrate.
Sodium sulfate, lead nitrate and sodium nitrate are soluble. Plumbous (Lead(II)) sulfate is insoluble.
2Na+(aq) + SO4=(aq) + Pb2+(aq) + 2NO3-(aq)  PbSO4(s) + 2Na+(aq) + 2NO3-(aq) Ionic eq’n. Sodium, ion &
nitrate ion are spectators.
SO4=(aq) + Pb2+(aq)  PbSO4(s) Net Ionic eq’n
c. Al2(SO4)3(aq) + 6NaOH(aq)  2Al(OH)3(s) + 3Na2SO4(aq)
aluminum sulfate + 6 sodium hydroxide --> 2aluminum
hydroxide + 3 sodium sulfate.
Aluminum sulfate, sodium hydroxide and sodium sulfate are soluble.
2Al3+(aq) + 3SO4=(aq) + 6Na+(aq) + 6OH-(aq)  2Al(OH)3(s) + 6Na+(aq) + 3SO4=(aq) Ionic eq’n. sodium ion
& sulfate ion are spectators.
2Al3+(aq) + 6OH-(aq)  2Al(OH)3(s) Net ionic eq’n.
26. When in doubt, convert to moles.
Data: MW zinc chloride = 136.30 g/mol
(5.00 g ZnCl2) (1 mol ZnCl2) (1000 mL) = 0.183 mol ZnCl2 = 0.183 M & we are entitled to 3 sig fig.
(200.00 mL ) (136.30 g
) (1 L
)
1L
27. When in doubt, convert to moles.
REM: 1 mole of anything is 6.022 x 1023 particles.
a. 500.0 mL (1 L
. ) (0.3750 mol ZnCl2) (136.30 g ) = 25.56 g ZnCl2 &we are entitled to 4 sig fig.
(1000 mL) ( 1 L
) (mol ZnCl2)
b. ZnCl2  Zn2+ + 2Cl- : 1 ZnCl2 has 1 Zn2+
500.0 mL (1 L
. ) (0.3750 mol ZnCl2) ( 1 mol Zn2+ ) (6.022 x 1023 particles) = 1.129 x 1023 Zn2+ ions.
(1000 mL) ( 1 L
) (1 mol ZnCl2) ( 1 mol
) & we get 4 sig fig.
c. ZnCl2  Zn2+ + 2Cl- : 1 ZnCl2 has 2 Cl500.0 mL (1 L
. ) (0.3750 mol ZnCl2) ( 2 mol Cl) (6.022 x 1023 particles) = 2.258 x 1023 Cl- ions.
(1000 mL) ( 1 L
) (1 mol ZnCl2) ( 1 mol
) & we get 4 sig fig.
28. When in doubt, convert to moles. REM: M is the abbreviation for moles solute/Liter of solution.
a. 100 mL (1 L
. ) (1.00 mol ZnCl2) (1 L
.)
(1000 mL) ( 1 L
) (0.500 moles ZnCl2)
= 0.2 L & we only get 1 sig fig!
100 mL of the 1 M ZnCl2  200
mL.
This piece converts mLs to
Liters
This tells you how many moles of
ZnCl2 you have in the original 1 M
sol’n
This tells you how many liters you
have to put those moles of ZnCl2 into
to get a 0.500 M sol’n.
28 b. . 100 mL (1 L
. ) (1.00 mol ZnCl2) (1 L
.)
(1000 mL) ( 1 L
) (0.250 moles ZnCl2)
c. . 100 mL (1 L
. ) (1.00 mol ZnCl2) (1 L
.)
(1000 mL) ( 1 L
) (0.100 moles ZnCl2)
= 0.4 L & we only get 1 sig fig!
(400 mL)
= 1.0 L & we only get 1 sig fig!
(1000 mL)
29. When in doubt, convert to moles. And start with a balanced equation!
ZnCl2 + 2NaOH  Zn(OH)2 + 2NaCl
First, how many moles of ZnCl2 do we kick off with?
300.0 mL (1 L
. ) (0.600 mol ZnCl2) = 0.180 moles ZnCl2 & we get 3 sig fig.
(1000 mL) ( 1 L
)
700.0 mL (1 L
. ) (0.250 mol NaOH) = 0.175 moles NaOH & we get 3 sig fig.
(1000 mL) ( 1 L
)
Now, which is the limiting reagent?
0.180 moles ZnCl2 (2 mol NaOH ) = 0.360 moles NaOH needed to react with all the ZnCl2.
(1 mol ZnCl2 )
Do we have enough NaOH? No, we’ve only got 0.175 moles.
NaOH is the limiting reagent.
OR you can do it this way:
0.175 moles NaOH (1 mol ZnCl2 ) = 0.0875 moles ZnCl2 needed to react with all the NaOH.
(2 mol NaOH )
Do we have enough ZnCl2? Yes, we’ve got 0.180 moles.
ZnCl2 is the excess reagent.
Now figure out how much zinc hydroxide you’ll make:
DATA: MW Zn(OH)2 = 99.40 g/mol
Working with the limiting reagent, NaOH, because that governs how much zinc hydroxide you can
make:
0.175 moles NaOH ( 1 mol Zn(OH)2 ) (99.40 g
.) = 8.70 g Zn(OH)2 made
( 2 moles NaOH ) (1 mole Zn(OH)2)
You will have also 0.175 moles NaOH ( 2 mol NaCl
) (58.44 g
( 2 moles NaOH ) (1 mole NaCl
.) = 10.2 g NaCl made
)
29 (Continued) The 10.2 g of NaCl will be present as spectator Na+ and Cl- ions in 300.0 mL + 700.0
mL = 1000.0 mL = 1.000 L of solution.
Their molar concentration will be:
10.2 g NaCl (1 mol NaCl ) (1 Mol Na+ ) = 0.175 M Na+
1.000 L
( 58.44 g
) (1 mol NaCl)
10.2 g NaCl (1 mol NaCl ) (1 Mol Cl- ) = 0.175 M Cl1.000 L
( 58.44 g
) (1 mol NaCl)
Remember also that you started with 0.180 moles of the xs reagent ZnCl2 and only used 0.0875 moles
ZnCl2. you will have 0.180 – 0.0875 = 0.0925 moles of ZnCl2 left over.
It will be present as 0.0925 M Zn2+ and 0.185 M Cl- .
If you could isolate the ZnCl2 away from the sodium chloride, water and zinc hydroxide, you’d have:
0.0925 moles ZnCl2 ( 136.30 g
) = 12.6 g of ZnCl2 left.
(mole ZnCl2 )

A word to the wise: I don’t do these stoichiometry problems for fun (Tho’
chemists do have an odd sense of fun) . I don’t even do them for my health.
They ARE IMPORTANT. They are the kinds of questions you’ll be doing
routinely for the rest of your scientific lives, and you should practice them a
lot. Get good at them! Practice!
30. When in doubt, convert to moles. And start with a balanced equation!
ZnCl2 + 2NaOH  Zn(OH)2 + 2NaCl
Data: MW Zn(OH)2 = 99.40 g/mol
IF we want 5.00 g of Zn(OH)2:
5.00 g Zn(OH)2 (1 mol Zn(OH)2) ( 2 mols NaOH ) ( 1 L NaOH
) =
(99.40 g
) (1 mol Zn(OH)2) ( 0.250 mol NaOH)
0.402 L of
0.250 M NaOH
needed.
30. (Continued) Is there enough ZnCl2?
5.00 g Zn(OH)2 (1 mol Zn(OH)2) ( 1mols ZnCl2 ) = 0.0503 mol ZnCl2 needed.
(99.40 g
) (1 mol Zn(OH)2)
We’ve got:
300.0 mL (1 L
. ) (0.600 mol ZnCl2) = 0.180 moles ZnCl2 & THAT’S PLENTY!
(1000 mL) ( 1 L
)
b.
0.180 mol ZnCl2 start
- 0.0503 moles ZnCl2 used
0.130 moles ZnCl2 left over.
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