Amounts of Reactants and Products
Amounts of Reactants and Products
** A basic
question raised in the chemical laboratory is “How much product will be formed
from specific amounts of starting materials (reactants)?” Or in some cases, we
might ask the reverse question: “How much starting material must be used to
obtain a specific amount of product?”
** To interpret
a reaction quantitatively, we need to apply our knowledge of molar masses and the
mole concept. Stoichiometry is the quantitative study of reactants and products
in a chemical reaction.
** Whether the
units given for reactants (or products) are moles, grams, liters (for gases),
or some other units, we use moles to calculate the amount of product formed in
a reaction. This approach is called the mole method, which means simply that the
stoichiometric coefficients in a chemical equation can be interpreted as the
number of moles of each substance.
For example,
industrially ammonia is synthesized from hydrogen and nitrogen as follows:
** The
stoichiometric coefficients show that one molecule of N2 reacts with
three molecules of H2 to form two molecules of NH3. It follows that
the relative numbers of moles are the same as the relative number of molecules:
** Thus, this
equation can also be read as “1 mole of N2 gas combines with 3 moles
of H2 gas to form 2 moles of NH3 gas.” In stoichiometric
calculations, we say that 3 moles of H2 are equivalent to 2 moles of
NH3, that is,
where the
symbol ≈ means “stoichiometrically
equivalent to” or simply “equivalent to.” This relationship enables us to write
the conversion factors
** Similarly,
we have 1 mol N2 ≈ 2 mol NH3 and 1 mol N2 ≈ 3 mol H2.
Let’s consider
a simple example in which 6.0 moles of H2 react completely with N2
to form NH3.
** To calculate
the amount of NH3 produced in moles, we use the conversion factor
that has H2 in the denominator and write:
** Now suppose
16.0 g of H2 react completely with N2 to form NH3.
How many grams of NH3 will be formed? To do this calculation, we
note that the link between H2 and NH3 is the mole ratio
from the balanced equation. So we need to first convert grams of H2 to
moles of H2, then to moles of NH3, and finally to grams
of NH3. The conversion steps are:
(1) we
convert 16.0 g of H2 to number of moles of H2, using the
molar mass of H2 as the conversion factor:
(2) we
calculate the number of moles of NH3 produced
(3) we
calculate the mass of NH3 produced in grams using the molar mass of NH3
as the conversion factor:
** These three
separate calculations can be combined in a single step as follows:
** Similarly,
we can calculate the mass in grams of N2 consumed in this reaction. The
conversion steps are:
By using the
relationship 1 mol N2 ≈ 3 mol H2 ,
we write
** The
following Figure shows the steps involved in stoichiometric calculations using
the mole method.
First: convert the quantity of
reactant A (in grams or other units) to number of moles.
Next: use the mole ratio in the
balanced equation to calculate the number of moles of product B formed.
Finally: convert moles of product to grams of product.
Finally: convert moles of product to grams of product.
![]() |
Figure (1) |
Solved problems
Problem (1): The
food we eat is degraded, or broken down, in our bodies to provide energy for growth
and function. A general overall equation for this very complex process represents
the degradation of glucose (C6H12O6) to carbon
dioxide (CO2) and water (H2O):
If 968 g of C6H12O6
is consumed by a person over a certain period, what is the mass of CO2
produced?
Strategy:
Looking at the balanced equation, how do we
compare the amount of C6H12O6 and CO2?
We can compare them based on the mole ratio from the balanced equation.
Starting with grams of C6H12O6, how do we
convert to moles of C6H12O6?
Once moles of
CO2 are determined using the mole ratio from the balanced equation, how
do we convert to grams of CO2?
Solution
We follow the preceding steps and Figure (1).
Step 1: The
balanced equation is given in the problem.
Step 2: To
convert grams of C6H12O6 to moles of C6H12O6,
we write
Step 3: From
the mole ratio, we see that 1 mol C6H12O6 ≈ 6
mol CO2. Therefore, the
number of moles
of CO2 formed is
Step 4: Finally,
the number of grams of CO2 formed is given by
After some
practice, we can combine the conversion steps
into one
equation:
Check
Does the answer
seem reasonable? Should the mass of CO2 produced be larger than the
mass of C6H12O6 reacted, even though the molar
mass of CO2 is considerably less than the molar mass of C6H12O6?
What is the mole ratio between CO2 and C6H12O6?
Problem (2): All
alkali metals react with water to produce hydrogen gas and the corresponding
alkali metal hydroxide. A typical reaction is that between lithium and water:
Strategy:
The question asks for number of grams of reactant
(Li) to form a specific amount of product (H2). Therefore, we need
to reverse the steps shown in Figure (1).
From the
equation we see that 2 mol Li ≈ 1 mol H2.
Solution:
The conversion
steps are
Combining these
steps into one equation, we write
Check:
There are roughly 4 moles of H2 in 7.79 g H2,
so we need 8 moles of Li. From the approximate molar mass of Li (7 g), does the
answer seem reasonable?
Reference: Organic chemistry / T.W. Graham Solomons , Craig B.Fryhle , Scott A.snyder , / ( eleventh edition) / 2014.
No comments