Chapter 3
Balancing Chemical Equations
What does mother nature lets you change and what doesn’t she?
H2 + O2  H2O
2H2 + O2  2 H2O
Coefficients are changeable (numbers appearing before a
substance)
Subscripts are not changeable: Changing the subscript changes the
chemical that is formed or used and you can’t do that!
Example:
The combustion of butane:
CH3CH2CH2CH3 + O2  CO2
+ H2O
Equivalent symbols:  (means “goes to”) can be replaced or
is equivalent to a mathematical “ = ” sign
CH3CH2CH2CH3 + O2 =
CO2
+ H2O
Application of the conservation of mass for carbon results in:
CH3CH2CH2CH3 + O2 =
4CO2
+ 5H2O
Balancing the oxygens:
CH3CH2CH2CH3 + 13/2O2 =
4CO2
+ 5H2O or
2CH3CH2CH2CH3 + 13O2 =
8CO2
+ 10H2O
Notice that we made an assumption here, that we had as much
oxygen as we needed. In reality, this may not be the case.
Therefore you also need to know the limiting reagent.
Sometimes by changing the coefficients in front of the
reactants, you can change the chemistry, but not control it
except superficially. Coefficients are change by changing the
relative amounts of reagents.
In the case of the reaction we just looked at, if O2 is the
limiting reaction, the product composition may change:
In a limited amount of air (O2)
CH3CH2CH2CH3 + O2 = CO + H2O
CH3CH2CH2CH3 + O2 =
4CO
CH3CH2CH2CH3 + 9/2O2 = 4CO
+ 5H2O
+ 5H2O
2CH3CH2CH2CH3 + 9O2 = 8CO + 10H2O
Less than the required amount of oxygen may lead to
incomplete combustion and a mixture of CO and CO2, carbon
soot, etc.
Fermentation of glucose by yeast to yield ethyl alcohol
1. C6H12O6  CO2 + CH3CH2OH
C6H12O6  2CO2 + 2CH3CH2OH
Fermentation of sucrose
2. C12H22O11  CO2 + CH3CH2OH
C12H22O11  4CO2 + 4CH3CH2OH ?
What else is present when we ferment grape juice or grains?
H2 O
C12H22O11 + H2O  4CO2 + 4CH3CH2OH
K + H2O  KOH + H2
K + H2O  KOH + 1/2H2
2K + 2H2O  2KOH + H2
Na + H2O  NaOH + H2
2Na + 2H2O  2NaOH + H2
Li + H2O  LiOH + H2
2Li + 2H2O  2LiOH + H2
movie cca 2-alkali
Review:
Atomic weight: sum of the number of protons and neutrons
Gram atomic weight =
weight in grams of 6.02x 1023 atoms
Molecular weight =
sum of the atomic weights of the elements in the
molecule
Gram molecular weight =
weight in grams of 6.02x 1023 molecules or
formula units (used for for ionic solids, for example: NaCl)
What is the gram molecular weight of aspirin = C9H8O4 ?
1 molecule has a mass of 9*(mass of C)+8*(mass of H) + 4*(mass of O)
Mass of C atom*6.02x1023 = 12 g/mol
Mass of H atom*6.02x1023 = 1 g/mol
Mass of O atom*6.02x1023 = 16 g/mol
Total = 9*12+8+4*16 = 180 g /mol
Aspirin is made by reacting salicyclic acid (C7H6O3) with acetic
anhydride (C4H6O3) according to the following reaction:
C7H6O3 + C4H6O3  C9H8O4 + CH3CO2H
How many grams of acetic anhydride are needed to react with 4.5 g
of salicylic acid?
gMW aspirin = 180
gMW acetic anhydride (C4H6O3) =
4*12+6+3*16 = 102 g/mol
gMW salicylic acid (C7H6O3) =
7*12+6+3*16 = 138 g/mol
gMW acetic acid (CH3CO2H) =
2*12+4+2*16 = 60 g/mol
Aspirin is made by reacting salicyclic acid (C7H6O3) with acetic
anhydride (C4H6O3) according to the following reaction:
C7H6O3 + C4H6O3  C9H8O4 + CH3CO2H
138 g/mol 102g/mol = 180g/mol
60g/mol
mass of reactants = mass of products
How many grams of acetic anhydride are needed to react with 4.5 g of
salicylic acid and how many grams of aspirin would be formed?
4.5g/138g/mol = 0.0326 mol
Since the coefficients in front of both salicyclic acid, acetic anhydride
and aspirin are 1, 0.0326 mol of aspirin should be theoretically formed;
this corresponds to x/180g/mol = 0.0326 mol or x = 5.87 g
5.87 is called the theoretical yield.
Suppose you actually isolated 3.0 g when you ran this reaction, the
actual yield would be 3.0/5.87x100=51%
How many grams of acetic anhydride are needed to react with 4.5 g
of salicylic acid?
C7H6O3 + C4H6O3 
138 g/mol 102g/mol
C9H8O4 + CH3CO2H
180g/mol
60g/mol
salicylic acid + acetic anhydride  aspirin
+ acetic acid
4.5g/138g/mol = 0.0326 mol C7H6O3
0.0326 mol C4H6O3 are also needed or 0.0326 mol x 102g/mol =3.33 g
Usually an excess of one reagent is used; the other reagent is called the
limiting reagent
Lets examine a series of reactions to make alum:
KAl(SO4)2.12H2O
What is Alum?
The most common form, potassium aluminum sulfate, or potash alum, is
one form that has been used in food processing. Another, sodium aluminum
sulfate, is an ingredient in commercially produced baking powder. (Have you
never noticed the faint metallic taste in baking powder? It comes from the alum.)
The potassium-based alum has been used to produce crisp cucumber and
watermelon-rind pickles as well as maraschino cherries, where the aluminum ions
strengthen the fruits' cell-wall pectins.
Alum is approved by the U.S. Food and Drug Administration as a food
additive, but in large quantities — well, an ounce or more — it is toxic to humans.
Alum's antibacterial properties contribute to its traditional use as an
underarm deodorant. It has been used for this purpose Today, potassium alum is
sold commercially for this purpose as a "deodorant crystal," often in a protective
plastic case.
Alum in powder or crystal form, or in styptic pencils, is sometimes
applied to cuts to prevent or treat infection.
Powdered alum is commonly cited as a home remedy for canker sores.
Lets examine the following series of reactions to make alum
Al + KOH + H2O  KAl(OH)4 + H2 
Let’s start by balancing the reaction:
Al and K are balanced as is; 2O vs 4O and 3H vs 6H
If we change the coefficient of KOH, the K is not balanced; changing the
coefficient on H2O doesn’t effect anything else
Al + KOH + 3H2O  KAl(OH)4 + 1.5H2 
2Al + 2KOH + 6H2O  2 KAl(OH)4 + 3H2 
Now suppose we add dilute sulfuric acid to this compound
KAl(OH)4 + H2SO4  KAl(SO4)2.12H2O
KAl(OH)4 + 2H2SO4  KAl(SO4)2.4H2O
Remember Alum is: KAl(SO4)2. 12H2O
KAl(OH)4 + 2H2SO4 +8H2O  KAl(SO4)2.12H2O
2Al + 2KOH + 6H2O  2 KAl(OH)4 + 3H2 
KAl(OH)4 + 2H2SO4 +8H2O  KAl(SO4)2.12H2O
Overall the reaction is:
2Al + 2KOH + 6H2O  2 KAl(OH)4 + 3H2 
2 KAl(OH)4 + 4H2SO4 + 16H2O  2KAl(SO4)2.12H2O
_______________________________________________
2Al +2KOH +22 H2O + 4H2SO4 = 2KAl(SO4)2.12H2O +3H2
This equations represents the overall stoichiometry of the reaction. It was
generated by a process of mass balance, so that we can use this reaction to
calculate the overall yield of the reaction. Suppose we started with 0.5 g
of Al and used an excess of all the remaining reagents. What would be the
theoretical yield of alum (KAl(SO4)2.12H2O)?
2Al +2KOH +22 H2O + 4H2SO4 = 2KAl(SO4)2.12H2O +3H2
Al = 27 g /mol
KAl(SO4)2.12H2O = 474 g/mol
H2SO4 = 98g/mol
0.5g/27g/mol = 0.0185 mol Al
How many mol of Al is required to produce 2 mol of alum?
2 mol Al produces 2 mol of alum
How many mol of alum is produced from 0.0185 mol Al?
0.0185 mol alum; What is the theoretical yield?
474g/mol x 0.0185 mol = 8.77 g alum
How much sulfuric acid is needed?
2Al +2KOH +22 H2O + 4H2SO4 = 2KAl(SO4)2.12H2O +3H2
Al +KOH +11H2O + 2H2SO4 = KAl(SO4)2.12H2O +1.5H2
1 mol of Al requires 2 mol of H2SO4
therefore: 0.0185 mol Al requires 0.037 mol H2SO4
0.037 mol * 98 g/mol = 3.63 g H2SO4