Government Property
NOT FOR SALE
NOT
General Chemistry 1
Quarter 1 - Module 4
Mass Relationships in Chemical Reactions
(design your own cover page)
Department of Education ● Republic of the Philippines
Senior High School
General Mathematics- Grade 11
Alternative Delivery Mode Quarter 1 - Module 4: Quantifying Chemical Reactions
First Edition, 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any
work of the Government of the Philippines. However, prior approval of the government
agency or office wherein the work is created shall be necessary for exploitation of such
work for profit. Such agency or office may, among other things, impose as a condition
the payment of royalty.
Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names,
trademarks, etc.) included in this book are owned by their respective copyright holders.
Every effort has been exerted to locate and seek permission to use these materials
from their respective copyright owners. The publisher and authors do not represent
nor claim ownership over them.
Published by the Department of Education – Division of Cagayan de Oro Schools Division Superintendent: Dr. Cherry Mae L. Limbaco, CESO V
Development Team of the Module
Author/s: Kim Charies L. Okit, Ma. Doris P. Napone Reviewers: Illustrator and Layout Artist:
Management Team Chairperson: Dr. Arturo B. Bayocot, CESO III Regional Director Co-Chairpersons: Dr. Victor G. De Gracia Jr. CESO V Asst. Regional Director
Cherry Mae L. Limbaco, PhD, CESO V Schools Division Superintendent
Alicia E. Anghay, PhD, CESE Assistant Schools Division Superintendent
Mala Epra B. Magnaong, Chief ES, CLMD Members Neil A. Improgo, EPS-LRMS Bienvenido U. Tagolimot, Jr., EPS-ADM Lorebina C. Carrasco, OIC-CID Chief Ray O. Maghuyop, EPS-Math Joel D. Potane, LRMS Manager Lanie O. Signo, Librarian II Gemma Pajayon, PDO II
Printed in the Philippines by
Department of Education – Division of Cagayan de Oro City
Office Address: Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro
Telefax: (08822)855-0048
E-mail Address: [email protected]
Senior High School
General Chemistry 1
Quarter 1 - Module 4
Mass Relationships in
Chemical Reactions
This instructional material was collaboratively developed and reviewed
by educators from public and private schools, colleges, and or/universities. We
encourage teachers and other education stakeholders to email their feedback,
comments, and recommendations to the Department of Education at action@
deped.gov.ph.
We value your feedback and recommendations.
Department of Education ● Republic of the Philippines
Senior High School
This page is intentionally blank
Table of Contents
What This Module is About ....................................................................................................................... i
What I Need to Know .................................................................................................................................. ii
How to Learn from this Module .............................................................................................................. iii
Icons of this Module ................................................................................................................................... iii
What I Know .................................................................................................................................................iv
Lesson 1: Determining the Amount of Reactant and Product in a Chemical Reaction ..................................................................................................................... 1
What’s In .................................................................................................. 1
What I Need to Know ................................................................................ 1
What’s New: Fishball anyone? .................................................................. 2
What Is It: Stoichiometry .......................................................................... 2
What’s More: Stoich In Action .................................................................. 4
What I Have Learned: Tell Me What You Know ........................................ 4
What I Can Do: Show Me What You Know ............................................... 5
Additional Activity: Let’s Go Online ........................................................... 5
Lesson 2:
Limiting and Excess Reagent ....................................................................................... 6
What’s In ................................................................................................................ 6
What I Need to Know ........................................................................................... 6
What’s New: Excess Baggage ........................................................................ ..7
What Is It: Limiting and Excess Reagent ................................................. 7
What’s More: More Than Enough ............................................................ 9
What I Have Learned: Tell Me What You Know ……………………………. 9
What I Can Do: How Big is the Balloon … ............................................... 10
Additional Activity: Let’s Go Online ........................................................... 10
Lesson 3:
Calculating Theoretical Yield and Percent Yield ................................... 11
What’s In ............................................................................................................................ 11
What I Need to Know..................................................................................................... 11
What’s New: Encircle Me ......................................................................................... . 11
What Is It: Yield of the Reaction ....................................................................... 12
What’s More: Solve, Solve, Solve ..................................................................... 13
What’s More: Let’s Go Online ............................................................................ 14
What I Have Learned: Tell Me What You Know … ....................................... 14
What I Can Do: Step by Step .................................................................. 14
Summary ................................................................................................................................................... 15
Assessment: (Post-Test) ...................................................................................................................... 16
Key to Answers ......................................................................................................................................... 18 References .................................................................................................................... 19
This page is intentionally blank
What This Module is About
Chemical reactions are apparent in the things around us or in our daily activities. The
quantitative relationship of reactants and products in a chemical reaction is manifested for
example, when we bake a cake or bread, or cook a dish. We make sure that all the ingredients
are present and in correct proportions based from a recipe to make the desired end-product.
Same concept applies to chemical reactions. After learning how to balance a chemical
equation in module 3, this module will help you understand the mass relationships of reactants
and products in a chemical reaction.
The lessons contained in this module are as follows:
• Lesson 1: Determining the Amount of Reactant and Product in a Chemical Reaction.
• Lesson 2: Limiting and Excess Reagent.
• Lesson 3: Calculating Theoretical Yield and Percent Yield in a Reaction.
i
What I Need to Know
At the end of this module, you should be able to:
1. 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. (STEM_GC11MRIg-h-38);
2. Determine mass relationship in a chemical reaction (STEM_GC11MRIg-h-42).
3. Explain the concept of limiting reagent in a chemical reaction; identify the excess
reagent(s) (STEM_GC11MRIg-h-40);
4. Calculate percent yield and theoretical yield of the (STEM_GC11MRIg-h-39);
ii
How to Learn from this Module
To achieve the objectives cited above, you are to do the following:
• Take your time reading the lessons carefully.
• Follow the directions and/or instructions in the activities and exercises diligently.
• Answer all the given tests and exercises.
Icons of this Module
What I Need to This part contains learning objectives that
Know are set for you to learn as you go along the
module.
What I know This is an assessment as to your level of
knowledge to the subject matter at hand,
meant specifically to gauge prior related
knowledge
What’s In This part connects previous lesson with that
of the current one.
What’s New An introduction of the new lesson through
various activities, before it will be presented
to you
What is It These are discussions of the activities as a
way to deepen your discovery and under-
standing of the concept.
What’s More These are follow-up activities that are in-
tended for you to practice further in order to
master the competencies.
What I Have Activities designed to process what you
Learned have learned from the lesson
What I can do These are tasks that are designed to show-
case your skills and knowledge gained, and
applied into real-life concerns and situations.
iii
What I Know
Multiple Choice. Read and understand each question and select the letter of the best answer from among the given choices.
1. Which of the following equations best represent the law of conservation of mass? A. 1𝐴 + 2𝐵 → 3𝐴𝐵 C. 6𝐴2𝐵 → 12𝐴 + 𝐵
B. 2𝐴2 + 8𝐵 → 4𝐴𝐵2 D. 2𝐴𝐵3 → 2𝐴 + 2𝐵
2. Which of the following statements is true about stoichiometric coefficient? A. It is written to make the number of each element in a chemical equation the same
in the reactant and product side. B. the sum of all coefficients in the left must be equal to the sum of all coefficients in
the right side of the equation. C. It is the subscript written after each element. D. None of the above.
3. In a chemical reaction, stoichiometry refers to:
A. Amount of materials consumed and products formed. B. The activation energy C. The rate or reaction D. Reaction in equilibrium
For items 4 to 7, base your answer from the combustion of butane (C4H10) as shown in the reaction:
2𝐶4𝐻10 (𝑔) + 13𝑂2(𝑔) → 8𝐶𝑂2(𝑔) + 10𝐻2𝑂 (𝑙)
4. What is the mole-mole factor of butane and water? A. 10:2 C. 8:10 B. 2:13 D. 2:10
5. which pair has the mole-mole factor of 8:10? A. Butane and oxygen C. Oxygen and carbon dioxide B. Carbon dioxide and water D. Water and butane
6. If 65.3 moles of oxygen gas are consumed, how many moles of carbon dioxide is produced?
A. 526.4 moles C. 40.2 moles B. 106.1 moles D. 8.16 moles
7. How many grams of C4H10 is needed to produce 37.8 moles of CO2?
A. 453.6 grams C. 2.607 grams B. 548.1 grams D. 8769.6 grams
For items 8 to 11. In the reaction 2H2 + CO → CH3OH, 6.8 g carbon monoxide gas was made
to react with 7.2 g of hydrogen gas. The reaction produced 5.2 g of methanol.
8. Which is the limiting reagent? A. Hydrogen C. methanol B. Carbon monoxide D. CO2
iv
9. What is the theoretical yield? A. 57.6 g C. 7.8 g B. 50.4 g D. 6.8 g
10. What is the percent yield? A. 76% C. 68% B. 78% D. 67%
v
This page is intentionally blank
Lesson
Determining the Amount of Reactant and Product in a
Chemical Reaction
1
What’s In
You have learned from the previous lesson that numerical coefficients are written before a chemical formula in the reactants or products side in order to balance a chemical equation. This numerical coefficient in a balanced chemical equation will be used to determine how much of the reactants are needed to produce a certain amount of product or how much product is produced given a certain amount of reactant. This lesson will introduce you to the concept of mass to mole or mole to mole ratio to determine the amount of reactants and products in a chemical reaction.
What I Need to Know
A balanced chemical equation indicates the number of moles required for each reactant to produce a certain number of moles of the product/s. When you know the ratio of the product to the reactants, it will be much easier to determine how much of the initial materials are you going to prepare to achieve a specific amount of product. For instance, you are selling fishballs in a stick for 10 pesos per stick. Each stick has 6 fish balls. If you plan to sell 100 sticks of fishballs to reach a profit of 1000 pesos, excluding all other materials like cooking oil and fuel, you will know that you needed to buy 100 pieces of bamboo sticks and 500 pieces of raw fish balls from the market. Similarly, when you have a balanced chemical equation, you will be able to predict the amount of product/s formed from knowing the amount of the reactants. In the same manner, knowing the amount of product/s formed will help you determine the amount of initial reactant used up in a single chemical reaction. In this lesson, you will learn to identify mole ratios of reactants and products from balanced chemical equations and be able to perform stoichiometric calculations related to chemical equations.
1
What’s New
Fishball anyone? Directions: Let us use the given example previously to understand the concept of determining the amount of reactants and products in a chemical reaction by answering the questions below. Susan is selling fishballs in a stick for 10 pesos per stick. Each stick has 5 fish balls.
1. Write the equation of fish balls, bamboo stick and fishballs in a stick.
__________________ + ________________________ → ______________________________ 2. If Susan has 100 fish balls, how many bamboo sticks will she need to consume all the
fish balls? 3. If Susan wants to make 50 fishballs in a stick, how many fish balls will she need?
What Is It
Stoichiometry
One of the requirements of a balanced chemical equation is that it follows the Law of
Conservation of Mass, which states that matter is neither created nor destroyed. The identity
and quantity of the elements in the reactants side, though they can change in pairing or
arrangement, must be equal to the identity and quantity of elements in the products side. To
do this, all elements in the left side of the equation must be reflected, and of the same number
in the right side of the equation. A stoichiometric coefficient is then added before each
element, ion or molecule to make the number of each element in the left side equal to the
number of the same element in the right side of the equation. This stoichiometric coefficient
denoted by a number, can be interpreted as the number of moles of each substance. The
mole method approach makes stoichiometry (the quantitative relationship between
reactants and products in a chemical reaction) more understandable. Let’s take for example
the formation of table salt or NaCl:
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g) The stoichiometric coefficients in the equation denotes that 2 elemental sodium (Na)
react with 2 molecules of hydrochloric acid (HCl) to form 2 molecules of sodium chloride (NaCl)
and 2 molecules of hydrogen (H2) gas. Following the mole method approach, the equation can
be interpreted and read as 2 moles of Na react with 2 moles of HCl to form 2 moles of NaCl
and 1 mole of H2. Since the equation is balanced, the stoichiometric coefficient for the
reactants and products can be used in a ratio or mole-mole factor:
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎 ≅ 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
The symbol ≅ means “equivalent to”. Hence, we can make the following mole-mole
factor: 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙 or
2 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙 or
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎
1 𝑚𝑜𝑙𝑒 𝐻2 or
2 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙
1 𝑚𝑜𝑙𝑒 𝐻2 or
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎 or
1 𝑚𝑜𝑙𝑒 𝐻2
2 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙
Let’s consider a simple example. Ammonia, NH3, is a leading industrial chemical used
in the production of agricultural fertilizers and synthetic fibers. It is produced by the reaction of
nitrogen and hydrogen gases.
3 𝐻2(𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)
2
As shown in the balanced equation, 3 moles of H2 are stoichiometrically equivalent to
1 mole N2 and to 2 moles NH3. The ratio of moles H2 to moles NH3 in 3:2; the ratio of moles
N2 to moles NH3 is 1:2. Using this ratio, we will be able to calculate the quantities of the product
or reactant like the example below:
a. How many moles of H2 are needed to produce 26.5 moles of NH3?
Solution:
26.5 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 𝑋 3 𝑚𝑜𝑙𝑒𝑠 𝐻2
2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 39.8 𝑚𝑜𝑙𝑒𝑠 𝐻2
b. How many moles of NH3 will be produced if 33.7 moles of N2 reacts completely with H2?
Solution:
33.7 𝑚𝑜𝑙𝑒𝑠 𝑁2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3
1 𝑚𝑜𝑙𝑒 𝑁2 = 67.4 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3
Note: In using the mole-mole factor or ratio, the arrangement of the numerator and
denominator is done in a way that there is a cancellation of similar units found in the numerator
and denominator. In first example above, moles of NH3 were cancelled while in the second
example, moles of N2 were cancelled.
The stoichiometric coefficients of the reactants and products can be used readily to
determine the mole-mole ratio of the substances involved. However, in laboratory setups, the
mass (in grams) of the substance are given or needed instead of moles for easy measurement.
In this type of problem, there is a need to convert the mass into number of moles using the
molar mass of the substance. Below are the steps used to convert grams to moles or vice
versa:
1. Convert the mass of the substance (A) to number of moles using its molar mass.
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴(𝑔) 𝑋 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐴
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴 (𝑔)= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴
2. Using the number of moles of substance A and the mole-mole ratio of substance A
and B from the balanced equation, compute for the number of moles of B.
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴 𝑋 𝑚𝑜𝑙𝑒𝑠 𝐵
𝑚𝑜𝑙𝑒𝑠 𝐴= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵
3. Convert the number of moles of substance B to mass using its molar mass.
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵 𝑋 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵 (𝑔)
1 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵= 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵 (𝑔)
The molar mass of the substance is used as a conversion factor to calculate the
number of moles of the substance given its mass and vice versa. The three steps can be
combined into a single step as follows:
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵(𝑔) = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴(𝑔) 𝑋 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐴
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐴(𝑔) 𝑋
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴𝑋
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵 (𝑔)
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐵
In every conversion, like the steps shown above, similar units found in the numerator
and denominator are cancelled out. Let’s have another example: Solid lithium hydroxide is
used to remove carbon dioxide and is called a CO2 scrubber. The reaction is:
2 𝐿𝑖𝑂𝐻(𝑠) + 𝐶𝑂2(𝑔) → 𝐿𝑖2𝐶𝑂3(𝑠) + 𝐻2𝑂(𝑙)
3
How many grams of CO2 can be absorbed by 236.1 g LiOH?
Solution:
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑂2(𝑔) = 236.2 𝑔 𝐿𝑖𝑂𝐻 𝑋 1 𝑚𝑜𝑙𝑒 𝐿𝑖𝑂𝐻
23.949𝑔 𝐿𝑖𝑂𝐻 𝑋
1 𝑚𝑜𝑙𝑒 𝐶𝑂2
2 𝑚𝑜𝑙𝑒𝑠 𝐿𝑖𝑂𝐻𝑋
44.01𝑔 𝐶𝑂2
1 𝑚𝑜𝑙𝑒 𝐶𝑂2
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑂2(𝑔) = 217.0 𝑔
Note: Since the given value (236.1g) has 4 significant figures, the answer will also have 4 significant figures.
What’s More
Stoich in Action! Direction: Use a separate sheet to show your computation for the following problems:
1. Rust, Fe2O3, form from the reaction of iron and oxygen in the following equation: 𝐹𝑒(𝑠) + 𝑂2(𝑔) → 𝐹𝑒2𝑂3(𝑠)
a. Write the balanced equation of the reaction. b. What is the mole-mole ratio of Fe to Fe2O3? c. How many moles of Fe2O3 is produced from 39.4 moles of Fe? d. What is the molar mass of Fe2O3? e. How many grams of O2 are needed to produce 29.8 g of Fe2O3?
2. Given the decomposition reaction: 2 𝐾𝐶𝑙𝑂3(𝑠) → 2 𝐾𝐶𝑙(𝑠) + 3 𝑂2(𝑔)
a. What is the mole-mole ratio of KClO3 to O2? b. What is the molar mass of KClO3? c. How many moles of KClO3 are needed to produce 36.6 moles of O2? d. How many grams of KCl are produced from an initial mass of 568.4 g KClO3?
What I have learned
Tell me what you Know Directions: In a separate sheet of paper, write your comprehensive understanding of this lesson by answering the questions.
1. Make a flowchart for determining the mass of the product from a given mass of the reactant.
2. How does an imbalanced chemical equation affect your calculation of the amount of reactant or product in a chemical reaction?
3. Describe how you would determine the mole-mole factor or ratio from a chemical equation.
4. Explain and give an example of how you would arrange the units and ratio in the conversion of mass to mole of reactants to products.
5. Site a situation in your daily life where you can apply the concept of stoichiometry.
4
What I Can Do
Show Me What You Know Directions: Solve the following problems. Write your complete solution and answers on a separate sheet of paper. 1. The fertilizer ammonium sulfate is prepared by the reaction between ammonia and
sulfuric acid: 2𝑁𝐻3(𝑔) + 𝐻2𝑆𝑂4(𝑎𝑞) → (𝑁𝐻4)2𝑆𝑂4
a. What is the molar mass of NH3? b. What is the molar mass of (NH4)2SO4? c. What is the mole-mole ratio of NH3 to (NH4)2SO4? d. How many moles of ammonium sulfate is produced when 638.4 g of NH3 reacts
completely with H2SO4? e. How many grams of NH3 is needed to produce 50.0 g of (NH4)2SO4?
2. In the fermentation process, ethanol is produced from decomposition of glucose.
𝐶2𝐻12𝑂6 → 𝐶2𝐻5𝑂𝐻 + 𝐶𝑂2
Glucose Ethanol Carbon dioxide
a. Write a balanced chemical equation of the decomposition of glucose. b. What is the mole-mole ratio of glucose (𝐶2𝐻12𝑂6) to ethanol (𝐶2𝐻5𝑂𝐻) c. What is the molar mass of glucose? d. What is the molar mass of ethanol? e. How much ethanol is produced from the starting material of 987.1 g of glucose?
Additional Activity
Let’s go online! Directions: Go to the link below and practice what you’ve learned from this lesson:
https://bit.ly/2Ce7nkm https://bit.ly/2Ce7z34 https://bit.ly/2ZbtkJD
5
Lesson
Limiting and Excess Reagent
2
What’s In
You have learned from lesson 1 that you can predict the amount of product produced given the initial amount of reactant, the mole-mole factor and the molar mass. The amount of initial reactant required to produce the desired amount of product can also be computed in the same manner. In a real scenario however, the amount of the reactants involved in the reaction are not exactly available according to the proportion stoichiometrically identified from the balanced equation. How to determine which of the reactants is present in excess or which reactant is used up first in the reaction will be discussed in this lesson.
.
What I Need to Know Suppose you are preparing cheeseburgers following strictly the proportions: 1 burger bun, 1 patty and 1 slice of cheese to produce 1 cheeseburger. The number of cheeseburgers produced depends largely on the availability of materials. There could be instances where all the ingredients are present in exact proportion, so there are no excess materials. Oftentimes, the number of buns, cheese slices and patties are not the same. You may have bought more buns than patties or there are more cheese slices than buns. The ingredients present in the least amount will determine how many cheeseburgers will be produced while the ingredient present in excess will have leftovers after making cheeseburgers. In the same analogy, in a chemical reaction, the amount of reactants are not present in the exact stoichiometrically determined ratio. In this lesson, you will learn how to determine which reactant is present in excess and which reactant is used up first in a reaction.
6
What’s New Excess Baggage
Directions: The table below shows the primary materials needed to produce the items in the first column. With the given available materials, identify how many items are produced, which materials are in excess and how many are left of the available materials.
Item Required Available materials Produced set
Excess materials
Bicycle 1 bike frame, 2 tires 2 pedals, 1 crank arm, 1 brake set
68 bike frames, 117 tires, 250 pedals, 72 crank arm, 93 brake set
Banana Cue
2 bananas, 1 barbecue stick, ½ cup brown sugar
236 bananas, 150 barbecue stick, 25 cups brown sugar
Cheese burger
1 burger bun, 1 beef patty, 1 slice of cheese
324 burger buns, 12 dozens beef patties, 261 slices of cheese
Table 1 table top, 4 legs, 8 nuts, 8 screws
20 table tops, 50 legs, 50 nuts, 50 screws
Milk tea 1 bag of black tea, 250 ml water, 1/8 cup milk, 2 tbsp sugar, ¼ cup tapioca pearls
15 tea bags, 3 liters water, 2 cups milk, 30 tbsp sugar, 4 cups tapioca pearls
What Is It
Limiting and Excess Reagent
When a chemical reaction is carried out in a flask, the amount of reactants are not always present in the exact proportion stoichiometrically determined from the balanced equation. To make sure that all of the more expensive reagent is completely used up and converted to the desired product, chemists usually add the cheaper reagent in excess quantity. The reagent that has completely reacted and used up in a reaction is called the limiting reagent. The excess reagent is the reactant that is present in quantity higher than what is required to react with the limiting reagent. For instance, you are preparing a ball dance for your 18th birthday consisting of pairs of male and female. Upon checking your list of friends, you found out that you have 18 male friends and 25 female friends. The number of male friends will limit the number of pairs to 18. All the males will have a partner while there will be 7 females who will not have a partner for the dance. Let’s use that context to the balanced chemical equation below:
3𝐻2(𝑔) + 𝑁2 (𝑔) → 2𝑁𝐻3 (𝑔)
7
Ammonia, NH3, is synthesized from the reaction of H2 and N2 gases. Suppose 6 moles of H2 was initially mixed with 4 moles of N2 gas at high pressure. To determine which of the 2 reactants is the limiting reagent, the amount of NH3 produced must be computed given the number of moles of H2 and N2 and the mole-mole factor from the balanced equation.
𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = # 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3
3 𝑚𝑜𝑙𝑒𝑠 𝐻2
𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3
3 𝑚𝑜𝑙𝑒𝑠 𝐻2
𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 4 𝑚𝑜𝑙𝑒𝑠
𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3
1 𝑚𝑜𝑙𝑒 𝑁2
𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 4 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3
1 𝑚𝑜𝑙𝑒 𝑁2
𝑚𝑜𝑙𝑒𝑠 𝑁𝐻3 = 8 𝑚𝑜𝑙𝑒𝑠
From the computed values we determined that if 6 moles of H2 completely reacts with N2, it can produce 4 moles of NH3 while 4 moles of N2 can produce 8 moles of NH3 when fully used up. Since there is less amount of NH3 produced with 6 moles of H2 than 4 moles of N2, H2 gas is the limiting reagent while the N2 gas is the reagent in excess. To determine how much of the 4 moles of N2 is in excess, we will use the mole-mole factor of N2 and H2.
𝑚𝑜𝑙𝑒𝑠 𝑁2 = 6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2 𝑋 1 𝑚𝑜𝑙𝑒 𝑁2
3 𝑚𝑜𝑙𝑒𝑠 𝐻2
𝑚𝑜𝑙𝑒𝑠 𝑁2 = 2 𝑚𝑜𝑙𝑒𝑠
The number of moles of N2 required to fully react to 6 moles of H2 is only 2 moles. Thus, the initial 6 moles of N2 has an excess of 4 moles.
𝑒𝑥𝑐𝑒𝑠𝑠 𝑁2 = 6 𝑚𝑜𝑙𝑒𝑠 − 2 𝑚𝑜𝑙𝑒𝑠 = 4 𝑚𝑜𝑙𝑒𝑠
Let’s have another example. The combustion of ethane produces carbon dioxide and water shown in the reaction below:
2𝐶2𝐻6 + 7𝑂2 → 4𝐶𝑂2 + 6𝐻2𝑂
1. How many moles of CO2 is produced with 56.2 moles C2H6 and 73.4 moles of O2? Solution:
𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 = 56.2 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶2𝐻6 𝑋 4 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2
2 𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻6
𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 = 112.4 𝑚𝑜𝑙𝑒𝑠
𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 = 73.4 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂2 𝑋 4 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2
7 𝑚𝑜𝑙𝑒𝑠 𝑂2
𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 = 41.9 𝑚𝑜𝑙𝑒𝑠
2. Which reagent is the limiting reagent? Which is the excess reagent? Limiting reagent is O2 while the excess reagent is 𝐶2𝐻6 .
3. How many moles of 𝐶2𝐻6 is in excess?
Moles of 𝐶2𝐻6 required for 73.4 moles on O2:
𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻6 = 73.4 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂2 𝑋 2 𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻6
7 𝑚𝑜𝑙𝑒𝑠 𝑂2
𝑚𝑜𝑙𝑒𝑠 𝐶2𝐻6 = 20.9 𝑚𝑜𝑙𝑒𝑠
Excess 𝐶2𝐻6 : 56.2 moles – 20.9 moles = 35.3 moles 8
Note: If the amount of the initial reactant is expressed in grams (mass) instead of moles, the number of moles must first be converted into grams using the molar mass of the reactant as you have learned in the previous lessons. In the same manner, if the problem asks to determine the mass (in grams) of the product produced by the reaction, the number of moles of the product must be converted into grams using the molar mass of the product.
What’s More
More Than Enough Directions: Read carefully the following problems. Find what is asked for each problem. Write your complete solution and answer in a separate sheet of paper. 1. In the production of Zinc sulfide, 36.8 g of zinc is made to react with 19.4 g of sulfur.
𝑍𝑛 + 𝑆 → 𝑍𝑛𝑆 a. How many moles of ZnS is produced when sulfur is completely used up? b. How many grams of ZnS is produced when zinc is completely used up? c. Which reactant is the limiting reagent? d. How many grams of the excess reagent is left?
2. An aqueous solution of NaOH can dissolve an aluminum sheet in the reaction: 2 𝐴𝑙 + 2 𝑁𝑎𝑂𝐻 + 2𝐻2𝑂 → 2 𝑁𝑎𝐴𝑙𝑂2 + 3 𝐻2
If 126.4 g of NaOH is 97.7 g of Al are made to react,
a. How many moles of NaAlO2 is produced if 126.4 g of NaOH reacts completely? b. How many grams of NaAlO2 is produced using the same amount of NaOH? c. Which reactant is the limiting reagent? d. How many grams of the excess reagent is left?
3. Ferric chloride, FeCl3, reacts with silver nitrate, AgNO3 to form ferric nitrate, Fe(NO3)3 and
silver chloride, AgCl. a. Write the balanced chemical equation of the reaction. b. If 108.5 moles of FeCl3 is combined with 76.8 moles of AgNO3, how many moles of
AgCl is produced? c. How many grams of AgCl is produced? d. Which reactant is the limiting reagent? e. Which reactant is the excess reagent?
What I Have Learned
Tell Me What You Know
Directions: Based on what you have learned from this lesson, write a concise explanation of the following questions. Write your answer on a separate sheet of paper.
1. What is the role of the stoichiometric coefficient in a balanced chemical equation to the determination of the limiting reactant? 2. Why is it that in most cases, the amounts of reactants and products are indicated in grams instead of the number for moles. 3. How do you determine the limiting reactant in a chemical reaction? 4. If a chemical reaction involves only one reactant, will there be a limiting reagent? 5. Site one (1) real-life situation where you added more than what is needed to produce something. Make justifications why you have to do it.
9
What I Can Do How big is the balloon! Directions: This activity is a simple application of the concept of limiting and excess reactants that you can do at home. Follow the instruction properly, observe what happens to the balloon after mixing the baking soda and vinegar then answer the guide questions below. For this simple experiment, you will need the following items:
• 3 350mL empty plastic water bottles • Vinegar • Baking soda • 3 balloons • funnel
Procedure:
1. Collect, clean, and dry three (3)1 350 ml water bottles. Label the bottles 1, 2 and 3 2. Pour 5 tablespoons vinegar into bottle 1, 10 tablespoons into bottle 2 and 15
tablespoons into bottle 3. 3. Fit the funnel into the opening of the balloon and carefully put 1 tablespoon of baking
soda into each balloon. 4. Carefully press the balloon to remove extra air inside without spilling the baking soda. 5. Fit the balloon snugly into the lip or opening of the water bottle. Make sure that the
baking soda does not fall into the bottle while doing this. If the balloon is loosely fit into the bottle, use rubber bands to ensure that air is trapped inside the bottle.
6. Repeat steps 3 to 5 for the other 2 balloons and water bottles. 7. Stretch the balloon over the top of the bottle to slowly pour the baking soda into the
vinegar. The baking soda will react with vinegar as shown in the equation:
𝐶2𝐻4𝑂2 + 𝑁𝑎𝐻𝐶𝑂3 → 𝑁𝑎𝐶2𝐻3𝑂2 + 𝐻2𝑂 + 𝐶𝑂2 Acetic acid Sodium Sodium water Carbon
bicarbonate acetate dioxide
8. Record your observations then answer the questions below. 9. Clean your work area and dispose the materials properly.
After performing the activity. answer the following questions. Write your answer on a separate sheet of paper.
1. When baking soda was mixed with vinegar, effervescence (fizzing or bubbling) was observed. What caused the effervescence?
2. How can you conclude that the reaction has gone to completion? 3. Which balloon has the least amount of air inside? 4. Which balloon has the most air inside? 5. Will the balloon size grow bigger if more vinegar is in the bottle? Why? 6. Will it take more time to complete the reaction if there were more vinegar in the bottle
with the same amount of baking soda? 7. Which is the limiting reagent? 8. Which is the reagent in excess?
Additional Activity
Let’s go online! Directions: Go to the link below and practice what you’ve learned from this lesson: https://bit.ly/309SojJ
https://bit.ly/3fGoF8g https://bit.ly/3fui4ht
10
Lesson
Calculating Theoretical Yield and Percent
Yield in a Reaction
3
What’s In
In your previous lessons, you were taught on how to calculate molecular formula given
molar mass, write and balance chemical equations and construct mole or mass ratios for a
reaction with stoichiometry in order to determine the amount of reactants needed to form the
products.
In this lesson, you will learn how to differentiate the theoretical, actual, and percent
yield as well as the process on how to calculate them using stoichiometry.
What I Need to Know
Ideally, in a chemical reaction, it is predicted to produce a 100% yield of the product
from the given reactants. But in reality, it is not that easy to achieve. Most of the chemical
reactions obtain are less than the 100% yield of the product/s due to several factors that could
affect the reaction process such as; experimental errors, incomplete reactions, unexpected
side reactions, amount of the reactant/s, undesirable by-product/s, other external factors, etc.
For you to be able to eliminate all of these contributory factors, you have to obtain an “ideal
environment” during the experimentation to achieve 100% yield, which is close to impossible
to attain under normal conditions.
In order to evaluate the success of a chemical reaction, you need to determine the
percent yield by calculating the theoretical yield and the actual yield of the reaction process.
What’s New
Encircle Me…
Direction: In this chemical equation, identify and encircle the following parts.
You may use another sheet of paper or you may answer directly on this page. That
is why the font is big so that it would be easier for you to encircle.
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
11
Direction of the
Reaction
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Product/s
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Reactant/s
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Subscripts
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
Coefficients
2NH4NO3(s) → 2N2(g) + 4H2O(g) + O2(g)
What Is It
Yield of the Reaction
Chemical reaction is a process in which substance/s interact chemically to produce a
new substance/s with different compositions. It is represented by a chemical equation.
Substance/s at the left side of the equation is/are called the reactant/s while the substance/s
located at the right side of the equation is/are called the product/s. Chemical properties of the
element or compound as reactant/s dictate the process in which an element or compound
undergoes changes during the reaction.
Evaluation of the percent yield is important to measure the success of a chemical
reaction. Percent yield is the ratio of the actual yield to the theoretical yield expressed as a
percentage. So, how are you going to compute for the percent yield of a chemical reaction?
You have to know the values of the theoretical yield and the actual yield of the reaction.
Following the formula of:
𝑷𝒆𝒓𝒄𝒆𝒏𝒕 𝒀𝒊𝒆𝒍𝒅 = 𝑨𝒄𝒕𝒖𝒂𝒍 𝒀𝒊𝒆𝒍𝒅
𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒀𝒊𝒆𝒍𝒅 𝑿 𝟏𝟎𝟎%
Theoretical yield is the amount of product that is expected to form based on
stoichiometry. It is the maximum amount of product produced from the given amount/s of
reactant/s. It is calculated based on the stoichiometry of the chemical reaction. While the
actual yield is the amount of product produced during the reaction. It is the amount of product
obtained after the actual reaction and it is normally lesser than the theoretical yield. It is
determined experimentally.
These are the steps in calculating the percent yield of a chemical reaction:
1. Balance the given chemical equation
2. Identify the limiting reactant and the excess reactant
12
3. Compute for the theoretical yield of the reaction
4. Calculate the percent yield
You can continue solving when asked:
a. Percent error
b. Amount of excess reactant
What’s More Activity 1: Solve, Solve, Solve…
Direction: Determine the theoretical yield and the actual yield, given the information in each
question. You must show your work, including units, through each step of the calculations.
Use separate papers for your answers for these set of problems. (These problems are adapted
and modified from https://bit.ly/2W18laC).
1. Cl2(g) + Al(s) AlCl3(s) 70.90 g/mol 26.98 g/mol 133.33 g/mol
a. Calculate the theoretical yield of aluminum chloride (in grams) that can be
produced from 10.00 grams of aluminum metal.
b. An experiment was performed and obtained 25.23 grams of aluminum chloride.
Determine the percent yield of aluminum chloride.
c. Determine the percent error.
d. Compute for the excess amount of the excess reactant.
2. V(s) + O2(g) V2O3(s) 50.94 g/mol 32.00 g/mol ∆ 149.88 g/mol
a. Calculate the theoretical yield of vanadium (III) oxide, assuming you begin with
200.00 grams vanadium metal.
b. After the experiment is performed, an experimental yield of 183.2 grams is
produced. Calculate the percent yield for this experiment.
c. Determine the percent error.
d. Compute for the excess amount of the excess reactant.
3. KI(aq) + H2O(l) + KMnO4(aq) I2(s) + MnO2(s) + KOH(aq) 166.0 /mol 18.02 g/mol 158.03 g/mol 253.80 g/mol 86.97 g/mol 56.11 g/mol
a. Calculate the mass of manganese (IV) oxide that can be synthesized from 15.00 grams of potassium iodide.
b. Calculate the percent yield of this experiment if a mass of 1.982 grams of manganese (IV) oxide is produced.
c. Determine the percent error.
d. Compute for the excess amount of the excess reactant.
13
Activity 2: Let’s Go Online…
Direction: Refer to the following sites/links for further discussions on how to calculate the
theoretical yield using stoichiometry and percent yield using the formula. Give your
comprehensive summary and personal reflection on what you have learn from the videos.
Write your answers on a separate paper.
https://bit.ly/3gEnN4f
https://bit.ly/2CjAYIS
https://bit.ly/3edzFIM
https://bit.ly/3faDK1W
What I Have Learned
Tell Me What You Know
Based on what you have learned from this lesson, briefly discuss the following questions. Limit
your answers to 3-5 sentences. Use a separate paper for your answers.
1. Why is it important to calculate the percent yield in a chemical reaction?
2. In a chemical reaction, explain the difficulty in obtaining the theoretical yield.
3. Discuss the reason why is it, that the actual yield is often lesser than the theoretical
yield?
4. Which is more important to determine in a chemical reaction, the limiting reactant or
the excess reactant? Elaborate your answer.
5. Why do you need to balance first the chemical equation before proceeding to the
calculations of theoretical yield and percent yield?
What I Can Do
Step-by-step…
Direction: Carefully read the following word problems. From the knowledge and skills in
computation that you have acquired in this lesson, do the following:
a. Balance the chemical equation;
b. Identify the limiting reactant and the excess reactant if applicable;
c. Compute for the theoretical yield;
d. Determine the percent yield of the reaction;
e. Calculate the percent error and
f. Compute for the excess amount of the excess reactant if applicable.
g. Show the complete solutions for your answers in a separate paper.
14
1. Eighty grams of Silver was obtained from one hundred and forty grams of Silver nitrate.
The Silver metal is prepared by reducing its nitrate. The chemical equation of the
reaction is: 14
Cu(s) + AgNO3(aq) Cu(NO3)2(aq) + Ag(s)
2. A hundred grams of Sulfuric acid yielded ten grams of water.
H2SO4(l) H2O(l) + SO3(l)
3. Fifty grams of Chlorine reacts completely with Phosphorus obtaining ninety grams of
the product.
P4(s) + Cl2(g) PCl3(l)
4. Aspirin is one of the products when you mix Salicylic acid and Acetic anhydride. One
kilogram of Salicylic acid produced one thousand and two hundred grams of Aspirin.
C7H6O3(s) + C4H6O3(l) C9H8O4(s) + CH3COOH(l)
Summary
The Law of Conservation of Mass provides that the amount of substances formed (product) in a chemical reaction must be equal to the amount of the initial materials (reactants). The stoichiometric coefficient balances the number of elements present in the left and right side of the equation. This coefficient can be used to determine the mole-mole factor or ratio between and among substances involved (both product and reactant). The mole method approach makes stoichiometric calculation easier by expressing the known and unknown quantities in moles then convert it to grams using the molar mass of the substance. In some reactions, an excess amount of a less expensive reagent (reagent in excess) is added to make sure that the more expensive reagent is completely converted to product. The chemical reaction will stop when all the limiting reagent is used up. Most of the time, the expected amount of product (theoretical yield) is not the same as the actual amount produced in the reaction (actual yield). The efficiency of the conducted reaction is determined by computing the percent yield which is the ratio of the actual yield and the theoretical yield multiplied by 100. The closer the value of the percent yield to 100, the more efficient is the reaction.
15
Assessment: (Post-Test)
Directions. Read and understand each question and select the letter of the best answer from among the given choices.
1. Which of the following equations best represent the law of conservation of mass? A. 𝐴 + 2𝐵 → 3𝐴𝐵 C. 6𝐴2𝐵 → 12𝐴 + 𝐵
B. 2𝐴2 + 8𝐵 → 4𝐴𝐵2 D. 2𝐴𝐵3 → 2𝐴 + 2𝐵
2. Which of the following statements is true about stoichiometric coefficient? A. It is written to make the number of each element in a chemical equation the same
in the reactant and product side. B. the sum of all coefficients in the left must be equal to the sum of all coefficients in
the right side of the equation. C. It is the subscript written after each element. D. None of the above.
3.. In a chemical reaction, stoichiometry refers to:
A. Amount of materials consumed and products formed. B. the activation energy C. the rate or reaction D. reaction in equilibrium
For items 4 to 6. In the reaction 2H2 + CO → CH3OH, 6.8 g carbon monoxide gas was made
to react with 7.2 g of hydrogen gas. The reaction produced 5.2 g of methanol.
4. Which is the limiting reagent? A. Hydrogen C. methanol B. Carbon monoxide D. CO2
5. What is the theoretical yield? A. 57.6 g C. 7.8 g B. 50.4 g D. 6.8 g 6. What is the percent yield? A. 76% C. 68% B. 78% D. 67%
For items 7 to 10, base your answer from the combustion of butane (C4H10) as shown in the reaction:
2𝐶4𝐻10 (𝑔) + 13𝑂2(𝑔) → 8𝐶𝑂2(𝑔) + 10𝐻2𝑂 (𝑙)
7. What is the mole-mole factor of butane and water? A. 10:2 C. 8:10 B. 2:13 D. 2:10
8. which pair has the mole-mole factor of 8:10? A. Butane and oxygen C. Oxygen and carbon dioxide B. Carbon dioxide and water D. Water and butane
16
9. If 65.3 moles of oxygen gas are consumed, how many moles of carbon dioxide is produced?
A. 526.4 moles C. 40.2 moles B. 106.1 moles D. 8.16 moles
10. How many grams of C4H10 is needed to produce 37.8 moles of CO2?
A. 453.6 grams C. 2.607 grams B. 548.1 grams D. 8769.6 grams
17
Key to Answers
Pretest Post Test
18
1. B 2. A 3. A 4. D 5. B 6. C 7. B 8. B 9. C 10. D
1. B 2. A 3. A 4. B 5. C 6. D 7. D 8. B 9. C 10. B
References
Chang, Raymond. Chemistry, 10th ed. New York, NY: McGraw-Hill
Science/Engineering/Math, 2009.
"Chapter 7.4: Stoichiometry." Chemistry LibreTexts. Last modified June 5, 2019.
https://chem.libretexts.org/Courses/Howard_University/General_Chemistry%3A_An_Atoms_
First_Approach/Unit_3%3A_Stoichiometry/Chapter_7%3A_Stoichiometry/Chapter_7.4%3A_
Stoichiometry.
Commission on Higher Education, General Chemistry 1: Teaching Guide for Senior High,
Manila, 2016.
Department of Education Central Office, Most Essential Learning Competencies (MELCS),
Manila, 2020.
Khan Academy. "Stoichiometry | Chemical reactions and stoichiometry | Chemistry | Khan
Academy." YouTube. August 27, 2009. https://www.youtube.com/watch?v=SjQG3rKSZUQ.
"Limiting Reagents Practice Problems". 2020. Chemistry.Wustl.Edu. Accessed July 12.
http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Plink/limreag/probsetlr.htm.
"Limiting Reagent Stoichiometry (Practice)". 2020. Khan Academy. Accessed July 12.
https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/limiting-
reagent-stoichiometry/e/limiting_reagent_stoichiometry.
Maribel, Melissa. "How to Calculate Percent Yield and Theoretical Yield The Best Way -
TUTOR HOTLINE." YouTube. April 30, 2019.
https://www.youtube.com/watch?v=MebTIQNRU5g.
"Quiz #2-4 PRACTICE: Balancing Equations & Mole Ratios | Mr. Carman's Blog".
2020. Kentschools.Net. Accessed July 15. https://www.kentschools.net/ccarman/cp-
chemistry/practice-quizzes/quiz-2-4/.
"Stoichiometry--Molar Mass, Mole Ratios - Quiz". 2020. Quizizz.Com. Accessed July 12.
https://quizizz.com/admin/quiz/58b47527b60c1ba227b6927c/stoichiometry-molar-mass-
mole-ratios.
The Organic Chemistry Tutor. 2015. How To Calculate Theoretical Yield And Percent Yield. Video. https://www.youtube.com/watch?v=jtAj0s203CI.
The Organic Chemistry Tutor. 2017. Stoichiometry Mole To Mole Conversions - Molar Ratio Practice Problems. Video. https://www.youtube.com/watch?v=3zmeVamEsWI. TheChemistrySolution. 2012. Theoretical, Actual And Percent Yield Problems - Chemistry Tutorial. Video. https://www.youtube.com/watch?v=mmsKDK9WXdE.
19
For inquiries and feedback, please write or call: Department of Education – Bureau of Learning Resources (DepEd-BLR) DepEd Division of Cagayan de Oro City Fr. William F. Masterson Ave Upper Balulang Cagayan de Oro Telefax: ((08822)855-0048
E-mail Address: [email protected]
