8.5 - Calculations based on Chemical Equations
8.5.4 - Limiting Reactant Calculations
Often in many chemical processes a reactant is present in excess.
The reactant that is present in excess is the chemical that remains after the
reaction is completed. The reactant that is fully consumed during the reaction
is called the limiting reagent. In combustion
reactions, oxygen gas is usually present in excess in order for the the reaction
to under complete combustion. Complete combustion
means that the products of the reaction are CO2 (g) and H2O
(l). Contrast to complete combustion is incomplete combustion. Incomplete combustion tends to form CO (g) as well as CO2 (g).
In the previous sections, we took a detailed look at how to carry
out Mole-mole, Mass-mass Mass-mole type calculations.
Using the same example from section
8.5.1, we now pose another question. This time you will be given the mass of both reactants (C2H5OH and O2) and
you will be asked to identify the limiting reagent and calculate the moles of the product (carbon dioxide gas) formed.
(a) How many grams of carbon dioxide gas is produced when
20.5 grams of ethyl alcohol, C2H5OH, burns with 100.0
grams of oxygen, O2?
(b) Identify the limiting reagent and how much of the other reactant is left
The steps involved to answer this question
is modified from the example in section
Step 1: Identify the chemical equation
involved: a combustion reaction.
Step 2: Write the balanced chemical equation
(review balancing equations, section 8.3).
Step 2a: Convert the mass of C2H5OH
to moles of C2H5OH.
Step 2b: Convert the mass of O2 to
moles of O2.
Step 3: Determine the stoichiometric ratio of
the substances that you are working with, specifically:
- between C2H5OH
- between C2H5OH
- between O2 and CO2.
Step 4: Using
the moles of C2H5OH and O2 calculated in Step
2a and 2b, compare the mole ratio of C2H5OH : O2 to that obtained in Step 3. Determine which is present as the limiting reagent
and which is present in excess.
Step 5: Calculate the amount of reactant
that is left over.
- Steps 1, 2, and 2a are the same as in sections 8.5.1, 8.5.2, and 8.5.3.
- We add Step 2b because we need to convert the given mass of O2,
100.0 g, to moles of O2.
- We modify Step 3 because the question is requires the knowledge of the stoichiometric
ratio of C2H5OH : O2 , and potentially, C2H5OH
: CO2 and O2 : CO2.
- In Step 4, we compare the stoichiometric ratio obtained from Step 2a and
2b and determine the limiting reagent. After that is determined, we use the
available number of moles of the limiting reagent to calculate the number
of moles of CO2 produced.
- In Step 5, we calculate the amount of the reactant that is left over. Because
it didn't specify the units that we need to report this quantity, we will
leave the amount of reactant left over in moles.
If you need to see the specific steps, click on the links
Refer to section
8.5.2 and section
8.5.3 for a detailed look at Step 1, 2, and 2a. Click for a detailed look
Mass of C2H5OH (20.5
moles of C2H5OH
limiting reagent = ?
grams of CO2 produced
Mass of O2
Moles of O2
reactant in excess = ?
moles of excess reactant left
Limiting Reagent Examples ..p254