Module 7

STOICHIOMETRY AND ITS APPLICATIONS

Limiting and Excess Reactants

Consider the following scenario. For instance, your mother asked you to help her
in placing all the eggs in the egg tray so that they will be delivered to your customers. You
counted the eggs and you found out that there were three hundred sixty eggs and fourteen
egg trays available. As shown in Figure 1, an egg tray can accommodate 30 eggs.
Considering the total number of eggs, only 12 egg trays will be utilized. In this situation,
eggs are limited, and egg trays are in excess.

Ideally, the chemical equation in a reaction shows the stoichiometric relationship

between reactants and products. It means that all the reactants are consumed during a
chemical reaction. However, not all chemical reaction goes to completion. Mostly, an
excess of one reactant is supplied to ensure that the more expensive reactant is
completely used up.

The reactant that is completely consumed in a chemical reaction is the limiting

reagent or reactant. On the other hand, the substance that is not completely used up
is called an excess reagent or reactant. This is the same with the eggs-egg tray analogy.
An excess of egg trays is supplied to make sure that all the 360 eggs (limiting component)
must be placed properly in a container.

To better understand the concept of limiting and excess reagents, consider the
reaction between N2 and H2 to yield NH3. The balanced chemical equation is:
N 2(g) + 3H 2(g) 2NH 3(g)

Figure 2. Chemical Reaction between N2 and H2 to produced NH3

As shown in Figure 2, the number of molecules in the reactant is not equal. There
are more H2 molecules than N2. Even though the amount of H2 gas is higher than N2, the
limiting reactant is not N2, it is H2. Why? Let us go back to the balanced equation on the
synthesis of ammonia. For every nitrogen atom, 3 hydrogen atoms are needed to form an
ammonia molecule. Thus, if a specific amount of one reactant is available, the reaction
will stop when that reactant is totally consumed whether the other reactant has been
used up or not.

Calculation of limiting, excess reactant and the amount left unreacted

Here are the steps in determining limiting, excess reactant, and the amount left
unreacted:
1. Balance the chemical equation for the chemical reaction
2. Convert the given information into moles.
3. Use stoichiometry for each individual reactant to find the mass of the product
produced.
4. The reactant that produces a lesser amount of product is the limiting reagent.
5. The reactant that produces a larger amount of product is the excess reactant.
6. To find the amount of remaining excess reactant, subtract the mass of excess
reagent
consumed from the total mass of excess reagent given.

Let us consider the following problem to calculate the limiting, excess, and amount of
reactant left after the reaction.

Sample Problem 1. A 4.00 g sample of ammonia (NH) is mixed with 8.00 g of molecular
oxygen (O2). Which is the limiting reactant and how much excess reactant remains after
the reaction has stopped? (N =14.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol)

✓ First, we need to balance the equation for the reaction.

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
✓ Next, convert the given information into moles.

✓ Use stoichiometry to calculate the mass of product produced by each

reactant. NOTE: If it is not specified in the problem, it does not matter which product is
chosen, but the same product must be used for both reactants so that the amounts can
be compared. (In this problem, we will calculate the amount of NO produced)

✓ The reactant that produces the lesser amount of product is the limiting
reagent. In this case, oxygen produced a lesser amount of NO, therefore, it is the limiting
reactant.
 To find the amount of excess reactant, we must calculate how much of the
excess reactant did react with the limiting reactant. (Since in this problem NH3 is the
excess reagent, we will calculate the amount of NH3 reacted during the reaction).

✓ We are not yet finished, the 3.41 g is the amount of ammonia that reacted,
not what is left over. To find the amount of excess reactant remaining, subtract
the amount that reacted from the amount in the original sample.

To sum it up, the limiting reactant is the reactant that limits the amount of
product that can be formed and is completely consumed during the reaction. The
excess reactant is the reactant that is left over once the reaction has stopped due
to the limiting reactant.

Actual, Theoretical, and Percent yield

Under the favorable condition, when two or more elements or compounds

are combined, or when a compound decomposed, a new set of chemical
compounds will be formed. Although transformation happened, ideally, according
to the Law of Conservation of Mass discovered by Antoine Lavoisier in 1789, in
any chemical reaction, mass is neither created nor destroyed but changed from
one form to another. It means that the mass of any elements or compounds at the
beginning of the reaction is equal to the mass at the end of the said chemical
reaction (Figure 3).

However, the Law of Conservation of mass is applicable to an isolated system

where there is no exchange of mass and energy from the system to the
surrounding. Chemical reactions do not always proceed exactly as planned on
paper. For instance, during the conduct of an experiment, many factors will
contribute to the formation of less product than would be predicted. These factors
include spills and other experimental errors, losses due to an incomplete reaction,
undesirable side reactions that generate undesirable products, and other factors.
To determine the efficiency of a chemical reaction and to make the most product
with the least waste, chemists need a measurement that indicates how successful
a reaction has been. This measurement is called the percent yield. Percent yield
is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
The following is the equation for calculating the percent yield:

What is percent yield?

To compute the percent yield, it is first necessary to determine the

theoretical and the actual yield.

Theoretical yield is the maximum amount of product formed from the

complete reaction of given amounts of reactants in a chemical reaction with the
assumption that product is 100% recovered. In addition, theoretical yield is a result
of a calculation from a balanced chemical equation. On the other hand, the
experimental or actual yield is the amount of product that is obtained when the
reaction is carried out in the laboratory.

Generally, percent yields are less than 100% due to the different factors
mentioned above that possibly occur during a chemical reaction. However, there
are instances when percent yields are greater than 100%. That is if the measured
product of the reaction contains impurities that cause its mass to be greater than
it would be.

How to calculate percent yield?

Sample Problem 2. Decomposition of Potassium chlorate

Potassium chlorate decomposes upon slight heating in the presence of a

catalyst according to the reaction below:
2KClO3(s) → 2KCl(s) + 3O2(g)
In a certain experiment, 20.0g KClO3 is heated until it completely
decomposes. The experiment is performed, the oxygen gas is collected, and its
mass is found to be 7.45g.

A. What is the theoretical yield of oxygen gas?

B. What is the percent yield for the reaction?
Solution:
A. Calculation of theoretical yield. (Calculate the theoretical yield based
on the stoichiometry).
 Step 1: Identify the given
Given: Mass of KClO3 = 20.0 g, Mass of O2 collected = 7.45 g
Find: Theoretical yield, g O2

Step 2: List other known quantities and plan the problem.

Step 3: Check if the chemical equation is balance.

2KClO3(s) → 2KCl(s) + 3O 2(g)
Step 4: Apply stoichiometry to convert from the mass of a reactant to the
mass
of a product:
Mass (g) KClO3 → mol KClO3 → mol O2 → mass (g) O2

Step 5. Solve:

Step 6: Think about your result.

The mass of oxygen gas must be less than the original amount (20.0 g) of
potassium chlorate.

B. Calculate the percent yield

Step 1: Identify the given

Theoretical yield = 7.83 g O2(calculated value in part A)
Actual yield = 7.45 g O2
Find % Yield of the reaction

Step2. Solve.

Stoichiometric Calculations Involving Gases

For a chemical reaction to proceed there must be collision, proper

orientation, and enough energy to overcome the activation energy. Activation
energy is the minimum amount of energy that is required to activate atoms or
molecules to a condition in which they can undergo chemical change. For gasses,
even though atoms or molecules are further apart from each other, under the right
conditions, they can still undergo chemical reactions. So, how to determine the
amounts in terms of volume, a number of moles, or mass of the reactants or the
products?

Avogadro’s Law on Gases

Avogadro’s Law (sometimes referred to as Avogadro’s hypothesis or Avogadro’s

principle) is a gas law; it states that under the same pressure and temperature
conditions, equal volumes of all gases contain the same number of moles or
molecules. The law is named after Amedeo Avogadro who, in 1811, hypothesized
that two given samples of an ideal gas - of the same volume and at the same
temperature and pressure - contain the same number of molecules; thus, the
number of molecules or atoms in a specific volume of an ideal gas is independent
of their size or the molar mass of the gas. This can be extended to a chemical
reaction involving gases. The coefficients in a balanced chemical reaction
involving gases can be treated as volumes.

Example 1: (Volume – Volume) at constant pressure and temperature

What volume of oxygen gas, O2 is needed to complete the combustion of 4 L
methane, CH4? Assume that the pressure and temperature remain constant.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Solution:
Based on the balance chemical reaction, 1 mole CH4 reacts with 2 moles O2
to form 1 mole CO2 and 2 moles H2O. Based on Avogadro’s Law, the coefficients of
the balance chemical reaction can be treated as volume ratio. In the case of
combustion of CH4, the volume ratio is 1L:2L:1L:2L of CH4, O2, CO2 and H2O
respectively. For volume of O2 needed for 4 L of CH4:
= 4 L CH4 x 2 L O2 = 8 L O2
1 L CH4

Example 2: (Volume – Mass) at constant pressure and temperature

What is the mass in grams of oxygen gas, O2 is needed to complete the combustion
of 6 L of methane, CH4? Assume that the pressure and temperature remain
constant.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Solution:
To answer this, we need volume to mole ratio of CH4 and O2 based on the balance
chemical reaction. The volume to mole ratio is 1L CH4: 2 moles O2. Then to convert
moles O2 we need molar mass of O2 = 16.00 g/mol.
For mass of O2 (MM = 16 g/mol) needed for 6 L of CH4:
= 6 L CH4 x 2 mol O2 x 16.00 g O2 = 192 g O2
1 L CH4 mol
Example 3: (Volume – Mass) with specific temperature and pressure
Ammonia, NH3 (MM = 17.04 g/mol) is synthesized from hydrogen and nitrogen
based on the reaction below.
N2(g) + 3H2(g) → 2NH3(g)

If 5.00 L of nitrogen reacts completely with hydrogen at a pressure of 3 atm and a

temperature of 298 K, how much ammonia, in grams, is produced?
Solution:

First, we need to convert 5.00 L of Nitrogen to L of NH3 by using the volume ratio
based on the balance chemical reaction. The volume ration of N2 and NH3 is 1L:2L.
= 5.00 L N2 x 2 L NH3 = 10.0 L NH3
1 L N2
Solving for the moles NH3, the ideal gas equation is used, PV = nRT. Rearranging to
get n:
n = PV
RT
n = ____3.00 atm x 10.0 L_____ = 1.23 mol NH3
(0.0821 L●atm/mol●K)(298 K)

To convert moles NH3, molar mass is needed.

1.23 mol NH3 x 17.04 g NH3 = 21.0 g NH3
1 mol NH3
Solution:

First, we need to convert 5.00 L of Nitrogen to L of NH3 by using the volume

ratio based on the balance chemical reaction. The volume ration of N2 and NH3 is
1L:2L.
= 5.00 L N2 x 2 L NH3 = 10.0 L NH3
1 L N2
Solving for the moles NH3, the ideal gas equation is used, PV = nRT.
Rearranging to get n:
n = PV
RT
n = 3.00 atm x 10.0 L____ = 1.23 mol NH3
(0.0821 L ● atm/mol ● K)(298 K)

To convert moles NH3, molar mass is needed.

1.23 mol NH3 x 17.04 g NH3 = 21.0 g NH3
1 mol NH3