# 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 N

_{2}reacts with three molecules of H2 to form two molecules of NH_{3}. 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 N

_{2}gas combines with 3 moles of H_{2}gas to form 2 moles of NH_{3}gas.” In stoichiometric calculations, we say that 3 moles of H_{2}are equivalent to 2 moles of NH_{3}, 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 N

_{2}≈ 2 mol NH3 and 1 mol N_{2}≈ 3 mol H_{2}.**Let’s consider a simple example in which 6.0 moles of H**

_{2}react completely with N_{2}to form NH_{3}.
** To calculate
the amount of NH

_{3}produced in moles, we use the conversion factor that has H_{2 }in the denominator and write:
** Now suppose
16.0 g of H

_{2}react completely with N_{2}to form NH_{3}. How many grams of NH_{3}will be formed? To do this calculation, we note that the link between H_{2}and NH_{3}is the mole ratio from the balanced equation. So we need to first convert grams of H_{2}to moles of H_{2}, then to moles of NH_{3}, and finally to grams of NH_{3}. The conversion steps are:**(1)**we convert 16.0 g of H

_{2}to number of moles of H

_{2}, using the molar mass of H

_{2}as the conversion factor:

**(2)**we calculate the number of moles of NH

_{3}produced

**(3)**we calculate the mass of NH

_{3}produced in grams using the molar mass of NH

_{3}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 N

_{2}consumed in this reaction. The conversion steps are:
By using the
relationship 1 mol N

_{2}≈ 3 mol H_{2 }, 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.

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 (C**

_{6}H_{12}O_{6}) to carbon dioxide (CO_{2}) and water (H_{2}O):

**If 968 g of C**

_{6}H_{12}O_{6}is consumed by a person over a certain period, what is the mass of CO_{2}produced?

**Strategy:**

Looking at the balanced equation, how do we
compare the amount of C

_{6}H_{12}O_{6}and CO_{2}? We can compare them based on the mole ratio from the balanced equation. Starting with grams of C_{6}H_{12}O_{6}, how do we convert to moles of C_{6}H_{12}O_{6}?
Once moles of
CO

_{2}are determined using the mole ratio from the balanced equation, how do we convert to grams of CO_{2}?**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 C

_{6}H_{12}O_{6}to moles of C_{6}H_{12}O_{6}, we write
Step 3: From
the mole ratio, we see that 1 mol C

_{6}H_{12}O_{6}≈ 6 mol CO_{2}. Therefore, the
number of moles
of CO

_{2}formed is
Step 4: Finally,
the number of grams of CO

_{2}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 CO

_{2}produced be larger than the mass of C_{6}H_{12}O_{6}reacted, even though the molar mass of CO_{2}is considerably less than the molar mass of C_{6}H_{12}O_{6}? What is the mole ratio between CO_{2}and C_{6}H_{12}O_{6}?**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 (H

_{2}). Therefore, we need to reverse the steps shown in Figure (1).
From the
equation we see that 2 mol Li ≈ 1 mol H

_{2}.**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 H

_{2}, 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.*

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