2. ii
General Chemistry 1
Alternative Delivery Mode
Quarter 1 – Module 9: Limiting and Excess Reagents
First Edition, 2020
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4. iv
Introductory Message
For the facilitator:
Welcome to the General Chemistry1 Alternative Delivery Mode (ADM)
Module on Limiting Excess Reagents!
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:
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.
Notes to the Teacher
In view to the new normal world we are facing,
this module was created to answer that
education should not stop for our learners.
This General Chemistry 1 Module for Quarter 1
is all about the Limiting and Excess
Reagents. With this we are trying to allow our
learners to work independently in discovering
through simple and enjoyable activities/
experimentation that are aligned to the
competencies that they should learn.
5. v
For the learner:
Welcome to the General Chemistry1 Alternative Delivery Mode (ADM)
Module on Limiting and Excess Reagent!
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; 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
6. vi
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.
Answer Key This contains answers to all activities in
the module.
At the end of this module you will also find:
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 What I Know before moving on to the other activities
included in the module.
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!
References This is a list of all sources used in
developing this module.
7. 1
What I Need to Know
This module was designed and written with you in mind. It is here to
help you explain the concept of limiting reagent in a chemical reaction and
identify the excess reagent. 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.
This module focuses on:
• Lesson 1 – Explaining the concept of limiting reagent in a chemical
reaction and identifying the excess reagent.
After going through this module, you are expected to:
1. Define limiting reagent and excess reagent;
2. Explain the concept of limiting reagent in chemical reaction; and
3. Identify the limiting reagent and excess reagent by solving, given the
mole/s and mass/es of a substance/s.
What I Know
Directions: Read the following sentence carefully. Encircle the letter of the
best answer.
1. What is a limiting reagent?
A. The reactant presents in quantities greater than necessary to react with
the quantity of the limiting reagent.
B. The reactant that is never used up.
C. The reactant that is completely consumed when a reaction is run to
completion.
D. The reactant that makes the product.
8. 2
2. You take a reactant A and calculate the amount of moles of another
reactant B required to use up all of A. How do you know which of two
reactants is the limiting one?
A. You compare the calculated amount of B to the actual amount
available. If more is required, then B is the limiting reagent. If less is
required, then A is the limiting reagent.
B. You compare the calculated amount of B to the actual amount
available. If more is required, then A is the limiting reagent. If less is
required, then B is the limiting reagent.
C. If the calculated value of B is larger than the amount of A, then A is the
limiting reagent.
D. If the calculated value of B is larger than the amount of A, then B is the
limiting reagent
3. The concept of the limiting reagent is analogous to the relationship
between men and women in a dance contest at a club. If there are 20 men
and only 10 women, then only 10 female/male pairs can compete. Ten men
will be left without partners. The number of women thus limits the number
of men that can dance in the contest, and there is an excess of men. This
analogous concept for limiting reagent is:
A.True B. False C. Somehow D. None sense
4. What is excess reagent?
A. The reactant presents in quantities greater than necessary to react
with the quantity of the limiting reagent.
B. The reactant that is never used up.
C. The reactant that is completely consumed when a reaction is run to
completion.
D. The reactant that makes the product.
5. In limiting and excess reagents, which of the two should be used in
computing for the value of the other substances in the chemical reaction?
A. Limiting reagent
B. Excess reagent
C. Both limiting reagent and excess reagent
D. Neither limiting reagent nor excess reagent
9. 3
For questions 6-8, refer to this chemical reaction.
6. 12 moles of Zn and 8 moles HCl.
A. Zn B. HCl C. H2 D. ZnCl2
7. 3 moles of zinc and 4 moles hydrochloric acid.
A. Zn B. HCl C. H2 D. ZnCl2
8. 40 grams of Zn and 56 grams of HCl.
A. Zn B. HCl C. H2 D. ZnCl2
For questions 9-11, consider this chemical reaction.
9. What is the excess reagent?
A. Al B. H2SO4 C. H2 D. Al2(SO4)3
10. How many grams of excess reagent is consumed in this reaction?
A. 0.010g B. 16.517g C. 14.683g D. 7.336g
11. What is the mass of excess reagent that remains after the reaction?
A. 72.664g B. 87.336g C. 32.664g D. 47.336g
For questions 12-14, consider the following reaction:
12. 65 grams of NH3 and 18 grams of O2.
A. NH3 B. O2 C. NO D. H2O
Zinc reacts with aqueous hydrochloric acid to produced
hydrogen gas and zinc chloride. Identify the limiting reagent
for each situation.
Zn + 2HCl H2 + ZnCl2
80 g of aluminum is placed in a solution that contains 40 g of
sulfuric acid. Hydrogen gas and aluminum sulfate were
produced in this reaction.
2Al + 3H2SO4 3H2 + Al2(SO4)3
Directions: Balance the chemical equation. Identify the excess
reagent for each situation.
NH3(g) + O2(g) NO(g) + H2O(g)
10. 4
13. 22 grams of NH3 and 44 grams of O2.
A. NH3 B. O2 C. NO D. H2O
14. 5 moles of NH3 and 10 moles of O2.
A.NH3 B. O2 C. NO D. H2O
15. Suppose that we combined 75 grams of magnesium with 45 grams of
nitrogen and heated them up. Predict the product of the reaction and
determined the limiting reagent.
A.MgN; N2 C. Mg3N2; Mg
B.Mg2N3; N2 D. Mg3N; Mg
This lesson will help you understand and explain the concept of
limiting reagent in a chemical reaction and identifying the excess reagent.
What’s In
In the previous lesson you have discussed the writing and balancing of
chemical equation and convert the amount of substance/s with a given
mass/es into moles. In this lesson we used this concept in a brief review to
link the current lesson. The next activity will help you check your prior
knowledge to understand the present lesson that focuses on the concept of
limiting reagents and identifying excess reagents.
Lesson
1
Limiting Reagent and
Excess Reagent
11. 5
ACTIVITY 1: WRITE AND BALANCE ME!
Let’s try this!
Directions: For each of the following problems, write and balance a chemical
equation.
1. When lithium hydroxide (LiOH) pellets are added to a solution of sulfuric
acid (H2SO4), lithium sulfate (Li2SO4) and water are formed.
2. Magnesium reacts with sodium fluoride (NaF) to produce magnesium
fluoride (MgF2) and elemental sodium.
3. If a copper coil is placed into a solution of silver nitrate (AgNO3), silver
crystals form and copper (I) nitrate (CuNO3) is generated.
4. When crystalline C6H12O6 is burned in oxygen gas (O2), carbon dioxide (CO2)
and water (H2O) vapor are formed.
5. Calcium carbonate (CaCO3) combines with hydrochloric acid (HCl) to
produce calcium chloride (CaCl2), water, and carbon dioxide gas.
Directions: Write and balance a chemical equation.
Aluminum bromide and chlorine gas react to form aluminum
chloride and bromine gas.
Answer:
2AlBr3 + 3Cl2 2AlCl3 + 3Br2
12. 6
ACTIVITY 2: CONVERT ME
Let’s try this!
Directions: Convert the mass of a substance into mole/s. For the mass/es,
use the attached periodic table of elements found at the back of this module.
1. How many moles are in 22 grams of argon?
2. How many moles are in 2.3 grams of phosphorus?
3. How many moles are in 68 grams of copper (II) hydroxide, Cu(OH)2?
2. How many moles are in 98.3 grams of aluminum hydroxide,
Al(OH)3?
(molar mass of Al(OH)3 is Al = 26.982 + (O = 3 x 15.999) +
(H = 3 x 1.008) = 78.003 g/mole)
Answer:
98.3 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟕𝟖.𝟎𝟎𝟑 𝒈𝒓𝒂𝒎𝒔
= 1.260 moles Al(OH)3
Examples
Directions: Convert the mass of a substance into mole/s. For the
mass/es, use the attached periodic table of elements found at the
back of this module.
1. How many moles are in 15 grams of lithium?
(molar mass of lithium is 6.941 g/mole from periodic table of
elements)
Answer:
15 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟔.𝟗𝟒𝟏 𝒈𝒓𝒂𝒎𝒔
= 2.161 moles Li
3. How many moles are in 65 grams of ammonium sulfate,
(NH4)2SO4?
(molar mass of (NH4)2SO4 is (N = 2 x 14.007) + (H = 8 x 1.008) +
S = 32.065 + (O = 4 x 15.999) = 132.139 g/mole)
Answer:
65 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟏𝟑𝟐.𝟏𝟑𝟗 𝒈𝒓𝒂𝒎𝒔
= 0.492 mole (NH4)2SO4
13. 7
4. How many moles are in 25 grams of calcium phosphate, Ca3(PO4)2?
5. How many moles are in 67.89 grams of manganese sulfate, Mn3(SO4)7?
“When a chemist carries out a reaction, the reactants are usually not
present in exact stoichiometric amounts, that is, in the proportions indicated
by the balanced equation. Because the goal of a reaction is to produce the
maximum quantity of a useful compound from the starting materials,
frequently a large excess of one reactant is supplied to ensure that the more
expensive reactant is completely converted to the desired product.
Consequently, some reactant will be left over at the end of the reaction. The
reactant used up first in a reaction is called the limiting reagent, because the
maximum amount of product formed depends on how much of this reactant
was originally present. When this reactant is used up, no more product can
be formed. Excess reagent are the reactants present in quantities greater than
necessary to react with the quantity of the limiting reagent.
The concept of the limiting reagent is analogous to the relationship
between men and women in a dance contest at a club. If there are 14 men
and only 9 women, then only 9 men/women pairs can compete. Five men will
be left without partners. The number of women thus limits the number of men
that can dance in the contest, and there is an excess of men” (Chang 103).
ACTIVITY 3: COMBO MEAL!
Directions: Analyze the example and answer the unfilled table and guide
questions.
What’s New
14 men + 9 women 9 men/women pairs
The one that limits: 9 women
Excess: 5 men
14. 8
Combo
Meals
No.
Components Available
Components
No. of
Combo
meals
will be
made?
The
one
that
limits
Excess
Components
Example 1 pc. chicken
joy
2 pcs. burger
steaks
1 plate
spaghetti
1 glass soft
drink
20 pcs.
chicken joy
20 pcs. burger
steaks
25 plates
spaghetti
21 glasses soft
drink
10
20
burger
steaks
10 pcs.
chicken
joy
15 plates
spaghetti
11 glasses
soft
drink
1 2 pcs. chicken
joy
1 pcs. burger
steak
1 cup rice
20 pcs.
chicken joy
10 pcs. burger
steaks
5 cups rice
2 2 pcs. chicken
joy
1 plate
spaghetti
1 glass soft
drink
3 pcs. lumpia
20 pcs.
chicken joy
10 plates
spaghetti
12 glasses
soft drink
27 pcs. lumpia
GUIDE QUESTIONS:
1. Why there is an excess component from available components in
producing combo meal?
_____________________________________________________________________
_____________________________________________________________________
____________________________________________________________________
2. Based on activity, what is limiting reagent?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
3. Based on activity, what is excess reagent?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
15. 9
What is It
The limiting reagent is the reactant that is completely used up in a
reaction, and thus determines when the reaction stops. The exact amount of
reactant needed to react with another element can be calculated. If the
reactants are not mixed in the correct stoichiometric proportions (as indicated
by the balanced chemical equation), then one of the reactants will be entirely
consumed while another will be left over. The limiting reagent is the one that
is totally consumed; it limits the reaction from continuing because there is
none left to react with the excess reagent.
Consider the production of sodium chloride (NaCl) from sodium and
chlorine gas. The balance equation is:
The coefficient is 1
2Na + Cl2 2NaCl
The coefficient is 2.
Suppose initially we have 4 moles of Na and 6 moles of Cl2. One way
to determine which of two reactants is the limiting reagent is to calculate the
ratio of a given mole/s and coefficient of the reactants
𝒎𝒐𝒍𝒆
𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕
For Na For Cl2
4 𝑚𝑜𝑙𝑒𝑠
2
= 2
6 𝑚𝑜𝑙𝑒𝑠
1
= 6
16. 10
Na has the lowest quantity per coefficient ratio; therefore, Na is the
limiting reagent. It means that the Na is the one that limits the reaction. Once
the Na is totally consumed, the reaction will stop. The excess reagent is Cl2.
Keep this in mind!
ACTIVITY 4: WHO AM I?
Directions: Identify the limiting reagent and excess reagent in the following
reactions.
1. 6 moles of zinc and 3 moles lead (II) nitrate (Pb(NO3)2) react to form zinc
nitrate (Zn(NO3)2) and lead.
Steps in determining limiting reagent and excess reagent using
mole/s as a given. Example Problem: 5 moles of hydrogen gas and 6 moles
of Nitrogen Monoxide react to form water and Nitrogen gas. What is the
limiting reagent? Excess reagent?
STEP 1: Write the chemical equation.
H2 + NO H2O + N2
STEP 2: Balance the chemical equation.
2H2 + 2NO 2H2O + N2
STEP 3: Solve the ratio of mole/s per coefficient of each reactants
(
𝑚𝑜𝑙𝑒/𝑠
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
).
For H2 For NO
5 𝑚𝑜𝑙𝑒𝑠
2
= 2.5
6 𝑚𝑜𝑙𝑒𝑠
2
= 3
STEP 4: Identify the limiting reagent.
H2 is limiting reagent
STEP 5: Identify the excess reagent.
NO is excess reagent
➢ The limiting reagent is the one that runs out first
between the reactants during reactions. A quick
and simple way to do that is to find which one has
the lowest quantity per coefficient ratio.
➢ Lower computed value = Limiting Reagent
➢ High computed value = Excess Reagent
17. 11
2. 5 moles aluminum bromide (AlBr3) and 3 moles chlorine gas (Cl2) react to
form aluminum chloride (AlCl3) and bromine gas (Br2).
3. 9 moles sodium phosphate (Na3PO4) and 5 moles calcium chloride (CaCl2)
react to form calcium phosphate (Ca3(PO4)2) and sodium chloride.
4. 3 moles potassium metal and 2 moles chlorine gas combine to form
potassium chloride (KCl).
5. 12 moles aluminum and 6 moles hydrochloric acid (HCl) react to form
aluminum chloride and hydrogen gas (H2).
18. 12
In this part, we use the given masses of the reactants for the chemical
reactions to identify the limiting reagent and excess reagent
What’s More
Steps in determining limiting reagent and excess reagent using
mass/es as a given.
Example Problem: Ammonium nitrate NH4NO3 reacts to sodium
phosphate Na3PO4 to produced ammonium phosphate (NH4)3PO4 and
sodium nitrate NaNO3. If you started with 30.0 grams of ammonium
nitrate and 50.0 grams of sodium phosphate:
a. Which of the reactants is the limiting reagent? Excess reagent?
STEP 1: Write the chemical equation.
NH4NO3 + Na3PO4 (NH4)3PO4 + NaNO3
STEP 2: Balance the chemical equation.
3NH4NO3 + Na3PO4 (NH4)3PO4 + 3NaNO3
STEP 3: Convert the given masses of reactants to moles.
For NH4NO3 (molar mass of NH4NO3 is (N = 2 x 14.007) +
(H = 4 x 1.008) + (O = 3 x 15.999) = 80.043 g/mole)
30.0 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟖𝟎.𝟎𝟒𝟑 𝒈𝒓𝒂𝒎𝒔
= 0.375 mole NH4NO3
For Na3PO4 (molar mass of Na3PO4 is (Na = 3 x 22.990) + P
= 30.974 + (O = 4 x 15.999) = 163.940 g/mole)
50.0 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟏𝟔𝟑.𝟗𝟒𝟎 𝒈𝒓𝒂𝒎𝒔
= 0.305 mole Na3PO4
STEP 4: Solve the ratio of mole/s per coefficient of each reactants
(
𝑚𝑜𝑙𝑒/𝑠
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
).
For NH4NO3
0.375 𝑚𝑜𝑙𝑒
3
= 0.125
For Na3PO4
0.305 𝑚𝑜𝑙𝑒
1
= 0.305
STEP 5: Identify the limiting reagent.
NH4NO3 is limiting reagent
STEP 6: Identify the excess reagent.
Na3PO4 is excess reagent
19. 13
ACTIVITY 5: IDENTIFY ME?
Directions: Identify the limiting reagent and excess reagent in the following
reactions.
1. 89.6 g of calcium hydroxide (Ca(OH)2) and 46.9 g of phosphoric acid (H3PO4)
react to form calcium phosphate (Ca3(PO4)2) and water.
2. 57 g of copper and 23.5 g of sulfuric acid (H2SO4) react to form copper (II)
sulfate (CuSO4) and water and sulfur dioxide (SO2).
3. Silver phosphate (Ag3PO4) and sodium nitrate (NaNO3) were produced when
200.0 grams of silver nitrate (AgNO3) and 200.0 grams of sodium phosphate
(Na3PO4) react.
4. 10 g of zinc metal reacts with 17.6 g of phosphoric acid (H3PO4) to produce
zinc phosphate (Zn3(PO4)2) and hydrogen gas.
5. Aluminum chloride (AlCl3) is formed when 5.3 grams of chlorine gas and
3.0 grams of aluminum react with each other.
What’s More Again!
To get the amount of excess reagent that consumed in the reaction and
amount of excess reagent that remains after the reaction the following
example will apply.
20. 14
Example:
160 grams of aluminum is placed in a solution that contains 80
grams of sulfuric acid. Hydrogen gas and aluminum sulfate are
produced in this reaction. (a) Identify the limiting reactant. (b) how
many grams of excess reagent is consumed in this reactant? (c) What
is the mass of excess reagent that remains after the reaction is
complete?
STEP 1: Write the chemical equation.
Al + H2SO4 H2 + Al2(SO4)3
STEP 2: Balance the chemical equation.
2Al + 3H2SO4 3H2 + Al2(SO4)3
STEP 3: Convert the given masses of reactants to moles.
For Al (molar mass of Al = 26.982 g/mole)
160 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟐𝟔.𝟗𝟖𝟐 𝒈𝒓𝒂𝒎𝒔
= 5.930 moles Al
For H2SO4 (molar mass of H2SO4 is (H = 2 x 1.008) +
S = 32.065 + (O = 4 x 15.999) = 98.077 g/mole)
80 grams x
𝟏 𝒎𝒐𝒍𝒆
𝟗𝟖.𝟎𝟕𝟕 𝒈𝒓𝒂𝒎𝒔
= 0.816 mole H2SO4
STEP 4: Solve the ratio of mole/s per coefficient of each reactants
(
𝑚𝑜𝑙𝑒/𝑠
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
).
For Al
5.930 𝑚𝑜𝑙𝑒
2
= 2.965
For H2SO4
0.816 𝑚𝑜𝑙𝑒
3
= 0.272
STEP 5: Identify the limiting reagent.
H2SO4 is limiting reagent
STEP 6: Identify the excess reagent.
Al is excess reagent
STEP 7: Use the formula LR x (coefficient x molar mass) to find the
amount in grams of excess reagent that consumed in the reaction.
Where:
LR = amount of mole per coefficient ratio of limiting reagent.
Coefficient = coefficient of excess reagent.
Molar Mass = molar mass of excess reagent.
0.272 x (2 moles x 26.982 g/mole) = 14.678 grams Al
It means, out of 160 grams of Al given, only 14.678 grams
of Al consumed during the reaction.
21. 15
STEP 8: Solve the mass of excess reagent that remains after the reaction.
Original amount of excess reagent MINUS the amount of excess reagent
consumed.
160 g Al – 14.678 g Al = 145.322 g Al remains after the reaction.
ACTIVITY 6: TRY ME?
Directions: Perform the following in each of the given sample reaction below:
(a) Identify the limiting reactant. Do this in a separate sheet of paper.
(b) how many grams of excess reagent is consumed in this reactant?
(c) What is the mass of excess reagent that remains after the reaction
is complete?
1. When 20 g of magnesium metal burns, it reacts with 40 g oxygen gas in the
air to form magnesium oxide (MgO), a white powder.
2. In the Blast Furnace, iron is made by reacting 35 g of iron (III) oxide (Fe2O3)
with 55 g of carbon monoxide (CO) gas. Carbon dioxide gas is also
produced.
3. Let say, in our bodies, 15 g of starch (C6H10O5), which we get from our food,
reacts with 10 g of water and breaks down to form glucose ( C6H12O6).
4. When 60 g of calcium metal is added to 80 g of water, a hydrogen gas is
given off and calcium hydroxide (Ca(OH)2) solution is formed. When tested
with a burning splint, the gas burns with a "pop".
5. 15 g of calcium carbonate (CaCO3) combines with 25 g of hydrochloric acid
to produce calcium chloride (CaCl2), water and carbon dioxide.
What I Have Learned
Directions: Fill in the missing words found in the box to complete the
paragraph.
excess reagent lower limiting reagent
left over high stops
22. 16
(1)_______________ is the reactant that is completely consumed during a
chemical reaction. Once this reagent is consumed, the reaction (2)___________.
(3)____________is the reactant that is (4)____________once the limiting reagent
is consumed. Using the balanced equation of a chemical reaction, molar mass
and a given mole or mass of the reactants, the limiting and excess reagent
can identify, the (5)_________ computed value is the limiting reagent and
(6)_____________ computed value is the excess reagent.
qqqqqqqqq What Can I Do
Identifying the limiting reagent has a big role in determining the amount
of product produced. Here is the sample application:
Ethane (C2H6) reacts with oxygen gas to produce carbon dioxide and
water. (a) If 5 moles of ethane is placed in a container with 16 moles of oxygen
gas, how many moles of CO2 will be produced? (b) If 30 g of ethane is mixed
with 84 g of oxygen gas, how many grams of water will be produced in this
reaction? Note: First, write and balanced the chemical reaction and identify
the limiting reagent, use the steps in previous activities.
2C2H6 + 7O2 4CO2 + 6H2O
In this example the oxygen gas is limiting reagent. (C2H6 5/2 = 2.50
and O2 16/7 = 2.29, the lower computed value between the reactants is O2,
therefore O2 is the limiting reagent)
Keep this in mind!
➢ To compute the other data such as the
amount of each products, the given value
either in mole or mass of limiting reagent is
used.
23. 17
Let’s solve the problem:
For letter (a) how many moles of carbon dioxide produced?
16 moles O2
𝟒 𝒎𝒐𝒍𝒆𝒔 CO2
𝟕 𝒎𝒐𝒍𝒆𝒔 𝑶𝟐
= 9.143 moles CO2
from balanced equation
It means 9.14 moles of CO2 produced out of 16 moles of oxygen gas
using the balanced chemical equation.
For letter (b) how many grams of water will be produced?
molar mass of product (H2O)
84 g O2
𝟏 𝒎𝒐𝒍𝒆 O2
𝟑𝟏.𝟗𝟗𝟖 𝒈 𝑶𝟐
x
𝟔 𝒎𝒐𝒍𝒆𝒔 H2O
𝟕 𝒎𝒐𝒍𝒆𝒔 𝑶𝟐
x
𝟏𝟖.𝟎𝟏𝟓 𝒈 H2O
𝟏 𝒎𝒐𝒍𝒆 𝑯𝟐𝑶
= 40.536 g H2O
molar mass of
limiting reagent (O2)
from balanced equation
It means 40.54 g of H2O produced out of 84 g of oxygen gas using the
balanced chemical equation and molar mass.
What I Can Do!
Directions: Compute the needed data in the following reactions. Do this in
your notes.
1. How many mole/s of hydrogen gas (H2) is produced from the reaction of 5
moles of magnesium (Mg) and 7.5 moles of hydrochloric acid (HCl)? The other
product is magnesium chloride (MgCl2).
2. How many grams of ammonia (NH3) can be produced from the reaction of
30.0 grams of nitrogen gas (N2) and 25.0 grams of hydrogen gas (H2)?
3. How much silver phosphate (Ag3PO4) and sodium nitrate NaNO3 is
produced when 200.0 grams of silver nitrate (AgNO3) and 200.0 grams of
sodium phosphate (Na3PO4) react?
24. 18
Assessment
Directions: Choose the letter of the best answer. Encircle the letter of the
correct answer.
1. A propane grill on a backyard patio is left burning for too long and
eventually goes out, the reaction is C3H8 + O2 CO2 + H2O. Which of
the following chemicals is the limiting reagent?
a. Carbon dioxide c. Oxygen gas
b. Propane d. Water vapor
2. Which of the following statement defines limiting reagent?
a. The reagent that weighs the least at the start of a chemical reaction.
b. The reagent that produces the least amount of product during a
chemical reaction.
c. The reagent least likely to become active in a chemical reaction.
d. The reagent that does not produce a product during a chemical reaction.
3. What information is needed to find the limiting reagent?
a. The balanced equation
b. The molecular weight of reactants and products
c. The amount of reactants used
d. All of the answer are correct
4. Which of the following describe the excess reagent?
a. Any reactant that is not completely consumed in the chemical reaction.
b. Have some amount unchanged, or left over, after the reaction.
c. Cannot be calculated without performing the reaction.
d. Has no effect in the amount of product formed.
5. When 12 moles of O2 reacts with 6 moles of C10H8, what is the limiting
reactant? C10H8 + 12O2 10CO2 + 4H2O
a. O2 b. H2O c. CO2 d. C10H8
For questions 6-8, refer to this chemical reaction.
Nitric acid can be neutralized by any base to form a salt and water, as in the
following equation.
Mg(OH)2 + 2HNO3 Mg(NO3)2 + 2H2O
25. 19
Identify the limiting reagent for each situation.
6. 2 moles of Mg(OH)2 and 9 moles HNO3.
a. Mg(OH)2 b. HNO3 c. Mg(NO3)2 d. H2O
7. 1 mole of Mg(OH)2 and 5 moles HNO3.
a. Mg(OH)2 b. HNO3 c. Mg(NO3)2 d. H2O
8. 6 grams of Mg(OH)2 and 3 grams of HNO3.
a. Mg(OH)2 b. HNO3 c. Mg(NO3)2 d. H2O
For questions 9-11, consider this chemical reaction.
In a small-scale experiment, 28.5 g silver nitrate in solution is reacted with 8
g copper wire.
2AgNO3 + Cu Cu(NO3)2 + 2Ag
9. In the reaction, what is the excess reagent?
a. AgNO3 b. Cu c. Cu(NO3)2 d. Ag
10. What is the mass of excess reagent consumed in the reaction?
a. 3.363 g b. 8.227 g c. 5.338 g d. 7.249 g
11. How many grams of excess reagent remains after the reaction?
a. 4.657 g b. 5.676 g c. 3.675 g d. 2.662 g
For questions 12-14, consider the following reaction:
Consider the following unbalanced chemical equation.
TiCl4 + O2 TiO2 + Cl2
Identify the excess reagent in each situation.
12. 65 grams of TiCl4 and 18 grams of O2.
a. TiCl4 b. O2 c. TiO2 d. Cl2
13. 22 grams of TiCl4 and 44 grams of O2.
a. TiCl4 b. O2 c. TiO2 d. Cl2
14. 10 moles of TiCl4 and 5 moles of O2.
a. TiCl4 b. O2 c. TiO2 d. Cl2
15. 145 grams of tin (IV) oxide (SnO2) combines with 3 grams of hydrogen
gas. Predict the products of the reaction and determined the limiting
reagent.
a. Sn + H2O; SnO2 c. Sn + H2O; H2
b. Sn2 + H2O; SnO2 d. Sn2 + H2O; H2
26. 20
Directions: Compute the needed data in the following reactions.
1. Nitrogen gas can be prepared by passing gaseous ammonia over solid
copper(II) oxide at high temperatures. The other products of the reaction
are solid copper and water vapor. If a sample containing 18.1 g of NH3 is
reacted with 90.4 g of CuO, which is the limiting reactant? How man grams
of N2 will be formed?
2. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. In
one process, 637.2 g of NH3 are treated with 1142 g of CO2. The other
product is water. (a) Which of the two reactants is the limiting reagent? (b)
Calculate the mass of (NH2)2CO formed. (c) How much excess reagent (in
grams) is left at the end of the reaction?
Additional Activities
30. 24
References
Chang, Raymond. Chemistry 10th ed. New York: McGraw-Hill, 2007. pages 103-106
Petrucci, Ralph H., et al. General Chemistry 9th ed. New Jersey: Pearsin Prentice
Hall, 2007. page 1
Zumdahl, Steven S. and Susan A. Zumdahl. Chemistry 7th ed. Boston, New York:
Houghton Mifflin, 2007. pages 110-111
31. 25
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