11
General
Chemistry 1
Quarter 2 – Module 1
Stoichiometry
What I Need to Know
This module was designed and written with you in mind. It is
here to help you master the concepts of stoichiometry. The scope of this module
permits it to be used in many different learning situations. The language used
recognizes the diverse vocabulary level of students. The lessons are arranged to
follow the standard sequence of the course. But the order in which you read
them can be changed to correspond with the textbook you are now using.
The module is about:
Lesson 10 – 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. Calculate percent yield
(STEM_GC11MRIg-h-39)
and
theoretical
yield
of
the
reaction.
3. Explain the concept of limiting reagent in a chemical reaction; identify the
excess reagent(s). (STEM_GC11MRIg-h-40)
4. (LAB) Determine mass relationship in a chemical reaction.
(STEM_GC11MRIg-h-41)
After going through this module, you are expected to:
1. Write the conversion factors for the mole and mass relationships between
reactants and products based on a balanced chemical equation;
2. Calculate the amount of moles or mass of a substance based on a
balanced chemical equation;
3. Calculate and identify the limiting reactant and the excess reagent in a
chemical reaction;
4. Calculate the percent yield and the theoretical yield of a chemical
reaction; and
5. Solve problems about stoichiometry happily.
Lesson
10
STOICHIOMETRY
What’s In
Law of Conservation of Mass states that matter is neither
created nor destroyed, it can only be transformed from one form to
another. This universal law can be observed from every chemical
reaction we see in our world, from rusting of iron, from cooking
rice, and from the metabolic pathways in our body. In this module,
we will determine quantitatively the relationship between the
application of the law of conservation of mass, and the law of
definite proportions to chemical reactions we see around us. This
is known as stoichiometry.
RUSTING OF IRON: A CHEMICAL REACTION
Image source: https://byjus.com/chemistry/rusting-iron-prevention/
What’s New
Activity 2
Ratio and Proportion
1.
Some learners are going on a road trip and they are to bring
some food to eat along the way. Lucy was asked to bring
hamburger sandwiches for which she will use two slices of
bread and one hamburger patty to make one sandwich. Show
the equation.
2.
Another student whose name is Geraldine brought 2 tetra packs
of orange juice and 6 liters of bottled water. The ideal mixture
to make a moderate fruity taste orange juice must contain 1 tetra
pack of juice per 1.2 liters of water. During the preparation, she
mixed 2 tetra packs of orange juice and 3 liters of water. Is the
orange juice bland, moderate fruity or too much concentrated?
Justify your answer.
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Stoichiometry is the study of the quantities of materials
consumed and produced in chemical reactions. It deals with
ratios and conversions involving the materials that are
consumed and produced. As a matter of fact, the balanced
chemical equations will help us:
Determine how much products will be produced from a
specific amount of reactants; and
b. Determine the amount of reactants needed to produce a
specific amount of
products
a.
Let us make hamburger sandwiches again. The equation is:
2 slices of bread + 1 hamburger patty = 1 hamburger
sandwich
1.
Suppose Lucy has 26 hamburger patties, how many slices of
bread will she need to consume all the patties?
The ratio of slices of bread to hamburger patty is 2:1.
Conversion factor is the numerical ratio that can be used as
a
multiplication ratio to convert a set of units with the same measurement to another
set of units. Conversion factors are enclosed in a box in the succeeding examples.
26 ℎ𝑎𝑚𝑏𝑢𝑟𝑔𝑒𝑟 𝑝𝑎𝑡𝑡𝑖𝑒𝑠 =
2 𝑠𝑙𝑖𝑐𝑒𝑠 𝑜𝑓 𝑏𝑟𝑒𝑎𝑑
= 52 ℎ𝑎𝑚𝑏𝑢𝑟𝑔𝑒𝑟 𝑠𝑎𝑛𝑑𝑤ℎ𝑖𝑐ℎ
1 ℎ𝑎𝑚𝑏𝑢𝑟𝑔𝑒𝑟 𝑝𝑎𝑡𝑡𝑦
Suppose that instead of plain burgers, Lucy is to make double
cheeseburger.
Show the equation so Karen can shop for enough ingredients.
2.
2 slices of bread + 1 hamburger patty + 2 slices cheese = 1
double cheeseburger
How many slices of cheese, hamburger patties, and slices of bread
will Lucy need to make 29 double cheeseburgers?
The ratio of double cheeseburger to slices of bread, hamburger
patty and slices of cheese is 1:2:1:2.
29 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐ℎ𝑒𝑒𝑠𝑒𝑏𝑢𝑟𝑔𝑒𝑟𝑠 𝑥
29 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐ℎ𝑒𝑒𝑠𝑒𝑏𝑢𝑟𝑔𝑒𝑟𝑠 𝑥
2 𝑠𝑙𝑖𝑐𝑒𝑠 𝑜𝑓 𝑏𝑟𝑒𝑎𝑑
= 𝟓𝟖 𝒔𝒍𝒊𝒄𝒆𝒔 𝒐𝒇 𝒃𝒓𝒆𝒂𝒅
1 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐ℎ𝑒𝑒𝑠𝑒𝑏𝑢𝑟𝑔𝑒𝑟
1 ℎ𝑎𝑚𝑏𝑢𝑟𝑔𝑒𝑟 𝑝𝑎𝑡𝑡𝑦
= 𝟐𝟗 𝒉𝒂𝒎𝒃𝒖𝒓𝒈𝒆𝒓 𝒑𝒂𝒕𝒕𝒊𝒆𝒔
1 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐ℎ𝑒𝑒𝑠𝑒𝑏𝑢𝑟𝑔𝑒𝑟
29 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐ℎ𝑒𝑒𝑠𝑒𝑏𝑢𝑟𝑔𝑒𝑟𝑠 𝑥
2 𝑠𝑙𝑖𝑐𝑒𝑠 𝑜𝑓 𝑐ℎ𝑒𝑒𝑠𝑒
= 𝟓𝟖 𝒔𝒍𝒊𝒄𝒆𝒔 𝒐𝒇 𝒄𝒉𝒆𝒆𝒔𝒆
1 𝑑𝑜𝑢𝑏𝑙𝑒 𝑐ℎ𝑒𝑒𝑠𝑒𝑏𝑢𝑟𝑔𝑒𝑟
Lucy will therefore have to buy 58 slices of bread, 29
hamburger patties, and 58 slices of bread.
Photosynthesis is the processes so that plants and all other
autotrophs can make their own food. It is also an anabolic process
that involves the reaction of carbon dioxide and water to make
glucose and oxygen gas. The balanced chemical reaction is as
follows:
6CO2 (g) + 6H2O (l) → C6H12O6 (s) + 6O2 (g)
The ratio of glucose to carbon dioxide, oxygen gas and water is
1:6:6:6.
1.
How many moles of CO2 are needed to produce 36.9 moles of
C6H12O6
36.9 moles C6 H12 O6 𝑥
6 moles CO2
= 𝟐𝟐𝟏. 𝟒 𝐦𝐨𝐥𝐞𝐬 𝐂𝐎𝟐
1 mole C6 H12 O6
Since carbon dioxide, oxygen gas and water have same molar
ratios, 221.4 moles of oxygen gas and 221.4 moles of water are
also needed to make 36.9 moles glucose.
2.
How many moles of C6H12O6 are produced from 2.6 moles of H2O?
2.6 moles of H2 O x
1 mole C6 H12 O6
= 𝟎. 𝟒𝟑 𝐦𝐨𝐥𝐞𝐬 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔
6 moles H2 O
3. How many grams of C6H12O6 are produced from 2.6 moles of H2O?
2.6 moles of H2 O x
1 mole C6 H12 O6 180.16 g C6 H12 O6
x
= 𝟕𝟕. 𝟒𝟕 𝐠 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔
6 moles H2 O
1 mole C6 H12 O6
4. How many grams of CO2 are needed to produce 36.9 moles of C6H12O6?
36.9 moles C6 H12 O6 𝑥
6 moles CO2
44 g CO2
𝑥
= 𝟗𝟕𝟒𝟏. 𝟔 𝐠 𝐂𝐎𝟐
1 mole C6 H12 O6 1 mole CO2
5. How many grams of oxygen will be produced from 72.4 grams of water?
72.4 g H2 O x
1 mole H2 O
6 moles O2
32 g O2
x
x
= 𝟏𝟐𝟖. 𝟕𝟏 𝐠 𝐎𝟐
18 g H2 O
6 moles H2 O 1 mole O2
Let’s go back to Geraldine’s orange juice. The ideal mixture to
make a moderate fruity taste orange juice must contain 1 tetra pack
of juice per 1.2 liters of water. She mixed 2 tetra packs of orange
juice and 3 liters of water. Is her orange juice bland, moderate fruity
or too much concentrated?
Let’s write first the equation of ideal mixture.
1 tetra pack orange juice + 1.2 liters water = 1 moderate fruity
taste orange juice
Let’s assess her version of orange juice by using water as one of the
ingredients
3 𝑙𝑖𝑡𝑒𝑟𝑠 𝑤𝑎𝑡𝑒𝑟 =
1 𝑡𝑒𝑡𝑟𝑎 𝑝𝑎𝑐𝑘 𝑜𝑟𝑎𝑛𝑔𝑒 𝑗𝑢𝑖𝑐𝑒
= 𝟐. 𝟓 𝒕𝒆𝒕𝒓𝒂 𝒑𝒂𝒄𝒌 𝒋𝒖𝒊𝒄𝒆𝒔
1.2 𝑙𝑖𝑡𝑒𝑟𝑠 𝑤𝑎𝑡𝑒𝑟
It means that in order to achieve a moderate fruity taste orange
juice using 3 liters of water, Geraldine still needs additional 0.5 or
half a tetra pack orange juice.
And now, let’s use her 2 tetra packs of
orange juice
2 𝑡𝑒𝑡𝑟𝑎 𝑝𝑎𝑐𝑘 𝑜𝑓 𝑜𝑟𝑎𝑛𝑔𝑒 𝑗𝑢𝑖𝑐𝑒 =
1.2 𝑙𝑖𝑡𝑒𝑟𝑠 𝑤𝑎𝑡𝑒𝑟
= 𝟐. 𝟒 𝒍𝒊𝒕𝒆𝒓𝒔 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓
1 𝑡𝑒𝑡𝑟𝑎 𝑝𝑎𝑐𝑘 𝑜𝑟𝑎𝑛𝑔𝑒 𝑗𝑢𝑖𝑐𝑒
It means that in order to achieve a moderate fruity taste
orange juice using 2 liters of water, Geraldine must remove the
excess 0.6 liters of water.
Since Geraldine is so clumsy, she made a juice with excess
water making the taste bland.
In the example, the ingredient that limits her juice are her
tetra packs while the excess is water. Both of them are reactants and
their product is a bland orange juice.
Let’s talk about limiting and excess reactants.
Limiting reactants/reagents - The reactants used up first in the
chemical reaction. It means that they have the lowest stoichiometric
amount in the given reactants.
Excess reactants/reagents - are reactants present in quantities
greater than what is needed by the reaction.
Methane, also known as a marsh gas is a colorless and it is
also known as the simplest hydrocarbon. Since it is combustible
gas, it reacts readily with oxygen gas to produce carbon dioxide
and water. The balanced chemical reaction is:
CH4(g) + 3O2(g) → CO2 (g) + 2H2O (l)
The stoichiometric ratio is 1 mole CH4: 3 moles O2: 1 mole CO2: 2
mole H2O
(1:3:1:2)
1. Supposed that we react 3 moles of methane and 6 moles of
oxygen gas, determine the limiting and excess reactant if there
is any.
1st Solution:
Use the amount of the reactants to determine the stoichiometric
amount of the other reagent.
In balanced reaction: Methane: oxygen (1:3)
In the given
: Methane: oxygen(3:9) using methane
It means that when we use 3 moles of methane, there must be 9
moles of oxygen gas so that there will be no excess and limiting
reactants.
In balanced reaction: Methane: oxygen (1:3)
In the given
: Methane: oxygen (2:6) using oxygen gas
It means that when we use 6 moles of oxygen, there must be 2
moles of methane gas so that there will be no excess and limiting
reactants.
In the given above, the first reagent that will be consumed is 6
moles of oxygen. It only needs the 2 moles in the 3 moles of
methane so there will be 1 mole of methane that is left.
Therefore, the limiting reagent is oxygen and the excess
reagent is methane.
2nd Solution:
Calculate the amount of reference product produced of each
amount of reactant and compare. Smaller product amount means
that the starting reactant is the limiting reactant. Larger product
amount means that the starting reactant is the excess reactant.
Let’s use carbon dioxide as the reference product:
In balanced reaction:
(1:3:2)
Methane: oxygen: carbon dioxide
3 moles of methane yield 6 moles of carbon dioxide while 6 moles
of oxygen gas yield 4 moles of carbon dioxide. Methane having
more carbon dioxide yields is the excess reactant while oxygen
having less carbon dioxide yields is the limiting reactant.
2. 96 grams of O2 reacts with 1.5 moles of CH4. (a) Determine the
limiting reactant. (b) How much (in grams) of the excess
reactant if left after the reaction proceeds to completion?
(a)
96 𝑔𝑟𝑎𝑚𝑠 𝑂2 𝑥
1 𝑚𝑜𝑙𝑒 𝑂2
2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2
𝑥
= 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2
32 𝑔𝑟𝑎𝑚𝑠 𝑂2
3 𝑚𝑜𝑙𝑒𝑠 𝑂2
1.
The limiting reactant is oxygen gas.
(b) To determine the amount of the excess reactant, determine the
difference on the amount of products yielded. (3 moles CO2 – 2
moles CO2 = 1 mole CO2)
Use the difference as the starting point.
1 𝑚𝑜𝑙𝑒 𝐶𝑂2 𝑥
1 𝑚𝑜𝑙𝑒 𝐶𝐻4
16 𝑔 𝐶𝐻4
𝑥
= 𝟖 𝒈 𝑪𝑯𝟒
2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 1 𝑚𝑜𝑙𝑒 𝐶𝐻4
3. 100 grams of oxygen reacts with 15 grams of methane. (a)
Determine the limiting reactant. (b) How much (in grams) of the
excess reactant if left after the reaction proceeds to completion?
(a)
15 𝑔𝑟𝑎𝑚𝑠 𝐶𝐻4 𝑥
1 𝑚𝑜𝑙𝑒 𝐶𝐻4
2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2
𝑥
= 1.875 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2
16 𝑔𝑟𝑎𝑚𝑠 𝐶𝐻4 1 𝑚𝑜𝑙𝑒 𝐶𝐻4
The limiting reactant is methane.
(b) The difference in product yield is:
(2.083 moles – 1.875 moles = 0.208 moles CO2 due to oxygen gas)
208 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 𝑥
3 𝑚𝑜𝑙𝑒𝑠 𝑂2
32 𝑔 𝑂2
𝑥
= 𝟗. 𝟗𝟖𝟒 𝒈 𝑶𝟐
2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑂2 1 𝑚𝑜𝑙𝑒 𝑂2
0.
The ideal reactions must be all reactants will be converted to
products. But most of the reactions cannot proceed to completion,
with lesser yield than is expected. It is best explained in the
concept of theoretical yield, actual yield and percent yield.
Theoretical yield is the maximum amount of product that would
result if the limiting reagent is completely consumed. It is the
amount of product predicted by stoichiometry.
Actual yield is the quantity of the desired product actually formed.
Percent yield is the ratio of actual yield and theoretical yield
multiplied by 100.
Example:
Silver metal (107.87 g/mol) reacts with sulfur (32.01 g/mol) to form
silver sulfide (247.80 g/mol) according to the following reaction:
2Ag(s) + S(s) → Ag2S(s)
a. Identify the limiting reagent if 60.0 g Ag reacts with 15.0 g S.
60 𝑔 𝐴𝑔 𝑥
15 𝑔 𝑆 𝑥
1 𝑚𝑜𝑙𝑒 𝐴𝑔
1 𝑚𝑜𝑙𝑒 𝐴𝑔2 𝑆 247.80 𝐴𝑔2 𝑆
𝑥
𝑥
= 68.92 𝑔 𝐴𝑔2 𝑆
107.87 𝑔 𝐴𝑔
2 𝑚𝑜𝑙𝑒 𝐴𝑔
1 𝑚𝑜𝑙𝑒 𝐴𝑔2 𝑆
1 𝑚𝑜𝑙𝑒 𝑆
1 𝑚𝑜𝑙𝑒 𝐴𝑔2 𝑆 247.80 𝑔 𝐴𝑔2 𝑆
𝑥
𝑥
= 116.12 𝑔 𝐴𝑔2 𝑆
32.01 𝑔 𝑆
1 𝑚𝑜𝑙𝑒 𝑆
1 𝑚𝑜𝑙𝑒 𝐴𝑔2 𝑆
The limiting reagent is Ag
b.
What is the theoretical yield in grams of Ag2S produced from the reaction?
Theoretical yield is dependent on the limiting reactant; therefore, the theoretical
yield is 68.92 g Ag2S.
c.
What is the amount in grams of the excess reactant expected to remain
after the reaction?
Get the difference of the yields per reactants used:
116.12 g Ag2S due to S – 68.92 g Ag2S due to Ag = 47.2 g Ag2S due to S
d.
When the reaction occurred, the amount of Ag2S obtained was 45.0 g.
What is the percent yield of the reaction?
% 𝐲𝐢𝐞𝐥𝐝 = 𝟔𝟓. 𝟐𝟗 %
What is It
This part of the module will strengthen your understanding of
the concepts by giving you the key ideas.
Stoichiometry is the study of the quantities of materials consumed
and produced in chemical reactions. From the balanced
chemical equation, we will be able to:
1. Determine how much products will be produced from a
specific amount of reactants
2. Determine the amount of reactants needed to produce a
specific amount of products
Conversion factor is the numerical ratio that can be used as a
multiplication ratio to convert a set of units with the same
measurement to another set of units.
Limiting reactants/reagents - The reactants used up first in the
chemical reaction. It means that they have the lowest
stoichiometric amount in the given reactants.
Excess reactants/reagents - are reactants present in quantities
greater than what is needed by the reaction.
To determine the limiting and excess reactants, here are the
possible ways: 1. Use the amount of the reactants to
determine the stoichiometric amount of the other reagent.
2. Calculate the amount of reference product produced of
each amount of reactant and compare. Smaller product
amount means that the starting reactant is the limiting
reactant. Larger product amount means that the starting
reactant is the excess reactant.
To determine the amount of excess reactant, determine the
difference on the amount of products yielded and do the
stoichiometric calculations that will determine the
amount of excess reactant.
Theoretical yield is the maximum amount of product that would
result if the limiting reagent is completely consumed. It
is the amount of product predicted by stoichiometry.
Theoretical yield is dependent on limiting reagent.
Actual yield is the quantity of the desired product actually formed.
Percent yield is the ratio of actual yield and theoretical yield
multiplied by 100.
What’s More
Activity 3
A. Direction: Write the balanced chemical equation describing
each of the following chemical reactions. Write your
answer on the space provided.
Given the following equation:
Al: 29.98 g/mol Cl: 35.453 g/mol
Al(s) + Cl2(g) →
AlCl3(s) 1. Balance the equation.
____Al(s) + ____Cl2(g) → ____AlCl3(s)
2.
If 6 grams of Al was used to react with excess Cl2, determine the
amount of AlCl3 formed. Show your solution and box the
conversion factor/s used.
3.
If 19 grams of Al reacts with 63 grams Cl2, determine the
limiting reactant. How much of the excess reactant will remain
after the reaction?
4.
What is the theoretical yield? If the product formed is 75 g
AlCl3, what is the percent yield?
Assessment
Identification. Write the proper conversion factor/s
needed to convert one stoichiometric quantity into another.
I.
A. Given the balanced chemical equation:
2KClO3(s) → 2 KCl(s) + 3O2(g)
Molar Masses: KClO3 : 122.55 g/mol
g/mol
KCl : 74.55 g/mol O2 :
1.
From 3 moles of KClO3 to grams of O2
2.
From 36 grams of KCl to moles of KClO3
3.
From 100 moles of O2 to moles of KClO3
32
II. Choose the letter of the best answer. Encircle the answer of your
choice.
For items 4-6 Given the unbalanced reaction:
Li(s) +
N2(g) →
Li3N(s)
Atomic masses: Li : 6.94 g/mol N : 14.01 g/mol
4. What are the coefficients of the Li, N2 and Li3N respectively to
make a balanced chemical equation?
6,1,2
B. 2,6,1
C. 3,1,2
D. 4,3,2
A.
5. What mass of lithium nitride could be formed from 104 g of
lithium and excess nitrogen gas?
35 g
B. 60 g
C. 105 g
D. 174 g
A.
6. If 6 moles of nitrogen gas react with excess lithium, how many
moles of lithium nitride will be formed?
A. 6 moles
B. 12 moles
C. 18 moles
D. 24 moles
7. The reactant that exist in smaller stoichiometric quantity is
called ___.
A. Limiting reactant
B. Excess reactant
C. Actual yield
D. Theoretical yield
8. If 5.0 g of each reactant were used for the the following
process, the limiting reactant would be:
2KMnO4 +5Hg2Cl2 + 16HCl → 10HgCl2 + 2MnCl2 +
2KCl + 8H2O
A. KMnO4
B. HCl
C. H2O
D. Hg2Cl2
9. What mass of ZnCl2 can be prepared from the reaction of 3.27
grams of zinc with 3.30 grams of HCl?
Zn +2HCl → ZnCl2 + H2
6.89 g
B. 6.82 g
C. 6.46 g
D. 6.17 g
A.
For items 10-11 Refer on the following balanced reaction:
SiO2 +3C → SiC + 2CO
10. Silicon carbide, an abrasive, is made by the reaction of
silicon dioxide with graphite. If 100 g of SiO2 and 100 g of C
are reacted as far as possible, which one of the following
statements will be correct? A. 111 g of SiO2 will be left over.
B. 44 g of SiO2 will be left over.
C. 82 g of C will be left over.
D. 40 g of C will be left over.
11. If 175 g of CO was produced, what is the percent yield?
A. 93. 21 g
B. 186.42 g
C. 83.21 g
D. 166.42 g
12. A commercially valuable paint and adhesive stripper,
dimethyl sulfoxide (DMSO), (CH3)2SO, can be prepared by
the reaction of oxygen with dimethyl sulfide, (CH3)2S, using
a ratio of one mole oxygen to two moles of the sulfide:
O2 + 2(CH3)2S→ 2(CH3)2SO
If this process is 83% efficient, how many grams of DMSO
could be produced from 65 g of dimethyl sulfide and excess
O2? A. 68 g
B. 75
g C.
83 g
D. 51 g
For questions no. 13 to 15.
The formation of aluminum oxide transpired through this
equation:
4Al +3O2 → 2Al2O3
13. How many moles of aluminum are needed to form 4.2 mol of
Al2O3?
A. 2.1
mol B.
4.6
mol
C. 7.4
mol
D. 8.4 mol
14. How many moles of oxygen are necessary to react
completely with
10.8 mol Al?
A. 8.10
mol B.
9.20
mol C.
12.7
mol
D. 14.4 mol
15. How many moles of Al2O3 are formed when 1.65 mol of O2
reacts with the aluminum?
A. 0.780
mol
B. 1.10
mol C.
1.35
mol
D. 1.96 mol