Based on the coefficients in a given a balanced equation, a certain ratio must
exist between reactants in order to obtain a certain amount of product. This
ratio is the mole
ratio we discussed
previously
However, now suppose that you simply want to make as much product as possible
given the amounts of reactant you currently have. Suppose also that the amount
of each reactant you have is not in the correct ratio. You find that you have
an almost endless supply of the first of your two reactants, but very little of
the second. What happens? Well, obviously, you are not going to be able to use
all of the first reactant. It will only combine in ratio with the second; the
excess will have nothing to react with. The amount of product you obtain will
therefore be in ratio with the second reactant. Confused? An example is long
overdue.
Problem: You have 10 g of solid carbon, but only 10 mL of pure oxygen
gas. Not all of the carbon will be used in the reaction. Given that the
reactants react according the following balanced equation, how many molecules of
CO2(g) can be produced? How many grams of C(s) will react?
How much C(s) is left over after all the oxygen is gone?
Solution: Let's answer the first question first. Start by converting to
moles
O2(g) [Note that oxygen is diatomic, so 10 mL of oxygen gas
means 10 mL of
O2(g)].
 × = 4.46×10-4 moles O2(g) |
|
Did you remember to convert milliliters to liters? If so, remember to carry
your units. Now, since our mole ratio between
O2(g) and
CO2(g) is 1:1, we can move to the last step.
 = 2.68×1020 molecules CO2(g) |
|
Now let's figure out how many grams of C(s) react. How do we figure this out?
Well, how many moles of C(s) will react with
4.46×10-4 moles of
O2(g)? Mole ratio tells us that an equal number of moles will
react. Therefore,
 = 5.352×10-3 grams C(s) |
|
It is a simple matter of subtraction to figure out how much C(s) is left over.
| 10 grams C(s) -5.352×10-3 grams C(s) = 9.995 grams C(s) |
|
There was not enough oxygen to react with all the carbon. For this reason, we
would call oxygen the limiting reagent of the reaction. As the name
implies, the limiting reagent limits or determines the amount of product that
can be formed. In contrast, carbon would be called the excess reagent.
There was more than enough of it to react with the other reactant(s).
Now, in the example problem, we were more or less told which reactant was the
limiting reagent. However, many times you will have to figure this out.
Problem: Salt (sodium chloride) is prepared by the reaction of sodium
metal with chlorine gas.
You have 71.68 L of
Cl2(g) and 6.7 moles of Na(s). What is the
limiting reagent? How many moles of salt are produced?
Solution: Did you check to make sure the equation was balanced? It
already is, but give yourself a congratulatory punch in the shoulder if you
remembered (don't hurt yourself). Our next step, as always is to convert to
moles.
 = 3.2 moles Cl2(g) |
|
So now we know there is 3.2 moles of
Cl2(g) and 6.7 moles of
Na(s).
Let's assume Na(s) is the limiting reagent and then prove our assumption right
or wrong.
How many moles of Na(s) are required to react completely with 3.2 moles of
Cl2(g)? Use the mole ratio.
 = 6.4 moles Na(s) |
|
Do you have 6.4 moles of Na(s)? Yes, you do. You have 6.7 moles of Na(s).
This is more than enough to react with all the
Cl2(g). Our
assumption was wrong, Na(s) is not the limiting reagent since there is more than
enough of the solid. Since this reaction features only two reactants, we
already know that
Cl2(g) is the limiting reagent. Let's prove
it.
 = 3.35 moles Cl2(g) |
|
3.35 moles of
Cl2(g) are needed to react completely with all
the Na(s).
You have only 3.2 moles.
Cl2(g) is obviously the limiting
reagent.
Now let's figure out how many moles of salt (NaCL) are produced. We want to use
the
mole ratio from
Cl2(g) since it has determined how many moles
of each
reactant will react.
 = 6.4 moles NaCl(s) |
|
