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GENERAL CHEMISTRY 115
USING BALANCING COEFFICIENTS
IN CHEMICAL REACTIONS
I.
Some Comments Concerning Balanced Chemical Reactions:
First, the balancing coefficients in a balanced chemical reaction tell us the ratios of the
NUMBERS of molecules that will react with one another – i.e., the ratios of MOLES of reagents
either being consumed, (reactants), or formed, (products), during the chemical reaction. For this
reason, the ONLY times we use these coefficients are when we want to determine how many moles
of one reactant or product will either react or be formed, when we know how many moles of
another reagent reacted or was formed.
NEVER multiply molecular weights (molar masses) by these coefficients!
The molecular weight (molar mass) is a fixed number – it tells us the weight in grams of 1 mole of
molecules, (i.e., 6.022 × 1023 molecules), of a particular substance. [For example, the molar mass
of H2O is 18.02 amu, (i.e., 18.02 g/mole). This tells us that every 18.02 gram sample of water will
contain 1 mole of molecules. This value is completely unaffected by any chemical reaction. It is
also completely independent by our choices for the balancing coefficients. For example, consider
the following two balanced chemical reactions involving water:
2 H2 + O2 → 2 H2O
and
H2 + ½ O2 → H2O .
Both forms are equally valid since they yield the identical ratios for the numbers of each substance
relative to the others. [For example, both forms tell us that the number of water molecules formed
will be identical to the number of H2 molecules that react and that half as much O2 will be
necessary to make the reaction go.] The mass of water in 1 mole remains 18.02 grams, no matter
which form is used to balance the chemical reaction.
Second, a balanced chemical reaction describes a process taking place over time. The left side, (the
side with the reactants), tells us which substances are initially put into the “pot”. In the reaction
above, H2 and O2 are the reactants – they are the chemical substances that are initially placed in the
reaction vessel. The right-hand-side shows us the products that will form, once the reaction
begins. [Water is the only product in the above reaction equation.] The products are the substances
that will form later in time – once reaction takes place. As reaction proceeds, we will lose reactant
molecules and gain product molecules, until such time as one or more of our reactant species have
been completely consumed. Once one of the reactants is completely “used up”, no more reaction
can occur – no more product can form.
Third, the balancing coefficients in a balanced chemical reaction describe how much of the reactant
species can potentially react together or how much of a product species can potentially be formed.
as a product – they do not tell us how much of each substance was actually put into the reaction
vessel – that is entirely up to us!
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III. Using Balancing Coefficients in Chemical Reaction Problems:
For the remainder of this handout, we will consider the following reaction:
N2 + 3H2  2NH3.
As balanced, it tells us the ratios of numbers of molecules reacting. Thus, if 1 molecule of N2
reacts, then we will need 3 molecules of H2 to react with it and 2 molecules of NH3 will be formed.
Or, if 100 molecules of N2 react, then we will need 300 molecules of H2 to react with it and 200
molecules of NH3 will be formed. Or, we could use a unit that involves a multiple of things. For
example, the reaction tells us that 1 dozen molecules of N2 will react with 3 dozen molecules of H2
to produce 2 dozen molecules of NH3.
Using dozens of molecules still involves quantities that are far too small to measure out in the lab.
However, if we use a very large unit of quantity, such as a mole, (i.e., 6.022 × 1023 things), then the
amounts involved become large enough that we can measure out such quantities in the lab. It now
becomes more convenient for us to interpret the ratios in terms of moles, as follows: If 1 mole of
N2 reacts, (i.e., 6.022 × 1023 molecules of N2), then we will need 3 moles of H2, (i.e., 3 ×
6.022 × 1023 molecules of H2), to react with them and we will produce 2 moles of NH3, (i.e., 2 ×
6.022 × 1023 molecules of NH3).
Now suppose that we measured out 0.200 moles of N2 and we wanted to know how much H2
would be needed to react with it and how much NH3 would form. The balancing coefficients can
be used as unit conversion factors in order to convert the known moles of N2 into moles of H2 and
NH3 as shown below:
3 mols H 2
0.200 mols N 2 
= 0.600 mols H 2 which must also react and
1 mol N 2
2 mols NH3
0.200 mols N 2 
= 0.400 mols NH 3 which must be produced.
1 mol N 2
[Notice how carefully we labeled each molar quantity with the substance to which it refers. You
cannot cancel moles of one reagent with moles of another – both the unit, (moles), and the
substance to which that unit belongs must cancel properly.]
The masses of the substances reacted or produced are then determined ONLY once we have
determined the numbers of moles of them reacted or produced. I.e., in the above example, if 0.200
mols N2 reacts, then the following masses will be involved:
mass of N2 reacted = 0.200 mol N2 × 28.02 g N2/mol N2 = 5.60 g N2 reacted
mass of H2 reacted = 0.600 mol H2 × 2.02 g H2/mol H2 = 1.212 g H2 reacted
mass of NH3 produced = 0.400 mol NH3 × 17.04 g NH3/mol NH3 = 6.82 g NH3 produced
[Once again – note that both the masses and moles are labeled with the substance involved. Note
also that we used units of g/mole for the molar masses when we performed our unit conversions.]
Note that the balancing coefficients are never used to convert masses into moles or vice versa! Nor
are they used to convert masses of one substance into masses of another. [To avoid these potential
pitfalls, always think of the balancing coefficients as having units of moles.] The balancing
coefficients should ONLY be used to convert moles of one reagent into moles of another. Always
label your units with the compound they belong to so that you do not end up mixing units. I.e.,
moles NH3 cannot be divided into moles N2 – the units must match perfectly! [Notice in the
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examples described earlier that moles N2 are cancelled by moles N2, but not by moles of any other
chemical substance.]
Since many of the problems you will be solving this semester will involve chemical reactions, I
strongly recommend you do the following steps before you even fully read the problem:
1)
Balance the chemical reaction equation, (if it isn’t already balanced). [If it contains any
coefficients other than 1, it is already balanced.]
2)
Convert any data you can into moles. [For example, if you are given a mass of a substance
and you know its chemical formula, use the molar mass of the substance to convert the mass
into moles. A couple of additional ways you can use to determining moles are to use molarity
× volume = moles solute, when concentrations are involved; or to use PV/RT = moles of gas
present, if a gas law problem is involved. We will discuss these methods in the relevant
chapters.]
IV. Limiting Reagent Problems:
When reactions are performed in the lab, we do not first carefully measure out exact proportions of
our reactants. Instead, we usually carefully measure how much we have of one of the reactants and
then enough of the others to ensure that they are in excess – i.e., that more moles of molecules of
those substance are present than are needed to completely react the first one. For example, suppose
we wanted to completely react exactly 1 mole of N2 molecules. As long as we have at least 3
moles of H2 molecules present, we can react all of the N2. However, if we have less than 3 moles
of H2 present, then not all of our sample of N2 can react. If we measured out 5 moles of H2
molecules, then only 3 moles of them will react with our 1 mole sample of N2, the remaining 2
moles of H2 will remain unreacted once all of the N2 has reacted. We say that the H2 is in excess in
this instance and the N2 is the limiting reagent, since it is the substance which determined how
much of the other reactants were consumed, (and also determines how much product is formed).
Thus the limiting reagent plays a very important role – its quantities, (and only its quantities), will
determine exactly how much of each reactant will be consumed and how much of each product
will be formed.
In limiting reagent problems, you will be given data concerning the quantities of 2 or more
reactants. Before you can determine how much product is formed, you will need to identify which
reactant is the limiting reactant. Once you have determined how many moles of each reactant are
present, you need to use that information to determine which is limiting and which will be in
excess, once the reaction is complete. Note that knowledge of the numbers of moles of each
reactant is insufficient information! For example, consider the reaction given above:
N2 + 3 H2  2 NH3.
If we started with 1.000 mole of N2 and 2.000 moles of H2, it would be the H2 that is the limiting
reagent – not the N2 – even though there are fewer moles of N2 present. That is because the H2 is
used up faster, according to the chemical reaction. For every mole of N2 that reacts, we need 3
times that amount of H2 to make the reaction go. I.e., In order to react all of the 1.000 mol of N2,
we need 3.000 moles of H2 – but we only have 2.000 moles of it available. So we cannot use up all
of the N2, since we don’t have enough H2 present. W will use up all of the limiting reagent, (H2 in
this example), but not all of the N2 – the N2 will be in excess at the end of the reaction.
There are several methods that can be used to determine the limiting reagent. Usually, the most
efficient approach is to calculate how much product is formed when each reagent is completely
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reacted. Whichever reactant produces the least amount of product, (in moles or grams), will be
the limiting reagent. [Once the limiting reagent is used up, no more product can be formed.] We
will demonstrate this using our previous example, (i.e., the reaction of 1.000 mole of N2 with 2.000
moles of H2):
1) Convert each of the moles reactants into moles of one of the products, using the balancing
coefficients. [Note that we must use moles of each reactant– never grams!!!]
2)
2 mol NH3
= 2.000 mol NH 3 produced
1 mol N 2
2 mol NH 3
= 1.667 mol NH 3 produced
If all H2 reacts: 2.000 mol H 2 
3 mol H 2
If all N2 reacts: 1.000 mol N 2 
2)
The reactant that produces the least amount of product will then be identified as the limiting
reagent. All other reactants will be in excess, once the reaction is complete. Thus in our
example above, the H2 is the limiting reagent.
3)
The amount of product formed is the amount predicted by the limiting reagent. In our
example, we see that when all of the 2.000 moles of H2 are used up during reaction, 1.667
moles of NH3 will be produced.
4)
In addition, we can also determine how much of the original 1.000 moles of N2 will remain
unreacted, after reaction is complete:
1 mol N 2
= 0.6667 mol N 2 reacted
Moles N2 actually reacted = 2.000 mol H 2 
3 mol H 2
Moles N2 remaining = Initial moles of N2 – # moles of N2 actually reacted = 1.000 mol N2 –
0.6667 mol N2 reacted = 0.3333 mol N2 remaining
Note that the N2 is indeed in excess, even though we had more molecules of N2 than H2 before the
reaction began!
Finally, not until we know how many moles of each reagent are involved can we calculate the
masses or concentrations of those reagents. I.e., we ALWAYS must first determine the numbers of
moles of each reagent that reacted or was produced before we can calculate the masses of those
reagents. Simply multiply the moles of each reagent by the molecular weight of that same reagent
to get its mass:
Mass of NH3 produced = 1.667 mol NH3 produced × 17.04 g NH3/mol NH3 = 28.40 g NH3
Mass of N2 remaining = 0.3333 mol N2 remaining × 28.08 g N2/mol N2 = 9.359 g N2
[Note that in each case, moles of the substance in the numerator were cancelled by moles of that
same substance in the denominator.]
Finally, remember that the balancing coefficients can only be used to convert MOLES of one
substance into another. Thus our steps are as follows:
1.
Convert all information about each reactant into moles of that reactant.
2.
Using the balancing coefficients from the balanced chemical reaction, convert the number of
moles of each reactant into moles of one of the products. The limiting reagent will be the one
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that produces the fewest moles of product. It will be the reactant that is used up during the
reaction – all others will be in excess when the reaction is complete.
3.
Using only the moles of the limiting reagent, we can then calculate the number of moles of
each product that is formed and the moles used up of each of the reactants, using the
balancing coefficients.
4.
Once we know how many moles have been formed of a particular product, we then convert
the moles of that substance into a mass of that substance, (or concentration or pressure or
whatever information we wish to find for that substance).
5.
Finally, remember that the above method will tell us the number of moles used up of
reactants, not the amount that remains after the reaction is complete. The number of moles
remaining is the difference between the amount we started with and the amount that reacted.
Schematically, the process we must use can be described as shown below. [For the masses below,
substitute the words “partial pressure” or “concentration”, as appropriate.]
Moles of
Moles of
balancing coefficients


Substance 1
Substance 2
Mass of
Substance 1
Mass of
Substance 2
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