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1st Semester - Module 3
Introductory Message
For the facilitator:
This module was collaboratively designed, developed and evaluated by the Development and
Quality Assurance Teams of SDO TAPAT to assist you in helping the learners meet the
standards set by the K to 12 Curriculum while overcoming their personal, social, and economic
constraints in schooling.
As a facilitator, you are expected to orient the learners on how to use this module. You also
need to keep track of the learners' progress while allowing them to manage their own learning.
Furthermore, you are expected to encourage and assist the learners as they do the tasks
included in the module.
For the learner:
This module was designed to provide you with fun and meaningful opportunities for guided
and independent learning at your own pace and time. You will be enabled to process the
contents of the learning resource while being an active learner.
The following are some reminders in using this module:
1. Use the module with care. Do not put unnecessary mark/s on any part of the module.
Use a separate sheet of paper in answering the exercises.
2. Don’t forget to answer Let’s Try before moving on to the other activities included in the
3. Read the instruction carefully before doing each task.
4. Observe honesty and integrity in doing the tasks and checking your answers.
5. Finish the task at hand before proceeding to the next.
6. Return this module to your teacher/facilitator once you are through with it.
If you encounter any difficulty in answering the tasks in this module, do not hesitate to
consult your teacher or facilitator. Always bear in mind that you are not alone.
We hope that through this material, you will experience meaningful learning and gain deep
understanding of the relevant competencies. You can do it!
Let’s Learn
Compounds react to one another just as how elements combine chemically and
through these interactions, different new compounds are formed. In this module, you will
discover how to write these chemical reactions to fully understand what happens during the
interactions of elements to compounds and/or compounds to another compounds. You will
also learn how to balance these equations in preparation for mass relationships. Hence, you
will use your knowledge in interpreting balanced chemical equations in terms of particles,
moles, and mass in making molar relationships from one substance to another.
This module has the following lessons:
 Lesson 1: Balancing Chemical Equations
 Lesson 2: Amounts of Substances in a Chemical Reaction
After going through this module, you are expected to:
1. write and balance chemical equations; and
2. construct mole or mass ratios for a reaction in order to calculate the amount of reactant
needed or amount of product formed in terms of moles or mass.
Let’s Try
Access the pre-assessment in the link below:
(Please note that an updated school link will be given by your subject teacher to access
the pre-assessment file.)
Writing and Balancing Chemical
Chemical reactions are happening every time everywhere, and for us to understand
what transpires during these interactions we must be able to write and balance chemical
Let’s Recall
1. What is the difference between an empirical formula and a molecular formula?
2. Dichloroethane, a compound that is often used for dry cleaning, contains carbon,
hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that
it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?
Let’s Explore
Spot the difference. Directions: Encircle the differences you were able to locate on these
Image 1
Image 2
1. What makes image 1 different from image 2?
2. What similarities do these two images have?
3. How can this be related to chemistry?
Let’s Elaborate
In the previous module, you learned that elements combine to form different
compounds. These elements and compounds may then be represented using symbols and
formulas, respectively. You also learned that each substance has its own mass which can be
calculated using important information seen in the periodic table. In this lesson you will learn
what happens when elements and compounds interact with one another and be able to
describe it using words and chemical symbols or formulas.
Before writing and/or classifying chemical reactions, you must know how to identify if
a chemical reaction occurred. There are several evidences that can help you prove its
occurrence. Here are some of the evidences and their examples (Brooks 2018;
Table 3.1 Evidences of chemical reactions
Change in color
a half-eaten apple turns brown after exposure to air
Change in odor
food slowly spoils and smells rotten
Formation of a solid or
the formation of deposits when calcium oxides mix with water
and clogs pipes
Evolution of gas (bubble
an antacid is dropped in a glass of water and begins to produce
Change in temperature
is when a burning fire produces heat
Production of light
is the chemical reaction in the bodies of fireflies that allows them
to produce light
Volume change
when magma inside a volcano expands rapidly due to chemical
reactions that happen inside the volcano
Here are the five major types of chemical reactions (Licuanan 2016; Helmenstine 2020):
A. Synthesis Reactions
When two or more elements or compounds combine to form a new substance, we call
this reaction as a synthesis reaction. It is also known as the combination reaction. The
general equation for this type of reaction is:
A + B → AB
2C(s) + O2(g) → 2CO(g)
B. Decomposition Reactions
In a decomposition reaction, a reactant (compound) breaks into two or more
products (an element or another compound). It is the opposite of a synthesis reaction. The
simplest decomposition reaction occurs when a compound breaks into its component element.
A complex decomposition, on the other hand, is resulting from the breaking down of
compounds into another compounds. The general equation for this reaction is:
AB → A + B
CaCO3(s) → CaO(s) + CO2(g)
C. Single Displacement Reaction
In a single displacement reaction, a free
element replaces another element in a compound.
This free element is more reactive than the element it
replaces. This reaction is also known as substitution
reaction. The general form of this reaction is:
A + BC → AC + B
Mg(s) + Cu(NO3)2(aq) →
Mg(NO3)2(aq) + Cu(s)
Note that not all single displacement reaction
Figure 3.1: Reactivity series of some
produces result. It depends on the reactivity of the
common metals [1]
elements involved in the reaction. Normally, a more
reactive element replaces the less reactive element in the reaction. This means that in cases
where the element to be replaced is more reactive than the free element, a reaction will not
occur. To help you determine whether a single displacement reaction will occur or not, a table
of reactivity series of some common elements is provided in Figure 3.1.
D. Double Displacement Reaction
A double displacement reaction, also known as metathesis reaction, happens when
two ionic compounds exchange ions. A positive ion from the first compound will combine with
the negative ion of the second compound, and the positive ion of the second compound
combines with the negative ion of the first. This reaction usually happens when the compound
is in an aqueous solution. The general form of this equation is:
AB + CD → AD + CB
2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s)
E. Combustion Reaction
This is an important classification of chemical reactions. In a combustion reaction a
compound or element reacts with O2 to form oxides of the element. A carbon-containing
compound will form carbon dioxide (CO2), when combusted. A sulfur-containing compound
that reacts with oxygen will form sulfur dioxide (SO2). The most common combustion reaction
is the combustion of hydrocarbons. Hydrocarbons contain C, and H, so when completely
combustion happens it will form carbon dioxide and water as byproducts. The general formula
for this reaction is
Hydrocarbon + O2 → CO2 + H2O
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
The process in which substances interact with one another to form new substances is
called a chemical reaction. The substance that undergoes a chemical reaction is called a
reactant while the newly formed compound is known to be the product. For us to be able to
understand chemical reactions, chemists have devised a way to represent them. They used
chemical equations. Chemical equations use chemical symbols to show what happens to
substances during chemical reactions. These equations come in different forms and can
convey a great deal of information. (Chang and Goldsby 2016; Ebbing and Gammon 2009).
A. Word Equations
A chemical reaction may be written using the names of the substances involved. This
equation is called a word equation. (Mustoe et al. 2011)
sodium reacts with chlorine to form sodium chloride
Word equation:
sodium + chlorine → sodium chloride
B. Skeleton Equations
A skeleton equation uses chemical formulas to simplify the chemical equation. It uses
different signs and symbols that are used to illustrate the chemical equation (Mustoe et al.
2011). Let us use again the example chemical reaction mentioned above:
sodium reacts with chloride to form sodium chloride
Skeleton equation:
Na + Cl2 → NaCl
In both equations, “+” means combine or reacts with and “→” means to form. The
formulas on the left side of the arrow represents the reactants or the starting substance in a
chemical reaction. The formulas on the right side refer to the products or results from a
reaction. However, this equation is incomplete, if you look closely the number of atoms of
chlorine in the reactant side is not the same as the number of its atoms on the product side.
To be consistent with the Law of Conservation of Mass which states that mass is neither
created nor destroyed, the number of atoms of each type of element on both sides must be
the same. Thus, we need to balance the equation, and this is done by placing the appropriate
coefficient in front of the compounds or elements. These coefficients give the relative number
of molecules or formula units involved in the reaction. (Chang and Goldsby 2016; Flowers et
al. 2018). It is understood that 1 is the coefficient when there is no number written in front of
the formula. When we place a coefficient in the equation above, it will look as:
2Na + Cl2 → 2NaCl
After placing the coefficients, you may see that each
atom on both sides are of the same number. Sometimes, it
is also useful to indicate the states of the substances in an
equation. There are appropriate labels to indicate them.
These are placed within parentheses following the chemical
formula (refer to Table 3.2 for the list of symbols). After
placing the phase labels, the previous equation will become:
2Na(s) + Cl2(g) → 2NaCl(s)
In the equation, you can also indicate the conditions
under which the reaction takes place. If the reaction is
heated, you may place ∆ (capital Greek delta) over the
arrow. If a catalyst, a substance that speeds up the reaction
without being used up in the reaction, is involved you may
write it over the arrow. See examples below:
Table 3.2: Symbols used in
Chemical Equations
(s) or ↓
(g) or ↑
2NaNO2(s) + O2(g)
2H2O(l) + O2(g)
reacts with
(reactant side)
and (product
to yield
solid or
in aqueous
a reaction is
Note that platinum, Pt, is not the only catalyst that is
used in chemical reactions. Other catalysts are iron, nitrogen (II) oxide, nickel, and many
others. In cases that both heat and catalyst are used, you may combine the two symbols by
placing one above the arrow and the other below the arrow.
When the coefficients in a chemical equation are correct, the number of atoms of each
element are equal on both sides, then the equation is said to be balanced. When writing a
chemical equation, one must identify first the identity of each of the substances involved. After
the identities are identified, chemical formulas are written and then assembled in a sequence
– reactants separated by an arrow from products. In this point, the equation that you wrote is
likely to be unbalanced. This means that the number of atoms in the reactant side is not the
same with the number of the same atoms in the product side. You must remember that in a
chemical reaction, the atoms simply recombine but none are created nor destroyed. Generally,
we can balance a chemical equation by following these steps (Chang and Goldsby 2016):
STEP 1: Write the skeleton equation by identifying all reactants and products.
STEP 2: Balance the equation by changing the coefficients of the reactants or products. You
may only change the coefficient and not the subscripts since it will change the identity of the
substance. You must also remember the following:
1. Use the smallest whole number coefficient.
2. Leave hydrogen, oxygen, and other elements until later.
3. Treat polyatomic ions as one unit, example phosphate, PO 4-3.
4. Balance last the elements that occur in its uncombined state, example Na or H 2.
STEP 3: Check your balanced equation. Make sure that the total number of each type of atoms
on both sides are equal.
Let us consider the example below:
Potassium nitrate and solid copper(II) hydroxide are formed from the reaction between
copper(II) nitrate and potassium hydroxide. Write the balanced chemical equation for this
(1) Identify which are the reactants and products then write the skeleton equation:
Cu(NO3)2(aq) + KOH(aq) → Cu(OH)2(s) + KNO3(aq)
(2) Plan your strategy. Since all formulas involve polyatomic ions (NO 3-, then OH-), you are to
balance them first. Then check if it is balanced, if not try to balance copper and potassium
(3) Act on your strategy. There are two NO3- ions on the reactant side and one on the product
side, you may place 2 in front of KNO3.
Cu(NO3)2(aq) + KOH(aq) → Cu(OH)2(s) + 2KNO3(aq)
Likewise, there are two OH- ions on the product side, so place 2 in front of KOH.
Cu(NO3)2(aq) + 2KOH(aq) → Cu(OH)2(s) + 2KNO3(aq)
(4) Check to see if the number of each type of atom on each side of the equation is balanced.
Cu(NO3)2(aq) + 2KOH(aq) → Cu(OH)2(s) + 2KNO3(aq)
Atoms or Ions
Reactant side
Product side
(5) Since the number of each type of atom on each side is balanced, therefore the chemical
equation is balanced.
Note that the steps in balancing mentioned above is balancing by inspection which is
also a trial-and-error method. There is no definite step in balancing, but the listed steps above
may be considered as hints or reminders to help us in balancing (Chang and Goldsby 2016;
Licuanan 2016; Flowers et al. 2018).
What is the purpose of balancing chemical equations? Balancing equations in
chemistry gives us the correct proportions of the reactants and products. It will tell you how
much product will be produced for a given chemical reaction using the given set of reactants.
Your knowledge in balancing is also essential when solving mass relationships and other
computations involving chemical reactions.
A balanced chemical equation can give us a great deal of information about the
chemical reaction it represents. Let us look at the example below:
One molecule
One mole
Three molecules
Three moles
Two molecules
Two moles
2(14.007 g)
= 28.014 g
3(2.016 g)
= 6.048 g
2(17.031 g)
= 34.062 g
34.062 g
34.062 g
Looking at the example above we may say that the balanced equation followed the
Law of Conservation of Mass which says that total mass of reactants is equal to the total mass
of the product after chemical reaction. This information will then be used for many purposes.
Let’s Dig In
Directions: Write a skeleton equation with proper labels for each reaction then classify which
type of chemical reaction it represents.
1. Solid zinc reacts with chlorine gas to form solid zinc chloride.
2. Aqueous lead (II) nitrate reacts with solid magnesium to yield aqueous
magnesium nitrate and solid lead.
3. Aqueous barium chloride reacts with aqueous sodium sulfate to yield an aqueous
sodium chloride and barium sulfate as precipitate.
4. Liquid ethanol reacts with oxygen gas to produce carbon dioxide and water
5. Solid iron reacted with oxygen gas to form solid ferric oxide
Directions: Balance the following equations:
1. __C2H6(g) + __O2(g)
2. __Be2C
+ __H2O
__Be(OH)2 +
3. __NH3(g) + __O2(g)
4. __Ca3(PO4)2(aq)
5. __Fe2(SO4)3 + __NH3 +
+ __(NH4)2SO4
Let’s Remember
Chemical reactions can be represented using chemical equations. These chemical
equations are then balanced to satisfy the Law of Conservation of Mass. When balancing only
the coefficients in front of each chemical formula are changed. Chemical equations can be
classified as synthesis, decomposition, single displacement, double displacement and
combustion reaction.
Let’s Apply
Directions: Accomplish the following, write your answers in a clean bond paper.
A. Look around inside or outside your house. Identify five (5) activities that involve a
chemical reaction. Once identified, describe what evidence/s you saw that helped
you in determining the occurrence of chemical reactions.
B. Using the activities that you wrote in A., write a chemical equation that may represent
this reaction. Balance and classify the chemical equation.
Amounts of Substances in a
Chemical Reaction
Balanced chemical equations are used by to solve mass relationships in chemical
reactions. There are different ways that we may find these balanced chemical equations
useful, and one of it is that it makes stoichiometric calculations easier.
Let’s Recall
1. How do we balance chemical equation?
2. Why do we need to balance chemical equations?
3. Balance the following equations:
(Example: B2Br6 + 6 HNO3 → 2 B(NO3)3 + 6 HBr)
a. __Na3PO4 + __HCl → __NaCl + __H3PO4
b. __TiCl4 + __H2O → __TiO2 + __HCl
c. __NH3 + __O2 → __NO + __H2O
d. __Fe + __HC2H3O2 → __Fe(C2H3O2)3 + __H2
e. __Hg(OH)2 + __H3PO4 → __Hg3(PO4)2 + __H2O
Let’s Explore
SCENARIO: You and your groupmates were tasked to make 100 cookies for a school event.
You searched on the internet for the recipe and here is what you got:
4 tbsp unsalted butter
1/3 cup brown sugar
1/3 cup white, granulated sugar
1 egg
1 cup All Purpose Flour
1 tbsp cornstarch
½ tsp baking soda
¼ tsp Kosher salt
1. Create an “equation” based on these
2. What is the ratio of each ingredient to another?
3. How many cookies will you be able to make if
you only have 6 cups of sugar?
4. How many ingredients will you need to make
all 100 cookies?
Makes 20 cookies
Let’s Elaborate
As you learned from the previous lesson that a balanced chemical equation gives a lot
of information about the reaction that took place. It relates the amounts of substances in a
reaction. The coefficients you write in front of the chemical formulas represents number of
moles as well as particles. It also provides the relative amounts of both reactant and product
allowing you to be able to do quantitative assessments on both chemical species. Thus, you
may use this information to relate one substance to another quantitatively. The quantitative
study of reactants and products in a chemical reaction is called stoichiometry (Chang and
Goldsby 2016; Licuanan 2016; Flowers et al. 2018).
In stoichiometry we use moles to calculate the amount of product formed whether the
units of reactants (or products) are moles, grams, or liters. This approach is called the mole
method, which means that the stoichiometric coefficients can be interpreted as the number of
moles of each substance. And we use these moles to relate reactants to products and vice
Basically, stoichiometry is a similar concept to how people go about different activities.
Let us say for example, when you prepare food. A simple recipe for one ham sandwich
requires two slices of bread, a slice of ham, a slice of cheese and a slice of tomato. Just as
what you did in Activity 1, the recipe may be represented by the following “equation”:
2 slices of bread + 1 slice of ham + 1 slice of cheese + 1 slice of tomato = 1 ham sandwich
Suppose that you have 12 slices of ham, how many slices of bread will you need to
use all the ham? Looking at the “equation” above, the ratio of bread to ham is 2:1, and we may
use it to compute as follows:
2 slices of bread
12 slices of ham x
= 24 slices of bread
1 slice of ham
Balanced chemical equations are used in the same manner, we use it to determine the
amount of one reactant required to react with a given amount of another reactant or needed
to produce a given amount of product, and so on. The coefficients are used as stoichiometric
factors or molar ratios to be able to compute for a desired quantity (Flowers et al. 2018). To
illustrate, consider the equation below:
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
The stoichiometric coefficients show that four molecules of Al react with three
molecules of O2 to produce 2 molecules of Al2O3. This equation can be interpreted as:
4 Al(s)
4 molecules
4 (6.022 x 1023 molecules)
3 O2(g)
3 molecules
3 (6.022 x 1023 molecules)
4 mol
3 mol
2 Al2O3(s)
2 molecules
2 (6.022 x 1023
2 mol
This equation may also be read as “4 moles of Al metal combines with 3 moles of O 2
gas to form 2 moles of solid Al2O3.” It shows that aluminum oxide molecules are produced
from oxygen molecules in a 2:3 ratio. It also says that 2 moles of Al2O3 are stoichiometrically
equivalent to 3 moles O2 and to 4 moles of Al. Looking at the balanced chemical equation we
may write molar ratios as:
2 Al2O3 molecules
2 mol Al2O3
3 O2 molecules
3 mol O2
2 Al2O3 molecules
4 Al molecules
2 mol Al2O3
4 mol Al
3 O2 molecules
4 Al molecules
3 mol O2
4 mol Al
These molar ratios can be used to compute the amount of aluminum oxide molecules
produced form a given amount of oxygen molecules or from a given amount of aluminum
molecules. Likewise, we may also use these molar ratios to compute for the required amount
of aluminum molecules to produce a given amount of aluminum oxide molecules and so on.
Let us use this as an example:
Using the balanced chemical equation:
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
Molar ratios
4 moles Al : 3 moles O2 : 2 moles Al2O3
a. How many moles of Al2O3 will be produced if 12.1 moles of O2 reacts completely with
Al? (moles O2 → moles Al2O3)
2 moles Al2O3
12.1 moles O2 x
= 8.07 moles Al2O3
3 moles O2
b. How many moles of Al are needed to produce 50.7 moles Al2O3?
(moles Al2O3 → mol Al)
4 moles Al
50.7 moles Al2O3 x
= 101 moles Al
2 moles Al2O3
c. How many grams of Al2O3 will be produced from 36.1 moles O2?
(moles O2 → moles Al2O3 → grams Al2O3)
2 moles Al2O3
101.96 g Al2O3
36.1 moles O2 x
= 2.45 x 103 g Al2O3
3 moles O2
1 mole Al2O3
d. How many grams of Al2O3 will be produced from 82 g of Al as it completely reacts with
O2? (grams Al → moles Al → moles Al2O3 → grams Al2O3)
1 mole Al
2 moles Al2O3
101.96 g Al2O3
82 g Al x
= 150 g Al2O3
26.98 g Al
4 mole Al
1 mole Al2O3
Note that to convert grams to moles requires the atomic weight of the element and/or
molar mass of the compound (sum of the atomic weights of each element in the compound).
Examples 1c and 1d, show us how these atomic weights are used to convert grams to moles
and vice versa.
Certain race cars use methanol, also called as wood alcohol, as a fuel. In a reaction, 28.83 g
of methanol are reacted with an excess of O2. Calculate the amount in grams of CO2 formed.
To solve this sample problem, remember to start by writing and balancing the chemical
equation first. This will always be your first step in every stoichiometric calculation.
(1) Balanced equation:
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l)
(2) Molar Ratios:
2 moles CH3OH : 3 moles O2 : 2 moles CO2 : 4 moles H2O
(3) Required:
Number of molecules of CO2
(4) Solution:
g CH3OH → moles CH3OH → mole CO2 → g CO2
1 mole CH3OH
2 moles CO2
44.01 g CO2
28.83 g
= 39.60 g CO2
32. 04 g CH3OH
2 moles CH3OH
1 mole CO2
From examples 1 and 2, you may see that you may convert moles of one substance
to moles of another substance, moles of the reactant to mass of the product, grams of the
reactant to the grams of the product and so on. All these conversions are made easy by the
stoichiometric factors or molar ratios. These molar ratios we construct from the coefficients of
the balanced chemical equation is like a gateway to converting one unit of a substance to the
desired unit of another substance. But these are not the only units that we could convert into
using stoichiometry, look at the example below:
The reaction of iron (III) oxide with powdered aluminum is known as the thermite reaction.
A. Calculate the mass of aluminum oxide, Al2O3, that is produced when 1.65 x 1022 atoms
of Al reacts with Fe2O3.
B. Find how many formula units of Fe2O3 are needed to react with 0.214 g of Al?
This example involves the conversion of mass to particles (atoms, molecules, ions, or
formula units) therefore we will be using our knowledge of the mole concept in this problem.
You learned that one mole of a substance is equal to 6.022 x 10 23 particle units. The number
6.022 x 1023 is also known as the Avogadro’s number. You can use this number or constant
to convert between mass and number of particles. Let us now solve the given problem.
(1) Balanced equation:
(2) Molar ratio:
(3) Required in A:
(4) Solution for A:
1.65 x 1022 atoms Al
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)
2 moles Al : 1 mole Fe2O3 : 1 mole Al2O3 : 2 moles Fe
mass of Al2O3
(atoms Al → moles Al → moles Al2O3 → g Al2O3)
1 mole Al
1 moles Al2O3
101.96g Al2O3
6.022 x 10 atoms x
2 moles Al
1 mole Al2O3
(5) Required in B:
(6) Solution for B:
0.214 g Al
1 mole Al
26.98 g Al
1.40 g Al2O3
formula units of Fe2O3
(g Al → moles Al → moles Fe2O3 → formula units Fe2O3)
1 moles Fe2O3
2 moles Al
6.022 x 1023 formula units Fe2O3
1 mole Fe2O3
= 2.39 x 1021 formula units Fe2O3
The general process for solving stoichiometric problems is summarized using the map in
Figure 3.2 below:
Figure 3.2: Stoichiometry Map
Retrieved from:
Let’s Dig In
Directions: Answer the following. Write the final answers in the correct significant figures.
1. Consider this reaction: Ba3N2 + 6 H2O → 3 Ba(OH)2 + 2 NH3
a. What is the ratio of Ba3N2 molecules to H2O molecules to Ba(OH)2 molecules to NH3
b. How many molecules of Ba3N2 is required to react with H2O to form 50 molecules of
NH3 based on your ratio in part (a)?
c. How many molecules of H2O are required to react completely with 6.022 x 10 23
molecules of Ba3N2?
d. How many moles of Ba(OH)2 will be formed from the complete reaction of 48.3 moles
2. Consider this reaction: 4 FeS + 7 O2 → 2 Fe2O3 + 4 SO2
a. What is the ratio of FeS molecules to O2 molecules to Fe2O3 molecules to SO2
b. How many molecules of SO2 will be produced from 16 molecules of FeS?
c. How many molecules of O2 is required to react with FeS to yield 14 molecules of
d. How many moles of FeS is required to produce 18.9 g of Fe2O3?
Directions: Write the balanced chemical equations of the following problems and compute for
the required values. Show your solutions for each item. Answers must be in the correct
significant figures.
1. Phosphorus pentachloride, PCl5, reacts with water to form phosphoric acid, H 3PO4, and
hydrochloric acid, HCl.
a. What mass of PCl5 is needed to react with an excess quantity of H2O to produce
33.9 g of H3PO4?
b. How many molecules of H2O are needed to react with 4.07 g of PCl5?
2. Butane, C4H10, burns with oxygen in air to give carbon dioxide and water.
a. What is the amount of water produced from 0.35 mol butane?
b. How many grams of carbon dioxide will be produced if 34.0 g of butane is burned
Let’s Remember
Stoichiometric computations are made easy through the molar ratios we can
obtain from the correct balanced chemical equations of a given reaction. We can use this ratio
to convert moles of reactants to moles of product, moles of reactant to grams of product, grams
of reactant to grams of product and more.
Let’s Apply
Directions: From the following list of household activities, state how stoichiometry is being
used. Give an example on how stoichiometric calculations are applied in each given activity
and briefly narrate how important these calculations are for the activity. Use the table below
to organize your answers. Write your output in a clean short bond paper.
A. Washing clothes
B. Making coffee
C. Preparing vegetable salad
Household activity
Washing clothes
Making coffee
vegetable salad
How stoichiometry
is used.
Example of
Importance of
Let’s Evaluate
Access the post assessment in the link below:
(Please note that an updated school link will be given by your subject teacher to access
the post assessment file.)
Let’s Extend
Objective: To inform other people about the relation of different types of chemical reactions
to daily life and how they can affect us.
1. Create an interesting poster about a type of chemical reaction of choice.
2. The poster must include the following parts:
a. Title of Poster
b. Definition of chemical reaction
c. Definition of the type of chemical reaction chosen.
d. The general form of the chemical reaction.
e. Give 2 real-life examples of this reaction
f. Images/Drawings of the example
g. Balanced chemical equations
h. Discussion of its effect to the body, the environment, the industry and/or
nature in general.
i. Your name and section
j. References/Sources
3. You may opt to choose to create:
a. an e-poster (that you will be sending to your teacher online)
b. a physical poster (handwritten or printed in a 17” by 28” paper / 2 long bond
papers taped together)
4. The finished product must be submitted at least a week after you finish this module.
5. You will be graded using the following rubric:
The poster includes
all required
elements as well as
All items of
importance on the
poster are clearly
labeled with labels
that are readable.
All required elements
are included on the
All but 1 of the
required elements are
included on the
Several required
elements were
Almost all items of
importance on the
poster are clearly
labeled with labels
that are readable.
Many items of
importance on the
poster are clearly
labeled with labels
that are readable.
Labels are too
small to view OR
no important
items were
Graphics –
All graphics are
related to the topic
and make it easier
to understand.
All borrowed
graphics have a
source citation.
All graphics relate to
the topic. One or two
borrowed graphics
a source citation.
Graphics do not
relate to the topic
OR several
graphics do not
have a source
The poster is
attractive in terms of
design, layout, and
All graphics are
related to the topic
and most make it
easier to understand.
borrowed graphics
have a source
The poster is
attractive in terms of
design, layout, and
The poster is
acceptably attractive
though it may be a bit
There are no
mistakes on the
There are 1-2
mistakes on the
There are 3-4
mistakes on the
The poster is
messy or very
poorly designed.
is not attractive.
There are more
than 4
mistakes on the
Reactivity Series of Metals. Accessed July 19, 2020
“Types of Chemical Reaction” Accessed June 23, 2020
Brooks, Rebekah. 2018. “Seven Things That Indicate a Chemical Change Is Occurring” Accessed
June 22, 2020.
Chang, R. & Goldsby, K. Chemistry. (12th ed.). New York: McGraw-Hill, 2016. PDF.
CK-12 Website. n.d. “Types of Chemical Reactions [Online Read]” Accessed June 23, 2020.
Ebbing, D. D. & Gammon, S. D. General Chemistry. (9th ed.). New York: Houghton Mifflin Company,
Flowers, P., K. Theopold, R. Langley & W.R. Robinson. 2018. Chemistry. Rice University: OpenStax
Helmenstine, Anne Marie, Ph.D. 2019. "Examples of 10 Balanced Chemical Equations." ThoughtCo.
Accessed June 22, 2020.
Helmenstine, Anne Marie, Ph.D. 2019. "What Is a Word Equation in Chemistry?" ThoughtCo.
Accessed June 23, 2020.
Helmenstine, Anne Marie, Ph.D. 2020. "Types of Chemical Reactions." ThoughtCo. Accessed June
23, 2020.
Licuanan, P.B. General Chemistry 1 - Teaching Guide for Senior High School. Quezon City:
Commission on Higher Education, 2016
Mustoe, F., C. Clancy, T. Doram, B. Heimbecker, M. Mazza and P. McNutly. McGraw-Hill Ryerson
Chemistry 11. Toronto: McGraw-Hill Ryerson, 2011.
Stoichiometry Flowchart. Accessed July 19 2020. (2017) "Chemical Change: Signs & Evidence." Accessed June 23, 2020
Cover photo is originally made by Victor G. Taleon ©2020
Other illustrations are originally made by Dioneda, M.A. J. ©2020
Development Team of the Module
Writer: Ma. Christina M. Dioneda – TSHS - Teacher II
Content Evaluators:
Elmer L. Belza Jr. – BNHS - Teacher II
Ian Luigie D. Ordoñez – GABHS – Teacher I
Jimmylin U. Sollano – UBNHS – Master Teacher I
Jennievive D. Dela Cruz – TSHS – Teacher II
Teresita L. Baltazar – SVNHS – Master Teacher II
Language Evaluator:
Wilhelmina C. Estrada – TSHS – Teacher II
Reviewer: Ruby N. Montefulca – TSHS – Head Teacher I
Illustrator: Ma. Alexandra Jade M. Dioneda and Victor G. Taleon
Layout Artists: Ma. Christina M. Dioneda, Elvira B. Bagacina and Jayson F. Antones
Management Team:
Dr. Margarito B. Materum - SDS
Dr. George P. Tizon – SGOD Chief
Dr. Ellery G. Quintia – CID Chief
Dr. Marivic T. Almo – EPS Science
Quinn Norman O. Arreza, J.D. SHS Focal
Dr. Daisy L. Mataac – EPS – LRMS/ALS
For inquiries, please write or call:
Schools Division of Taguig City and Pateros Upper Bicutan Taguig City
Telefax: 8384251
Email Address: [email protected]