Chemistry 161
Lecture 11: Writing Balanced Chemical Equations, Combustion Reactions
Sections 3.3, 3.4 in Gilbert, Kirss, Foster, and Davies 2nd edition
10/23/08
Learning Objectives
1. Develop a method for writing balanced chemical equations
2. Use balanced chemical reactions and molar mass (mass to mole
conversions) to determine amount of reagents needed and products
formed
3. Define the C3 (critical chemistry concept) of limiting reagent,
theoretical yield, and percent yield
4. Use balanced chemical reactions and molar mass (mass to mole
conversions) to determine limiting reagent, theoretical yield, and
percent yield
3.4: Writing and Balancing Chemical Equations
 Please refer to the general scheme outlined on page 89 of your text
o Bear in mind that in this class step 4 is optional
A Brief Word on Chemical Equations and Arrows
52
Balancing Chemical Reactions
All you need to remember is the total number of each type of atom must be
the same on either side of a chemical equation; since you cannot change the
number of atoms within a given molecule, you must change the number of
molecules – you do this with the stoichiometric coefficient. As an example,
CH4 + O2  CO2 + H2O
is “balanced” by placing a 2 in front of water, then a 2 in front of oxygen,
since there must be the same number of total atoms of each element
present
CH4 + 2O2  CO2 + 2H2O

Note that is a good idea to leave elemental species until last when
balancing

Bear in mind subscripts following an atom indicate the number of that
atom within a given compound; the stoichiometric coefficient indicates
the number of that compound (really the ratio in which the species
combine). Thus, the total “number” of a given atom is the
stoichiometric coefficient multiplied by the number within compound
o For the reaction above, the 4 H needed on the product side are
contained within the 2H2O
To formalize:
1. Write the chemical equation without stoichiometric coefficients
(reactants on the left proceeding to products on the right, of course)
2. In this class, it will be important to indicate the phase for a given
reactant or product. Convention is to place (s), (l), or (g) following the
compound (or s, l, or g as a subscript)1. Notice (aq) is used to indicate
a species in water
1
In solution chemistry it is not unusual (and perfectly acceptable) to use  for a gas phase
product and  to indicate a precipitate
53
3. Write the appropriate stoichiometric coefficient in front of the
reactant or product that allows you to balance one of the atoms in a
compound. For example, in the reaction
C4H10 (g) + O2 (g)  CO2 (g) + H2O (g)
Place a 4 in front of CO2, and 5 in front of H2O
 Clearly, we arbitrarily assign a coefficient of 1 to one of the
species involved – in the case of combustion of organic (C
based) molecules, leave the organic molecule alone and match
the others to it
4. Leave elemental species until the end, and then look at the number of
that atom required. In the example just cited, I would need 13 O in
total or 13/2 O2.
5. While the generally accepted form is to have integer stoichiometric
coefficients, it is perfectly acceptable to leave it as a simple fraction,
since we are mixing large (molar) ratios of species.
C4H10 (g) + (13/2)O2 (g)  4CO2 (g) + 5H2O (g)
Once we have the balanced chemical equation, and we can readily convert
mass to moles using molecular weight, and as such, we can determine

Mass needed to carry out a reaction, given a desired amount of
product or amount of one of the reactants

Amount of product that can be formed from a given amount of
reactants
Question: What is the minimal amount of oxygen needed to produce 44 g of
CO2 in the combustion of butane (C4H10)?
54
Question: What is the minimal amount of oxygen needed to consume 29 g of
butane?
The above allows us to further develop the concepts of limiting reagent and
theoretical yield, answering the question what is the maximum amount of
product that can be produced for a given mass of reactants?
Question: What is the maximum amount of CO2 that can be produced from
29 g of butane and 32 g of O2?
HW # 3, Question 3 (2 points) : Given pentane (C5H10) has a density of .62
g/mL, 1 mole of ideal gas will occupy 22.4 L under usual atmospheric
conditions, and O2 is 20% of our atmosphere, which will be the limiting
reagent if 1 L of pentane is burned in a closed container of air with volume
112 L?
Finally, if not sadly, side and/or incomplete reactions occur under various
conditions. As a result, one of the most important pieces of data you can
relay to other experimenters is percent yield:
% yield = 100(actual yield/theoretical yield)
Question: What is the % yield for a reaction of butane with O2 if 29 g of
butane and 208 g of O2 yield 44 g CO2?
55
Consider the combustion of natural gas (mainly methane, CH4) vs. the
combustion of propane (C3H8). Combustion of methane liberates 212
kcal/mol, and propane liberates 530 kcal/mol.
CH4
+
O2
C3H8
+
O2 

Question: Which gives the most energy on a gram basis?
Question: Which gives more H2O on a gram basis?
Question: Which gives more CO2 on a gram basis?
56