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CHEMISTRY 104
CHAPTER 6: CHEMICAL REACTIONS
Homework problems: 37, 38, 40, 41, 43, 46, 49, 52, 54, 56, 58, 63, 64, 66, 68, 70, 72,
77, 80, 82, 83, 95, 97, 104, 105, 106
Outline
I.
Terms
A. Reactions
1. reactant(s)  product(s)
B. Molecular mass revisited
1. also called molecular weight, formula mass, formula weight
2. same concept as atomic mass, but for molecules instead of atoms
C. Mole
1. Avagadro’s number = 6.02214199… x 1023
2. just a number, like a dozen or a gross or a quarted
3. a very large number, but exactly what you need to go from amu
to grams
1.5 moles of CCl4
how many molecules of CCl4 in 1.5 moles of CCl4?
1.5 moles x (6.02 x 1023 molecules/mole) = 9.0 x 1023 molecules
How many atoms of C in 1.5 moles of CCl4?
1.5 moles x (6.02 x 1023 molecules/mole) x 1 C/ CCl4 = 9.0 x 1023 atoms C
How many of Cl in 1.5 moles of CCl4?
1.5 moles x (6.02 x 1023 molecules/mole) x 4 Cl/ CCl4 = 3.6 x 1024 atoms C
How many grams of CCl4 in 1.5 moles of CCl4?
1.5 moles x 153.8 g/mole = 230g
How many molecules of oxygen in 14.6 g of O2?
14.6 g x (1 mole/32.0g) x (6.02 x 1023 molecules/mole) =
2.75 x 1023 molecules
II.
Chemical Equations
A. Reactants and Products
B. Balanced by atoms AND charge AND mass
1. Coefficients
2. implied “1” if nothing written
a. like you to write it anyway for now
3. lowest whole number ratio
C. How to balance
1. method on p137 or…
2. another way
a. find biggest, ugliest molecule
b. put a “1” down as its coefficient
c. work your way back and forth from reactant to
product to get ratio with ugliest molecule
d. when all done, multiply through (if necessary) to
get whole number ratio
4. Cannot change formulas in eqn to balance it!
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
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
2 C4H10
1 H3PO4
H2O
O2
O2
O2
NaOH





+
+
+
+
+
2 H2O 
3 O2 
2 O2 
13 O2 
3 NaOH
NaOH +
Al2O3
H2O +
H2O +
Na3PO4
H2
CO2
CO2
+
H2O
2 NaOH
+
1 H2
2 Al2O3
2 H2O +
1 CO2
10 H2O
+
8 CO2

1 Na3PO4
+
3 H2O
D. Symbols of state (s, l, g, aq)
E. Stoichiometry
1. balanced eqn coefficients = moles (or molecules), NOT units mass (g)
2. grams A  moles of A  moles B  grams B
a. THIS is how you compare mass
b. This is stoichiometry
3. limiting reagent
a. what you run out of first
4. percent yield
a. actual yield
b. theoretical yield
c. % yield = actual yield/theoretical yield
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) =
96.7 g oxygen gas is needed
Example 2: You have 1.20 g of sodium carbonate and 0.365 g of hydrogen chloride
(hydrochloric acid). The reaction is written below.
HCl (aq)
+
Na2CO3 → H2O (l) + CO2 (g) + NaCl (aq)
How many grams of water are produced?
First, balance the eqn:
2 HCl (aq)
+
1 Na2CO3 → 1 H2O (l)
+ 1 CO2 (g)
+ 2 NaCl (aq)
Next, find the limiting reactant:
1.20 g Na2CO3 x (1 mole/106 g) x (1 H2O/1 Na2CO3) x (18.0 g/mole) = 0.204 g H2O
0.365 g HCl x (1 mole/36.5 g) x (1 H2O/2 HCl) x (18.0 g/mole) = 0.0900 g H2O
So, HCl is the limiting reagent and 0.0900g of water is produced
How much sodium carbonate is leftover?
The amount of Na2CO3 leftover is the amount that is not used. Figure out how much
is used by the limiting reagent, then subtract that from the original amount
original Na2CO3 - reacted Na2CO3 = leftover Na2CO3
0.365 g HCl x (1 mole/36.5 g) x (1 Na2CO3/ 2 HCl) x (106 g/mole)= 0.53 g Na2CO3 used
1.20 g
0.53 g
=
0.67 g Na2CO3 leftover
If the experiment is run and the actual yield was 0.0871 g, what is the percent yield?
% = actual/theoretical
= 0.0871 g/ 0.0900g = 97.1%
III.
Reactions
A. Aqueous solutions (stuff is dissolved in water)
1. what happens when ionic stuff dissolved in water
2. polyatomic ions do NOT separate into individual atoms
3. Table of solubilities
a. how to tell if a reaction occurs using table
4. Equations
a. molecular equation
(written as molecules: what we have been doing)
b. total ionic equation (break apart into ions whatever
is both ionic and aqueous. This takes into account that ions come apart in water)
c. net ionic equation
(remove spectator ions (see below) from reactant and product sides)
d. 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) +
2 Cl- (aq)
Net ionic: Ca2+ (aq) + CO32- (aq)  CaCO3 (s)
Spectator ions are sodium and chloride in this reaction.
B. Redox reactions
1. what is oxidation and reduction
a. loss and gain of electrons
b. oxidation number (O.N.) rules
i. any element by itself, O.N. = 0
ex. H2: H = 0, Fe: Fe = 0
ii. any elemental (simple) ion, O.N. = charge
on ion
ex. Br-: Br = -1, Fe2+: Fe = 0
iii. for H in a molecule, O.N. = +1
ex. SH2: H = +1
iv. for O in a molecule, O.N. = -2
ex. CO2, H2O: O = -2
v. sum of O.N. in any molecule or ion = charge
ex. H2O: each H = +1, O = -2
total O.N. = +1 +1 -2 = 0
an each total box may be helpful other molecules like for CO2
C
each
total
O
-2
C
O
-2
-4
C
+4
O
-2
-4
C
+4
+4
O
-2
-4
So, knowing the O.N. of O, we can calculate the O.N of C in CO2
Example: determine the oxidation numbers for each atom in sodium phosphate (Na3PO4)
We know that the O.N. on oxygen is –2 (O.N. for oxygen in a non-peroxide cmpd) and
for sodium, it is +1 (O.N. 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 O.N.’s must be five for this to balance out (3 + x – 8= 0),
so x = 5. Since there’s only one P, it’s O.N. is five
2. Mnemonics
a. OIL RIG
Oxidation Is Loss, Reduction Is Gain (of electrons)
b. LEO says GER
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.
3. redox reaction examples
a. combustion
b. breathing/respiration
c. rust
d. bleach
e. battery
4. Determining whether a rxn is redox
So, when do we have a redox rxn? When the O.N. numbers on an atom change
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
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
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 +
+1 -1
Ca(OH)2
+2 -2 +1

HOH +
+1 -2 +1
CaBr2
+2 -1
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
5. 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).
C. Heat of reaction
1. exothermic/exergonic
2. endothermic/endergonic
Chapter Objectives
Knowledge
Memorize the rules for determining oxidation number (outline III.B.1.b)
Recall number of items in a mole to 3 sig figs (6.02 x 1023)
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 # of electrons) 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 if told heat of reaction is positive or negative
Identify a limiting reagent
Application
Use mw to convert back and forth between mass and moles
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