Balancing Chemical Equations

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Chapter 3
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Balancing Chemical Equations What does mother
nature lets you change and what doesnt 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 cant do that! Example The
combustion of butane CH3CH2CH2CH3 O2 ? CO2
H2O
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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 CH3CH2CH2CH
3 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.
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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 5H2O CH3CH2CH2CH3 9/2O2 4CO
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.
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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? H2O C12H22O11 H2O ? 4CO2 4CH3CH2OH
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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
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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)
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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
atom6.02x1023 12 g/mol Mass of H
atom6.02x1023 1 g/mol Mass of O atom6.02x1023
16 g/mol Total 9128416 180 g /mol
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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) 4126316 102
g/mol gMW salicylic acid (C7H6O3)
7126316 138 g/mol gMW acetic acid (CH3CO2H)
2124216 60 g/mol
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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
3.0/5.87x10051
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How many grams of acetic anhydride are needed to
react with 4.5 g of salicylic acid?
C7H6O3 C4H6O3 ? C9H8O4
CH3CO2H 138 g/mol 102g/mol
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
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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.
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Lets examine the following series of reactions to
make alum Al KOH H2O ? KAl(OH)4 H2
? Lets 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 doesnt
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
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Overall the reaction is 2Al 2KOH 6H2O
? 2 KAl(OH)4 3H2 ? 2 KAl(OH)4 4H2SO4
16H2O ? 2KAl(SO4)2.12H2O ________________________
_______________________
2Al 2KOH 6H2O ? 2 KAl(OH)4 3H2 ?
KAl(OH)4 2H2SO4 8H2O ? KAl(SO4)2.12H2O
2Al 2KOH 22 H2O 4H2SO4 2KAl(SO4)2.12H2O
3H2
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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
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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