This is the Physical Science Quarter 1, Module 6 self-learning module, “Stoichiometry: Magical but Real!”, an Alternative Delivery Mode (ADM) module for Grade 11 Senior High School learners, first edition 2020, from DepEd National Capital Region. It teaches how to measure the amounts of reactants and products in a chemical reaction. By the end, learners can balance equations, convert between mass and moles, and find the limiting and excess reagents in a reaction.
On this page 8 sections
What learners will learn
The module states that to master stoichiometry, learners should be able to:
- Define what stoichiometry is
- Balance chemical equations correctly
- Convert mass to mole and vice versa
- Compute for the mass or mole of the reactants or products using stoichiometric ratio
- Identify the limiting or excess reagent in a chemical reaction
Key terms
The module’s “What I Have Learned” part sums up four terms:
| Term | Meaning |
|---|---|
| Stoichiometry | The quantitative study of reactants and products in a chemical reaction |
| Mole (molar) ratio | The ratio of moles of reactants and products, taken from the coefficients in the balanced chemical equation |
| Limiting reagent | The substance that controls or limits the maximum amount of product formed |
| Excess reagent | The reactant present in quantities greater than needed to react with the limiting reagent |
Solving stoichiometric problems step by step
The module uses four steps. The first step is always to balance the chemical equation. Then:
- Convert the mass of the given substance (A) to moles of A.
- Convert moles of A to moles of the wanted substance (B) using the mole ratio from the balanced equation.
- Convert moles of B to mass of B.
The module asks learners to write final answers in the correct number of significant figures.
In Activity 2, “Cook like a Chemist,” learners practice these steps on everyday reactions. Examples include finding how much carbon dioxide is released when 42.0 g of baking soda (NaHCO3) is heated, which is what makes bread and pastries rise. Another asks how much titanium(IV) oxide, a white paint pigment, forms from 1000.0 g of the mineral ilmenite (FeTiO3).
Limiting and excess reagents
The module explains this idea with chicken adobo. If you have 10 pounds of chicken but only one onion, you can only cook about one pound of chicken so the dish does not turn bland. The onion is the limiting ingredient and the chicken is in excess. In a reaction, the reactant in limited amount likewise decides how much product can form.
To find the limiting reagent, assume one reactant is used up completely, then compute how much of the other reactant it needs. The module’s worked example uses N2 + 3H2 → 2NH3 with 10 grams of each reactant:
- 10 g of N2 (0.36 mol) needs only 2.2 g of H2. Since there are 10 g of H2, hydrogen is in excess.
- 10 g of H2 (5.0 mol) would need 48 g of N2. Since there are only 10 g of N2, nitrogen is the limiting reagent.
Activity 3, “Am I too much for you?”, gives three more problems. Learners find the limiting reagent and the amount of product formed, for example for Mg(OH)2 reacting with HCl, and MnO2 reacting with HCl.
Real-life application
In “What I Can Do,” learners look at ammonia for fertilizer, made through the Haber–Bosch process (N2 + 3H2 → 2NH3). They compute how many grams of hydrogen gas are needed for 2500 grams of nitrogen gas, and how much ammonia that amount can produce.
The Additional Activities ask learners to list five activities at home that apply stoichiometric problem solving and limiting and excess reagents. For each one, they take a photo or make a drawing and explain the connection.
How to use this module
The module follows the standard ADM parts: What I Need to Know, What I Know (pretest), What’s In, What’s New, What is It, What’s More, What I Have Learned, What I Can Do, Assessment, Additional Activities, and Answer Key.
- Learners: Answer the 15-item What I Know pretest first. If you get 100%, you may skip the module. Write your answers on a separate sheet, not on the module. Finish each task before going to the next, and check yourself honestly with the answer key. Return the module to your teacher when you are done.
- Parents: Help your child keep to the order of the parts. Encourage them to work out each problem step by step before looking at the answer key.
- Teachers and facilitators: Orient learners on how to use the module and keep track of their progress. Look for the “Notes to the Teacher” boxes for tips.
The answer key at the end gives the answers to the What I Know pretest and the 15-item Assessment.
Frequently asked questions
What grade and quarter is this Physical Science module for?
It is for Grade 11 Senior High School learners and covers Quarter 1, Module 6 of Physical Science.
What is stoichiometry?
According to the module, stoichiometry is the quantitative study of reactants and products in a chemical reaction.
What is the first step in solving a stoichiometric problem?
Balance the chemical equation first. Then convert the mass of the given substance to moles, use the mole ratio to get moles of the wanted substance, and convert those moles to mass.
What is the difference between a limiting reagent and an excess reagent?
The limiting reagent controls the maximum amount of product that can form. The excess reagent is present in a greater amount than needed to react with the limiting reagent.
Does the module have an answer key?
Yes. The answer key at the end gives the answers to the 15-item What I Know pretest and the 15-item Assessment.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Physical Science Quarter 1 – Module 6: Stoichiometry: Magical but Real!
Senior High School
[illustration: students reading and studying under a tree, with a teacher at a desk, a child writing numbers, a globe, and a lamp]
Physical Science Alternative Delivery Mode Quarter 1 – Module 6: Stoichiometry: Magical but Real 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 royalties.
Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module 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 Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
Development Team of the Module
Writers: Maria Ana A. Dionisio
Editors: Joey H. Villanueva, Arlene C. Malaybalay
Reviewers: Tommy R. Rico, Emiterio D. Macarubbo, Maricris N. Surigao
Illustrator: Maria Ana A. Dionisio
Layout Artist: Justine C. Montoya
Management Team: Wilfredo E. Cabral, Director IV Jennifer Vivas, CLMD Chief Dennis M. Mendoza, Regional EPS in-charge of LRMS Micah S. Pacheco, Regional ADM Coordinator Jocelyn M. Aliñab, CID Chief Tommy R. Rico, Division in-charge of LRMS and Division ADM Coordinator
Printed in the Philippines by ______________________
Department of Education-National Capital Region
Office Address: Misamis St., Bago Bantay, Quezon City
Telefax: 02-929-0153
Email Address: depedncr@deped.gov.ph
Physical Science Quarter 1 – Module 6: Stoichiometry: Magical but Real!
Introductory Message
For the facilitator:
Welcome to the Physical Science Grade 11 Alternative Delivery Mode (ADM) Module on Stoichiometry!
This module was collaboratively designed, developed and reviewed by educators both from public and private institutions to assist you, the teacher or facilitator in helping the learners meet the standards set by the K to 12 Curriculum while overcoming their personal, social, and economic constraints in schooling.
This learning resource hopes to engage the learners into guided and independent learning activities at their own pace and time. Furthermore, this also aims to help learners acquire the needed 21st century skills while taking into consideration their needs and circumstances.
In addition to the material in the main text, you will also see this box in the body of the module:
Notes to the Teacher This contains helpful tips or strategies that will help you in guiding the learners.
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:
Welcome to the Physical Science Grade 11 Alternative Delivery Mode (ADM) Module on Stoichiometry!
The hand is one of the most symbolized part of the human body. It is often used to depict skill, action and purpose. Through our hands we may learn, create and accomplish. Hence, the hand in this learning resource signifies that you as a learner is capable and empowered to successfully achieve the relevant competencies and skills at your own pace and time. Your academic success lies in your own hands!
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.
This module has the following parts and corresponding icons:
What I Need to Know This will give you an idea of the skills or competencies you are expected to learn in the module.
What I Know This part includes an activity that aims to check what you already know about the lesson to take. If you get all the answers correct (100%), you may decide to skip this module.
What’s In This is a brief drill or review to help you link the current lesson with the previous one.
What’s New In this portion, the new lesson will be introduced to you in various ways such as a story, a song, a poem, a problem opener, an activity or a situation.
What is It This section provides a brief discussion of the lesson. This aims to help you discover and understand new concepts and skills.
What’s More This comprises activities for independent practice to solidify your understanding and skills of the topic. You may check the answers to the exercises using the Answer Key at the end of the module.
What I Have Learned This includes questions or blank sentence/paragraph to be filled in to process what you learned from the lesson.
What I Can Do This section provides an activity which will help you transfer your new knowledge or
skill into real life situations or concerns.
Assessment This is a task which aims to evaluate your level of mastery in achieving the learning competency.
Additional Activities In this portion, another activity will be given to you to enrich your knowledge or skill of the lesson learned. This also tends retention of learned concepts.
Answer Key This contains answers to all activities in the module.
At the end of this module you will also find:
References This is a list of all sources used in developing this module.
The following are some reminders in using this module:
- 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.
- Don’t forget to answer What I Know before moving on to the other activities included in the module.
- Read the instruction carefully before doing each task.
- Observe honesty and integrity in doing the tasks and checking your answers.
- Finish the task at hand before proceeding to the next.
- 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!
What I Need to Know
This module is all about stoichiometry: a topic that is necessary in order to master more advanced topics in chemistry. In order to master stoichiometry, you need to achieve the following objectives:
- Define what stoichiometry is;
- Balance chemical equations correctly;
- Convert mass to mole and vice versa;
- Compute for the mass or mole of the reactants or products using stoichiometric ratio;
- Identify the limiting or excess reagent in a chemical reaction.
What I Know
DIRECTIONS: Read the questions very carefully and write the letter of the correct answer on the space provided before each number.
_____1. Stoichiometry is defined as the quantitative study of _____________ in a chemical reaction.
a. moles and mass b. elements and compounds c. reactants and products d. matter and energy
_____2. What is the first step in solving stoichiometric problems?
A. convert mass of the given substance to mole B. determine the mole ratio of the involved substance C. convert the moles of the wanted substance to the desired unit D. balance the chemical equation
_____3. In the reaction, 2Al + 6HCl → 2 AlCl3 + 3 H2, the mass of aluminum was given and you were asked to compute how much H2 will be produced. What stoichiometric ratio are you going to use?
A. 3NthHt / tNthAh [?] B. tNthHt / 3NthAh [?] C. tNthAh / 3NthHt [?] D. 3NthAht / NthHt [?]
[transcriber’s note: the four answer choices are printed as stacked fractions whose text is garbled in the image; the readings above are best approximations and the fraction layout may not be exact]
_____4. What are the correct coefficients to balance the equation below?
_____KClO3 → _____KCl + _____O2
A. 2-2-3 B. 2-3-2 C. 1-2-3 D. 3-2-2
For questions 5-10:
Nitric acid is formed from the reaction of nitrogen dioxide and water as shown in the equation, 3NO2 + H2O → 2HNO3 + NO. Initially, you were given 10 grams of NO2 and 10 grams of H2O.
_____5. How many moles of NO2 are there in the actual reaction?
A. 0.19 mol B. 0.20 mol C. 0.21 mol D. 0.22 mol
_____6. How many moles of H2O are there in the actual reaction?
A. B. C. D.
[transcriber’s note: the answer choices for question 6 appear blank in the image]
I can’t transcribe pages 9 to 16 because they aren’t in your message. The only pages attached are labeled 1 to 8, and the earlier pages were already done separately, so I haven’t re-transcribed them.
Please re-upload the PDF pages 9 to 16. I’ll number their page markers 9 to 16 and keep the same structure.
The attached pages also don’t match the filename. They’re a chemistry stoichiometry module with a quiz at the start, not a Physical Science Q1 Module 6 opening. Please check that the file you’re sending is the right one. If it is, the pages 9 to 16 you send should come from that same document.
What is It
Activity 2: Cook like a Chemist
Imagine you are a chemist. How do you know how much reactants you are going to use and how much products are going to be formed? Try solving these problems to practice your skills.
Objectives:
- Solve the following stoichiometric problems correctly; and
- Follow the steps in solving stoichiometric problems carefully
Procedure:
- Follow the four steps in solving stoichiometric problems to solve for what is being required.
- Write you final answer in the correct number of significant figures.
- When baking soda (NaHCO3) is heated, carbon dioxide (CO2) is released. This makes bread, cookies and other pastries to rise. If 42.0 g of NaHCO3 was used, how much CO2 was released? (NaHCO3 = 84.00 g/mol, CO2 = 44.01, Na2CO3 = 105.99 g/mol, H2O = 18.02 g/mol)
2NaHCO3 → Na2CO3 + H2O + CO2
- How much potassium iodide (166.00 g/mol) is required to get 1.78 g of mercury(II) iodide precipitate (454.39 g/mol). The unbalanced chemical equation is:
KI + Hg(NO3)2 → HgI2 + KNO3
- Titanium (IV) oxide (TiO2) is a pigment used for white paint. It is formed from the reaction of the mineral ilmenite (FeTiO3) with sulfuric acid. How much TiO2 will be produced if 1000.0 g of FeTiO3 is reacted with sufficient amount of sulfuric acid?
(FeTiO3 = 151.71 g/mol, TiO2 = 79.87 g/mol)
FeTiO3 + H2SO4 → TiO2 + FeSO4 + H2O
There are times when the ingredients that we use when cooking dictates how much food we are going to make. In the case of our chicken adobo, if you have 10 pounds of chicken but you only have one onion, you will be forced to cook only one pound of chicken to avoid making your dish go bland. In the same manner, certain reactants with limited amount dictate the amount of product we are going to form. This is where the concept of limiting reagent and excess reagent comes in.
A limiting reagent is the substance in a chemical reaction that controls or limits the maximum amount of product formed while an excess reagent is the reactant present in quantities greater than necessary to react with the quantity of the limiting reagent. In the example I gave earlier, the limiting reagent(ingredient) is the onion while the excess reagent(ingredient) are the chicken.
To identify which of the reactant is the limiting reagent and which one is the excess, you have to assume that one substance is limited, then compute for the required amount of the other reactant. Let’s us use again the equation
N2 + 3H2 → 2NH3
10 g 10 g
If you have 10 grams each of N2 and H2, which one is the limiting reagent and which one is the excess reagent? To answer this, you have to assume on reactant to be a limiting reagent, then compute for the required amount of the other reactant to completely consume the other. Let us assume that N2 is the limiting reagent. How much H2 do you need to completely use up all the N2?
[chart: flowchart; steps of the stoichiometry calculation, in order]
- Convert Mass A to Mole A
- Convert Mole A to Mole B using Mole Ratio
- Convert Mole B to Mass B
10 g N2 [illegible] = 0.36 mol N2 [illegible] 1.1 = [illegible] mol H2 [illegible] = 2.2 g H2
[transcriber’s note: the fraction labels and the operator in “1.1 =” are not legible in the page image; the diagonal cancellation strokes are omitted]
As you can see from the calculation, 10 grams of N2 will only require 2.2 grams of H2. Since you have 10 grams of H2, it means that it is in excess while N2 is in limited quantity. But what if you assumed that H2 was the limiting reagent? How much N2 do you need to consume all 10 grams of H2?
[chart: flowchart; steps of the stoichiometry calculation, in order]
- Convert Mass A to Mole A
- Convert Mole A to Mole B using Mole Ratio
- Convert Mole B to Mass B
10 g H2 [illegible] = 5.0 mol H2 [illegible] 1.7 = [illegible] mol N2 [illegible] = 48 g N2
[transcriber’s note: the fraction labels and the operator in “1.7 =” are not legible in the page image; the diagonal cancellation strokes are omitted]
If all 10 grams of H2 get used up in the reaction, you will be needing 48 grams of N2. Since you only have 10 grams of N2, it means that its quantity is limited and, therefore, H2 is in excess.
What’s More
Activity 3: Am I too much for you?
Objectives:
- Follow the steps in solving for the limiting and excess reagents carefully; and
- Identify the limiting and excess reagents.
Procedure:
- Solve the following problems to identify the limiting and excess reagent using a single thread of solution, just like in the example
- Report your answer in the correct number of significant figures
In the reaction Mg(OH)2 + 2HCl → MgCl2 + 2H2 O, you were given 15 grams of Mg(OH)2 and 25 grams of HCl. Which of the two reactants is the limiting reagent? How much MgCl2 will be produced in the reaction? (Mg(OH)2 = 58.32 g/mol, HCl = 36.46 g/mol, MgCl2 = 95.32 g/mol)
Nitric oxide (NO) is a substance that can improve heart health and enhance your performance during exercise or workout. It can be formed through the reaction:
NH3 + O2 → NO + H2O
If 25 g NH3 is reacted with 20 g O2, which of the two is the limiting reagent? How much NO will be formed in the reaction? (NH3 = 17.03 g/mol, O2 = 32.00 g/mol, NO = 30.01 g/mol)
- Consider the reaction MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O. If the both the reactants have a mass of 100 grams, which of the two will be used up first? How many grams of MnCl2 will be produced after the reaction? (MnO2 = 86.94 g/mol, HCl = 36.46 g/mol, MnCl2 = 125.8 g/mol)
What I Have Learned
Stoichiometry – the quantitative study of reactants and products in a chemical reaction
Mole (molar) Ratio – the ratio of moles of reactants and products according to the coefficients in the balanced chemical equation.
Limiting reagent – the substance in a chemical reaction that controls or limits the maximum amount of product formed
Excess reagent – the reactant present in quantities greater than necessary to react with the quantity of the limiting reagent.
What I Can Do
Fertilizers are one of the most important substances in the world since it is used to make food for 7.8 billion world population as of March 2020. Hundreds of millions of tons of fertilizer are made each year just to support our growing need for food. The most common material in making fertilizer is ammonia, which is made through the Haber–Bosch process, is synthesized through the following chemical reaction:
[photo: a worker in protective gear spraying liquid from a backpack sprayer onto trees]
N2 + 3H2 → 2NH3
This ammonia is then used create other fertilizers, but sometimes, ammonia can be directly sprayed on to the soil or plant as seen in the picture above.
Guide Questions:
- If you were to synthesize ammonia, how many grams of hydrogen gas are you going to need if you have 2500 grams of nitrogen gas?
- How much ammonia can you produce from the given amount of nitrogen gas?
Assessment
DIRECTIONS: Read the questions very carefully and write the letter of the correct answer on the space provided before each number.
_____1. Stoichiometry is defined as the quantitative study of _______________ in a chemical reaction.
- A. moles and mass
- B. elements and compounds
- C. reactants and products
- D. matter and energy
_____2. What is the first step in solving stoichiometric problems?
- A. convert mass of the given substance to mole
- B. determine the mole ratio of the involved substance
- C. convert the moles of the wanted substance to the desired unit
- D. balance the chemical equation
_____3. In the reaction, 2Al + 6HCl → 2 AlCl3 + 3 H2, the mass of aluminum was given and you were asked to compute how much H2 will be produced. What stoichiometric ratio are you going to use?
[transcriber’s note: the answer choices for item 3 are printed as fractions whose text is not legible in the page image]
- A. [illegible]
- B. [illegible]
- C. [illegible]
- D. [illegible]
_____4. What are the correct coefficients to balance the equation below?
_____KClO3 → _____KCl + _____O2
- A. 2-2-3
- B. 2-3-2
- C. 1-2-3
- D. 3-2-2
For questions 5-10:
Nitric acid is formed from the reaction of nitrogen dioxide and water as shown in the equation, 3NO2 + H2O → 2HNO3 + NO. Initially, you were given 10 grams of NO2 and 10 grams of H2O.
_____5. How many moles of NO2 are there in the actual reaction?
- A. 0.19 mol
- B. 0.20 mol
- C. 0.21 mol
- D. 0.22 mol
_____6. How many moles of H2O are there in the actual reaction?
- A. 0.53 mol
- B. 0.55 mol
- C. 0.57 mol
- D. 0.59 mol
_____7. Which of the reactants is in excess?
- A. NO2
- B. H2O
- C. HNO3
- D. NO
_____8. Which of the reactants is limited in amount?
- A. NO2
- B. H2O
- C. HNO3
- D. NO
_____9. If you were asked to look for the amount of HNO3, what stoichiometric ratio are you going to use?
[transcriber’s note: the answer choices for item 9 are printed as fractions whose text is not legible in the page image]
- A. [illegible]
- B. [illegible]
- C. [illegible]
- D. [illegible]
_____10. How many moles of HNO3 will be produced in this chemical reaction?
- A. 0.13 mol
- B. 0.15 mol
- C. 0.17 mol
- D. 0.19 mol
_____11. Which statement below is false when 10 g of nitrogen reacts with 5.0 g of hydrogen to produce ammonia?
N2 (g) + 3 H2 (g) → 2 NH3 (g)
- A. Hydrogen is the excess reactant.
- B. Nitrogen is the limiting reactant.
- C. 2.8 grams of hydrogen are left over.
- D. The theoretical yield of ammonia is 15 g.
_____12. Which of these interpretations of the following balanced equation is TRUE?
2S (s) + 3O2 (g) → 2SO3 (g)
- A. 2 atoms of S and 3 atoms of O2 form 2 atoms of SO3
- B. 2 grams of S and 3 grams of O2 form 2 grams of SO3
- C. 2 moles of S and 3 moles of O2 form 2 moles of SO3
- D. 2 L of S and 3 L of O2 form 2 L of SO3
_____13. Which of the following quantities is conserved in every chemical reaction?
- A. Mass
- B. Moles
- C. Molecules
- D. Formula units
A. B. C. D.
[transcriber’s note: the line “A. B. C. D.” above is printed on the page below item 13 with no visible question; it appears to be a stray label from the answer-choice images]
[transcriber’s note: the document title does not appear on these pages; the cover and first pages were transcribed separately]
_____14. Which type of stoichiometric calculation does not involve the gram formula mass?
a. Mass-mass problems b. Mass-volume problems c. Mass-particle problems d. Volume-volume problems
_____15. An excess amount of Zn (s) is added to 500 mL of 0.800 M HCl. What mass in grams of H2 (g) is produced?
a. 0.100 g b. 0.200 g c. 0.400 g d. 0.600 g
Additional Activities
[illustration: gears and a hand, decorative icon beside the heading]
How is the concept of stoichiometry applicable in your everyday life? List five (5) activities that you do in your house the applies the concept of stoichiometric problem solving and limiting and excess reagents. Take a picture and attach it below or draw these activities and explain how it relates to stoichiometric problem solving and limiting and excess reagents.
[transcriber’s note: two empty boxes for attaching a photo or drawing, each with blank writing lines beside it]
[transcriber’s note: continuation of the Additional Activities page; one empty box for attaching a photo or drawing, with blank writing lines beside it, followed by two more empty boxes with blank writing lines beside them]
Answer Key
[illustration: key and hand, decorative icon beside the heading]
What I Know
[transcriber’s note: the answer column is printed upside down in the original]
- C
- D
- C
- A
- C
- B
- B
- A
- A
- B
- D
- C
- A
- D
- C
Assessment
[transcriber’s note: the answer column is printed upside down in the original]
- C
- D
- C
- A
- C
- B
- B
- A
- A
- B
- D
- C
- A
- D
- C
References
Chicken https://www.pngitem.com/pimgs/m/408-4084569_chicken-meat-png-background-image-raw-chicken-with.png
Pepper corn https://pngimg.com/uploads/black_pepper/black_pepper_PNG11.png
Garlic clove https://thumbs.dreamstime.com/b/peeled-garlic-clove-white-illustrations-background-83689303.jpg
Onion https://image.shutterstock.com/image-illustration/red-onion-on-white-background-260nw-581247784.jpg
Vinegar https://www.pinclipart.com/picdir/big/70-702159_cashier-clipart-casher-bottle-of-vinegar-clipart-png.png
Soy Sauce – https: // images. clipartlogo. com/ files/ istock/ previews/ 1043/ 104356229-soy-sauce.jpg
Cup of water – https://www.pngitem.com/pimgs/m/485-4852817_water-cup-png-graphic-royalty-free-stock-cup.png
Chicken adobo – https://i.pinimg.com/originals/8f/2c/a8/8f2ca87a34b58096b380dba06b104efd.png
Ammonia sprayed as fertilizer – https://l.hdnux.com/350x235p/s3-us-west-1.amazonaws.com/contentlab.studiod/getty/0936f3bdf8b34a739e7ddc34c3b0f80b.jpg
https://www.edu.gov.mb.ca/k12/dl/iso/practice_exams/gr11_chem_fpe_key.pdf
http://mrayton.com/uploads/3/4/2/3/34236437/multiple_choice_sample_stoichiometry.pdf
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