CHEMISTRY 104
CHAPTER 5: CHEMICAL REACTIONS
Homework problems: 8, 14, 18, 26, 34, 38a-c, 42, 54, 60
Outline
I. Balanced Equations
A. Reactants and Products
B. Balanced by atoms AND charge AND mass
C. Symbols of state (s, l, g, aq)
D. Lowest possible whole number coefficients
1. Cannot change formulas in eqn!
Start by picking a molecule. Put a one down as its coefficient. Then work back and
forth, balancing each element in turn. If you end up with a fractional coefficient,
multiply the whole thing through to get a whole number. It doesn’t matter which
molecule you pick to put a one as the coefficient. Usually, you pick the most
complicated molecule to make things easier. But it doesn’t matter which one you start
with, you will get the same answer.
Examples:
Balance the reaction below
Mg
+
HCl 
MgCl2
H2
Start by putting a one in front of hydrogen
Mg
+
HCl 
MgCl2
1 H2
one hydrogen molecule on the product side means two hydrogen atoms on the product
side, so we need two hydrogen atoms on the reactant side
Mg
+
2 HCl 
MgCl2
1 H2
Two chlorides on the reactant side means we need two chlorides on the product side, so
one magnesium chloride.
Mg
+
2 HCl 
1 MgCl2
1 H2
Finally, one Mg on the product side means one Mg on the reactant side
1 Mg
+
2 HCl 
1 MgCl2
1 H2
But what if I hadn’t picked hydrogen to put a “1” down as the coefficient? Would I get
the same answer? Let us try it.
The biggest, ugliest molecule is the magnesium chloride. Start by putting a one there.
Mg
+
HCl 
1 MgCl2
H2
One magnesium chloride means one Mg on the product side, so one Mg on reactant side
1 Mg
+
HCl 
1 MgCl2
H2
Now, we have gone down a dead end, so we need to go back to an earlier molecule that
already has a coefficient in front of it. You CANNOT put another random “1” down now
to continue. You HAVE to go back to something that is already got a coefficient. One
1
magnesium chloride also means two chlorides on the product side, so we need two on the
reactant side
1 Mg
+
2 HCl 
1 MgCl2
H2
Finally, two hydrogen atoms on the reactant side means 2 hydrogen atoms on the product
side. Two hydrogen atoms make one hydrogen molecule
1 Mg
+
2 HCl 
1 MgCl2
1 H2
Okay, but what if I accidentally pick the “wrong” molecule to start and put a one in front
of it? Let us try that. Start with a one in front of hydrogen monochloride.
Mg
+
1 HCl 
MgCl2
H2
One HCl means we have one hydrogen atom on the reactant side. That means we can
only have one hydrogen atom on the product side. So, how many hydrogen molecules
can we make with one hydrogen atom? We can only make half a hydrogen molecule.
Take the number of hydrogen atoms (1) and divide by the number of hydrogen atoms in
the molecule (2) to get one half (1/2). Just like when you go to the store and apples are
two dollars per pound, and you only have one dollar, so you can only get half a pound of
apples. It isn’t a whole number, but we can fix that later. For now, it is the correct ratio
of HCl to H2.
Mg
+
1 HCl 
MgCl2
½ H2
Now, were at another cul-de-sac, so we have to go back to another molecule that already
has a coefficient in front of it. We absolutely CANNOT pick another molecule and put a
one in front of that. If you do that, the equation will probably not balance. So, one HCl
means one chloride. One chloride atom means half a magnesium chloride
Mg
+
1 HCl 
½ MgCl2
½ H2
Finally, half a magnesium chloride means half a Mg on the product side, so we need half
a Mg on the reactant side.
½ Mg
+
1 HCl 
½ MgCl2
½ H2
Okay, so now we have the ratio, but it is not in whole numbers. Multiply both sides of
the equation by two.
1 Mg
+
2 HCl 
1 MgCl2
1 H2
Regardless of where you start, you will get the same answer. But you do need to start
somewhere.
Balance these equations:
Na
Al
CH4
C4H10
H3PO4
2 Na
4 Al
1 CH4
+
+
+
+
+
H2O
O2
O2
O2
NaOH





+
+
+
2 H2O 
3 O2 
2 O2 
NaOH +
Al2O3
H2O +
H2O +
Na3PO4
H2
CO2
CO2
+
2 NaOH
2 Al2O3
2 H2O +
H2O
+
1 H2
1 CO2
2
2 C4H10
1 H3PO4
+
+
13 O2 
3 NaOH
10 H2O
+

1 Na3PO4
8 CO2
+
3 H2O
II. Oxidation Numbers (Ox #)
A. Rules (p 141)
After applying rules, do each/total box like we did earlier to figure out the charge on a
transition metal ion, only this box is for oxidation numbers.
Example: determine the oxidation numbers for each atom in sodium phosphate (Na3PO4)
We know that the ox # on oxygen is –2 (ox # for oxygen in a non-peroxide cmpd) and for
sodium, it is +1 (ox # for simple ions is same as the charge), but to get the phosphorus’ #,
we use the other two and the fact that the total of all oxidation numbers is equal to the net
charge, which is zero in this case
Na3PO4
Each +1 ? -2
Total +3 ? -8
The total of all phosphorus ox #’s must be five for this to balance out (3 + n – 8= 0),
so n= 5. Since there’s only one P, it’s ox # is five
B. Redox (Reduction/Oxidation) Reactions
1. Can’t have oxidation without reduction
Something has to lose electrons in order for something else to gain, so at least two ox #’s
have to change, one goes up and another goes down
2. Oxidizing agents and reducing agents
What is oxidized is the reducing agent, what is reduced is the oxidizing agent. You can
look at it from the perspective of what happened to it (it was oxidized), or from the
perspective of what it did to the other atom (it was the reducing agent since it caused the
other molecule to be reduced).
3. LEO says GER or OIL RIG
Loss Electrons is Oxidation, Gain Electrons is Oxidation. Since electrons are negatively
charged, losing electrons (oxidation) makes your oxidation number go up, while gaining
electrons makes your oxidation number go down.
Also, Oxidation Is Loss, Reduction Is Gain (of electrons)
C. In polyatomic ions, Ox. # is not same as ionic charge
Examples:
1 CH4
ea. –4 +1
tot. -4 +4
+
2 O2
0

2 H2O +
+1 –2
+2 -2
1 CO2
+4 –2
+4 –4
Ox. # for hydrogen unchanged, so it was neither oxidized nor reduced. The Ox # of
oxygen went from 0 to –2, so it was reduced (it was also the oxidizing agent) while the
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ox # of carbon went from –4 to +4 so it was oxidized (you could also say it was the
reducing agent.
You don’t even need a balanced equation to figure out oxidation numbers, because ox #’s
are entirely within the molecule
Ox #: each
H2O2

+1 -1 (peroxide)
O2
0
+
H2O
+1 –2
This is a funny case where the oxygen in the hydrogen peroxide is both oxidized (in the
case of the diatomic oxygen) AND reduced (the case of the O in water).
Not every reaction is a redox reaction. In the reaction of hydrobromic acid and calcium
hydroxide, all oxidation numbers are the same before and after the rxn
HBr
+
Ca(OH)2

HOH +
CaBr2
III. Reactions (by type – differ from text to text, so don’t memorize these)
A. Decomposition (start with one thing, end with several)
A  B
+
C
+
…
H2O2

O2
+
H2O
B. Combination (opposite of decomposition: start with many, end with one)
A + B
+
….
C
4 Al
+
3 O2 
2 Al2O3
C. Replacement (also called displacement)
1. Single
A + BC

B
+
AC
CuCl2 +
Zn

ZnCl2 +
Cu
2. Double
AB + CD

AD
+
CB
2 HBr +
Ca(OH)2

2 HOH +
CaBr2
IV. Ionic Equations (Eqns) (p149)
A. Total or Full Ionic Eqn
We have been writing molecular equations, were everything is written as a molecule. It
turns out that when ionic cmpds are dissolved in water, they come apart into separate
cations (positive ions) and anions (negative ions) which can go their separate ways. A
total ionic equation illustrates this by breaking apart everything that is BOTH aqueous
AND ionic.
B. Net Ionic Eqn
Any ion or molecule that does not change in state (s, l, g, aq) does not react and is
removed from the total ionic to give the net ionic eqn
C. Spectator Ions (ions not changing in the reaction, remain same charge and state)
Write balanced total and net ionic eqations for :
Molecular:
CaCl2 (aq) + Na2CO3 (aq) 
CaCO3 (s) + 2 NaCl (aq)
2+
+
Total ionic: Ca (aq) + 2 Cl (aq) + 2 Na (aq)+ CO32- (aq) CaCO3 (s) + 2 Na+ (aq) +
4
2 Cl- (aq)
Net ionic: Ca2+ (aq) + CO32- (aq)  CaCO3 (s)
Spectator ions are sodium and chloride in this reaction.
V. Energy
A. Exothermic (heat as product)
B. Endothermic (heat as reactant)
SI Units of energy are Joules, but a lot of times calorie is used (1 cal =4.184 J). This is
different from the Nutritional Calorie (spelled with a capital “C,” as in Calorie, to
differentiate) in that a nutritional Calorie is actually a 1000 smaller calories (or one
kilocal)
VI. Moles in Eqns
A. Converting
1. Grams of X  moles of X  moles of Y  grams of Y
Balanced equations are given in mole ratios, not mass ratios, so if we want to compare
how much of Y we need to react with X, we need to translate grams into moles, and then
compare amounts.
Example: How many grams of oxygen to rust 2250 g iron?
First, you need a balanced eqn
4 Fe
+
3 O2 
2 Fe2O3
Then, convert 2250 g Fe moles Fe  moles oxygen  g oxygen
The 1st and 3rd conversions (grams into moles or moles to grams) is done using mw. The
middle conversion is done using the balanced eqn coefficients
2250 g Fe x (1 mole/55.847 g) x (3 oxygen molecules/ 4 iron atoms) x (32.00g/mole) =
967 g oxygen gas is needed
B. Limiting Reagent What you run out of first
C. Yields
1. Theoretical
2. Actual
3. % yield = actual/theoretical
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Chapter Objectives
Knowledge
Memorize the rules for determining oxidation number (p141)
Know the terms exothermic, endothermic, reducing and oxidizing agents, oxidation
number, percent and theoretical yields, limiting reagent
Comprehension
Be able to identify a reaction as redox (change in Ox #) or not
Identify oxidizing and reducing agents in a redox reaction
Identify spectator ions in a Full Ionic Eqn
Identify a rxn as exo- or endothermic
Identify a limiting reagent
Application
Balance a chemical eqn
Determine Oxidation numbers using rules
Write Net Ionic Eqns from Total Ionic Eqns
Analysis
Convert freely between Grams of X  moles of X  moles of Y  grams of Y
Calculate theoretical and percent yield
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