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Resources
Chapter Presentation
Bellringer
Transparencies
Sample Problems
Visual Concepts
Standardized Test Prep
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Chapter 9
Stoichiometry
Table of Contents
Section 1 Calculating Quantities in Reactions
Section 2 Limiting Reactants and Percentage Yield
Section 3 Stoichiometry and Cars
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Bellringer
Basic pancake
Yield: Serves 4
Ingredients:
2 cups all-purpose flour, stirred or sifted before measuring
2 1/2 teaspoons baking powder
3 tablespoons granulated sugar
1/2 teaspoon salt
2 large eggs
1 1/2 to 1 3/4 cups milk
2 tablespoons melted butter
Serve with maple syrup and butter !!
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Stoichiometry" is derived from
the Greek words (stoicheion,
meaning element) and (metron,
meaning measure.)
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Objectives
• Use proportional reasoning to determine mole ratios
from a balanced chemical equation.
• Explain why mole ratios are central to solving
stoichiometry problems.
• Solve stoichiometry problems involving mass by
using molar mass.
• Solve stoichiometry problems involving the volume
of a substance by using density.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Objectives, continued
• Solve stoichiometry problems involving the number
of particles of a substance by using Avogadro’s
number.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Balanced Equations Show Proportions
• The proportions of the ingredients in a muffin recipe
let you adjust the amounts to make enough muffins
even if you do not have balanced amounts initially.
• A balanced chemical equation is very similar to a
recipe in that the coefficients in the balanced equation
show the proportions of the reactants and products.
• Consider the reaction for the synthesis of water.
2H2 + O2
2H2O
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Balanced Equations Show Proportions,
continued
2H2 + O2  2H2O
• The coefficients show that two molecules of hydrogen
react with one molecule of oxygen and form two
molecules of water.
• Calculations that involve chemical reactions use the
proportions from balanced chemical equations to find
the quantity of each reactant and product involved.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Balanced Equations Show Proportions,
continued
• For each problem in this section, assume that there is
more than enough of all other reactants to completely
react with the reactant given.
• Also assume that every reaction happens perfectly,
so that no product is lost during collection.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Balanced Equations Show Proportions,
continued
Relative Amounts in Equations Can Be Expressed
in Moles
• The coefficients in a balanced equation also
represent the moles of each substance.
• For example, the equation below shows that 2 mol of
C8H18 react with 25 mole O2 to form 16 mol CO2
and 18 mol H2O.
2C8H18 + 25O2  16CO2 + 18H2O
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Balanced Equations Show Proportions,
continued
Relative Amounts in Equations Can Be Expressed
in Moles, continued
• You can determine how much of a reactant is needed
to produce a given quantity of product, or how much
of a product is formed from a given quantity of
reactant.
• The branch of chemistry that deals with quantities of
substances in chemical reactions is known as
stoichiometry.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key
• The coefficients in a balanced chemical equation
show the relative numbers of moles of the substances
in the reaction.
• As a result, you can use the coefficients in conversion
factors called mole ratios.
• Mole ratios bridge the gap and can convert from
moles of one substance to moles of another.
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Chapter 9
Visual Concepts
Stoichiometry
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Converting Between Amounts in Moles
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Chapter 9
Visual Concepts
Conversion of Quantities in Moles
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Section 1 Calculating Quantities in
Reactions
Chapter 9
Using Mole Ratios
Sample Problem A
Consider the reaction for the commercial preparation
of ammonia.
N2 + 3H2  2NH3
How many moles of hydrogen are needed to prepare
312 moles of ammonia?
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Using Mole Ratios, continued
Sample Problem A Solution
amount of NH3 = 312 mol
amount of H2 = ? mol
3 mol H2 = 2 mol NH3
The mole ratio must cancel out the units of mol NH3
given in the problem and leave the units of mol H2.
Therefore, the mole ratio is
3 mol H2
2 mol NH3
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Using Mole Ratios, continued
Sample Problem A Solution, continued
3 mol H2
? mol H2  312 mol NH3 

2 mol NH3
468 mol H2
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Sample Problem A,pg.304
Practice 1
Calculate the amounts requested if 1.34 mol H2O2
completely react according to the following
equation.
2H2O2 → 2H2O + O2
a. moles of oxygen formed
b. moles of water formed
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Sample Problem A,pg.304
Practice 2
Calculate the amounts requested if 3.30 mol Fe2O3 completely
react according to the following equation.
Fe2O3 + 2Al → 2Fe + Al2O3
a. moles of aluminum needed
b. moles of iron formed
c. moles of aluminum oxide formed
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Homework
1. How many moles of H2CO3 can form when 2.57
mol CO2 reacts with excess H2O ?
H2O (l) + CO2 (g)  H2CO3 (aq)
A: (2.57molH2CO3)
2. In the combustion of propane how many moles
of C3H8 are needed to combine completely with
2.96 mol O2? How many moles of CO2 form? How
many moles of H2O form? A: (0.592mol C3H8 ; 1.78molCO2 ;
2.37mol H2O)
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key, continued
Getting into Moles and Getting out of Moles
•
All stoichiometry problems have three basic steps.
1. First, change the units you are given into moles.
1. Second, use the mole ratio to determine moles of the
desired substance.
1. Third, change out of moles to whatever unit you need for
your final answer.
•
If you are given moles, just skip to the first step.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key, continued
Solving Stoichiometry Problems
1. Gather information.
•
Determine the balanced equation for the reaction.
•
Write the information provided for the given
substance.
•
•
Determine the information you need to change the given
units into moles.
Write the units you are asked to find for the
unknown substance.
•
Determine the information you need to change moles into
the desired units.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key, continued
Solving Stoichiometry Problems, continued
• Write an equality using substances and their
coefficients that shows the relative amounts of the
substances from the balanced equation.
2. Plan your work.
•
•
Think through the three basic steps :
•
change to moles
•
use the mole ratio
•
change out of moles
Find the conversion factors for each step.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key, continued
Solving Stoichiometry Problems, continued
•
Write the mole ratio: moles of unknown subs tance
moles of given subs tance
3. Calculate.
•
Write a “?” with the units of the answer followed by
“=“ and the quantity of the given substance.
•
Write the conversion factors—including the mole
ratio—in order so that you change the units of the
given substance to the units needed.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key, continued
Solving Stoichiometry Problems, continued
• Cancel units and check that the remaining units are
the required units of the unknown substance.
• When you have finished your calculations, round off
the answer to the correct number of significant figures.
• In the examples in this book, only the final answer is
rounded off.
• Report your answer with correct units and with the
name or formula of the substance.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
The Mole Ratio Is the Key, continued
Solving Stoichiometry Problems, continued
4. Verify your result.
•
•
Verify your answer by estimating.
•
Round off the numbers in the setup in step 3 and make a
quick calculation.
•
Or, compare conversion factors in the setup and to predict
the size of an answer.
Make sure your answer is reasonable.
•
Large differences in predicted quantities should alert you
that there may be an error and that you should doublecheck your work.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, Volume, or
Particles
•
Stoichiometric calculations are used to determine
how much of the reactants are needed and how
much product is expected.
•
Calculations don’t always start and end with moles.
•
Mass, volume, or number of particles can all be
used as the starting and ending quantities of
stoichiometry problems.
•
The key to all these problems is the mole ratio.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, Volume, or
Particles, continued
For Mass Calculations, Use Molar Mass
•
The conversion factor for converting between mass
and amount in moles is molar mass. A mass-mass
problem is a three-step process.
1. Convert the given mass in grams into moles.
2. Use the mole ratio to convert into moles of the desired
substance.
3. Convert this amount in moles into grams.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Solving Mass-Mass Problems
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Chapter 9
Visual Concepts
Mass-Mass Calculations
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Section 1 Calculating Quantities in
Reactions
Chapter 9
Problems Involving Mass
Sample Problem B
What mass of NH3 can be made from 1221 g H2 and
excess N2?
N2 + 3H2
2NH3
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, continued
Sample Problem B Solution
mass of H2 = 1221 g H2
molar mass of H2 = 2.02 g/mol
mass of NH3 = ? g NH3
molar mass of NH3 = 17.04 g/mol
3 mol H2 = 2 mol NH3
To change grams to moles, use the molar mass of H2.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, continued
Sample Problem B Solution, continued
The mole ratio must cancel out the units of mol H2
given in the problem and leave the units of mol NH3.
Therefore, the mole ratio is
2 mol NH3
3 mol H2
To change moles of NH3 to grams, use the molar mass
of NH3.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass
Sample Problem B Solution, continued
? g NH3  1221 g H2 
2 mol NH3 17.04 g NH3
1 mol H2



2.02 g H2 3 mol H2
1 mol NH3
6867 g NH3
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Sample Problem B , pg. 307
Practice 1 to 4
Use the equation below to answer the questions that
follow. Fe2O3 + 2Al → 2Fe + Al2O3
1) How many grams of Al are needed to
completely react with 135 g Fe2O3?
2 )How many grams of Al2O3 can form when
23.6 g Al react with excess Fe2O3?
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Practice continued
Fe2O3 + 2Al → 2Fe + Al2O3
3 )How many grams of Fe2O3 react with excess Al
to make 475 g Fe?
4 ) How many grams of Fe will form when 97.6 g
Al2O3 form?
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Homework
1. What mass of H2O is produced if 65.2 g CaCO3
reacts with excess H3PO4, to form Ca3(PO4)2 , H2O,
and CO2 ?
(A: 11.7 g H2O)
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2. What mass of O2 forms when 49.89 g KClO3
decomposes? (KCl also forms. )
(A: 19.54 g O2)
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, Volume, or
Particles, continued
For Volume, You Might Use Density and Molar Mass
• To do calculations involving liquids, you add two more
steps to mass-mass problems—converting volume to
mass and mass to volume.
• One way to convert between the volume and mass of
a substance is to use the density.
• If a substance in the problem is a gas at standard
temperature and pressure (STP), use the molar
volume of a gas (22.41 L/mol for any gas) to convert
the volume of the gas to moles.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, Volume, or
Particles, continued
For Volume, You Might Use Density and Molar Mass,
continued
• If a substance in the problem is in aqueous solution,
use the concentration of the solution to convert the
volume of the solution to the moles of the substance
dissolved.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Solving Volume-Volume Problems
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Volume
Sample Problem C
What volume of H3PO4 forms when 56 mL POCl3
completely react? (density of POCl3 = 1.67 g/mL;
density of H3PO4 = 1.83 g/mL)
POCl3(l) + 3H2O(l)
H3PO4(l) + 3HCl(g)
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Volume, continued
Sample Problem C Solution
volume POCl3 = 56 mL POCl3
density POCl3 = 1.67 g/mL
molar mass POCl3 = 153.32 g/mol
volume H3PO4 = ?
density H3PO4 = 1.83 g/mL
molar mass H3PO4 = 98.00 g/mol
1 mol POCl3 = 1 mol H3PO4
To change milliliters of POCl3 to moles, use the density
of POCl3 followed by its molar mass.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Volume, continued
Sample Problem C Solution, continued
The mole ratio must cancel out the units of mol POCl3
given in the problem and leave the units of mol H3PO4.
1 mol H3PO4
Therefore, the mole ratio is
1 mol POCl3
To change out of moles of H3PO4 into milliliters, use
the molar mass of H3PO4 followed by its density.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Volume, continued
Sample Problem C Solution, continued
? mL H3PO4
1.67 g POCl3
 56 mL POCl3 

1 mL POCl3
1 mol POCl3
1 mol H3PO4 98.00 g H3PO4



153.32 g POCl3 1 mol POCl3
1 mol H3PO4
1 mL H3PO4

1.83 g H3PO4
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33 mL H3PO4
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Sample Problem C, pg.309
Practice 1 to 2
Use the densities and balanced equation provided to answer the
questions that follow. (density of C5H12 = 0.620 g/mL; density of
C5H8 =0.681g/mL; density of H2 = 0.0899 g/L) C5H12(l) →
C5H8(l) + 2H2(g)
1) How many milliliters of C5H8 can be made from
366 mL C5H12?
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2 )How many liters of H2 can form when 4.53 ×103 mL
C5H8 form?
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, Volume, or
Particles, continued
For Number of Particles, Use Avogadro’s Number
• You can use Avogadro’s number,
6.022  1023 particles/ mol, in stoichiometry problems.
• If you are given particles and asked to find particles,
Avogadro’s number cancels out.
• For this kind of calculation you use only the coefficients from
the balanced equation.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Solving Particle Problems
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Particles
Sample Problem D
How many grams of C5H8 form from 1.89 × 1024
molecules C5H12?
C5H12(l)
C5H8(l) + 2H2(g)
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Particles, continued
Sample Problem D Solution
quantity of C5H12 = 1.89  1024 molecules
Avogadro’s number = 6.022  1023 molecules/mol
mass of C5H8 = ? g C5H8
molar mass of C5H8 = 68.13 g/mol
1 mol C5H12 = 1 mol C5H8
Set up the problem using Avogadro’s number to
change to moles, then use the mole ratio, and finally
use the molar mass of C5H8 to change to grams.
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Particles, continued
Sample Problem D Solution, continued
? g C5H8  1.89  1024 molecules C5H12 
1 mol C5H12
1 mol C5H8 68.13 g C5H8



23
6.022  10 molecules C5H12 1 mol C5H12
1 mol C5H8
214 g C5H8
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Chapter 9
Section 1 Calculating Quantities in
Reactions
Problems Involving Mass, Volume, or
Particles, continued
Many Problems, Just One Solution
•
Each stoichiometry solution will have three steps.
1. Take whatever you are given, and find a way to change it
into moles.
1. Then, use a mole ratio from the balanced equation to get
moles of the second substance.
1. Finally, find a way to convert the moles into the
units that you need for your final answer.
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Chapter 9
Sample Problem D pg. 311
Practice 1 to 2
Use the equation provided to answer the questions that
follow. Br2(l) + 5F2(g) → 2BrF5(l)
1) How many molecules of BrF5 form when 384 g Br2
react with excess F2?
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Use the equation provided to answer the questions
that follow. Br2(l) + 5F2(g) → 2BrF5(l)
2 )How many molecules of Br2 react with 1.11× 1020
molecules F2?
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Chapter 9
Section 1 Calculating
Quantities in Reactions
Homework
1) How many molecules of carbon dioxide
are produced when 79.5 g of K2CO3
decompose ?
A:( 3.46 X10 23 molecules CO2)
2) If 2.46X10 25 molecules of chlorine react
completely, how many grams of NaCl will
form ?
A: ( 4.77X10 3 g NaCl )
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Homework
Section Review pg. 311,
# (1 to 5)
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4. Use the equation below to answer the questions
that follow.
Br2 + Cl2 → 2BrCl
a. How many moles of BrCl form when 2.74 mol
Cl2 react with excess Br2?
b. How many grams of BrCl form when 239.7 g Cl2
react with excess Br2?
c. How many grams of Br2 are needed to
react with 4.53 × 1025 molecules Cl2?
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5. The equation for burning C2H2 is
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
a. If 15.9 L C2H2 react at STP, how many moles of
CO2 are produced? (Hint: At STP, 1 mol = 22.41 L
for any gas.)
b. How many milliliters of CO2 (density = 1.977 g/L)
can be made when 59.3 mL O2 (density = 1.429 g/L)
react?
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Bellringer
• Discuss how the amount of product calculated from a
given amount of reactant is an ideal amount.
• Discuss how the amount of product formed from a
real reaction might differ from the amount calculated.
• Offer reasons for the difference.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Objectives
• Identify the limiting reactant for a reaction and use it
to calculate theoretical yield.
• Perform calculations involving percentage yield.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield
• In the previous section, we assumed that 100% of the
reactants changed into products.
• That is what should happen theoretically.
• In the real world, other factors can limit the yield of a
reaction.
• the amounts of all reactant
• the completeness of the reaction
• product lost in the process
• Whatever reactant is in short supply will limit the
quantity of product made.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
The Limiting Reactant Forms the Least Product
• Reactants of a reaction are seldom present in ratios
equal to the mole ratio in the balanced equation, so
one of the reactants is used up first.
• For example, If you combine 0.23 mol Zn and 0.60 mol HCl,
would they react completely?
Zn + 2HCl  ZnCl2 + H2
• The coefficients indicate that 0.23 mol Zn forms 0.23 mol H2,
and 0.60 mol HCl forms 0.30 mol H2.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
The Limiting Reactant Forms the Least Product,
continued
• Zinc is called the limiting reactant because it limits
the amount of product that can form.
• It gets used up.
• The HCl is the excess reactant because there is
more than enough HCl present to react with all of the
Zn.
• There will be some HCl left over.
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Chapter 9
Visual Concepts
Limiting Reactants and Excess Reactants
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
Determine Theoretical Yield from the Limiting
Reactant
• The maximum quantity of product that a reaction
could theoretically make if everything about the
reaction works perfectly is called the theoretical yield.
• The theoretical yield of a reaction should always be
calculated based on the limiting reactant.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield
Sample Problem E
Identify the limiting reactant and the theoretical yield of
phosphorous acid, H3PO3, if 225 g of PCl3 is mixed
with 125 g of H2O.
PCl3 + 3H2O  H3PO3 + 3HCl
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
Sample Problem E Solution
mass PCl3 = 225 g PCl3
molar mass PCl3 = 137.32 g/mol
mass H2O = 125 g H2O
molar mass H2O = 18.02 g/mol
mass H3PO3 = ? g H3PO3
molar mass H3PO3 = 82.00 g/mol
1 mol PCl3 = 1 mol H3PO3
3 mol H2O = 1 mol H3PO3
Set up problems that will calculate the mass of H3PO3 you
would expect to form from each reactant.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
Sample Problem E Solution, continued
? g H3PO3
1 mol PCl3
1 mol H3PO3
 225 g PCl3 

137.32 g PCl3 1 mol PCl3
82.00 g H3PO3

 134 g H3PO3
1 mol H3PO3
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
Sample Problem E Solution, continued
? g H3PO3
1 mol H2O 1 mol H3PO3
 123 g H2O 


18.02 g H2O 3 mol H2O
82.00 g H3PO3
 187 g H3PO3
1 mol H3PO3
PCl3 is the limiting reactant.
The theoretical yield is 134 g H3PO3.
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Sample Problem E, practice pg. 314
Using the given reaction , identify the limiting
reactant and the theoretical yield (in grams) of HCl
for each pair of reactants.
PCl3 + 3H2O  H3PO3 + 3HCl
1) 3.00 mol PCl3 and 3.00 mol H2O
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PCl3 + 3H2O  H3PO3 + 3HCl
2 )75.0 g PCl3 and 75.0 g H2O
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PCl3 + 3H2O  H3PO3 + 3HCl
3 )1.00 mol of PCl3 and 50.0 g of H2O
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Homework
1) Hydrogen sulfide gas is formed when HCl reacts
with FeS. FeCl2 is the other product. 130.5 g of FeS is
mixed with 70.4 g of HCl in solution. (A: 33.06 g H2S)
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
Limiting Reactants and the Food You Eat
• In industry, the cheapest reactant is often used as the
excess reactant so that the expensive reactant is
more completely used up.
• In addition to being cost-effective, this practice can be
used to control which reactions happen.
• For example, cost is used to choose the excess reactant
when making banana flavoring, isopentyl acetate.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Limiting Reactants and Theoretical Yield,
continued
Limiting Reactants and the Food You Eat, continued
• Acetic acid is the excess reactant because it costs
much less than isopentyl alcohol.
CH3COOH + C5H11OH → CH3COOC5H11 + H2O
acetic acid + isopentyl alcohol → isopentyl acetate + water
• When a large excess of acetic acid is present, almost
all of the isopentyl alcohol reacts.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Actual Yield and Percentage Yield
• Sometimes reactions do not make all of the product
predicted by the theoretical yield.
• In most cases, the actual yield, the mass of product
actually formed, is less than expected.
• There are several reasons why the actual yield is
usually less than the theoretical yield.
• Many reactions do not use up the limiting reactant.
• Some of the products turn back into reactants, resulting
in a mixture of reactants and products.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Actual Yield and Percentage Yield, continued
• In many cases the main product must go through
additional steps to separate it from other chemicals:
• A liquid may need to be distilled, or isolated based on its
boiling point.
• Solid compounds must be recrystallized. Some of the
product may be lost in the process.
• There also may be other reactions, called side reactions,
that can use up reactants without making the desired
product.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Actual Yield and Percentage Yield, continued
Determining Percentage Yield
• The ratio relating the actual yield of a reaction to its
theoretical yield is called the percentage yield.
• A percentage yield describes the efficiency of a
reaction.
• The percentage yield is
actual yield
percentage yield 
 100
theoretical yield
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Chapter 9
Visual Concepts
Comparing Actual and Theoretical Yield
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Chapter 9
Visual Concepts
Percent Yield
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Section 2 Limiting Reactants and
Percentage Yields
Chapter 9
Calculating Percentage Yield
Sample Problem F
Determine the limiting reactant, the theoretical yield,
and the percentage yield if 14.0 g N2 are mixed with
9.0 g H2, and 16.1 g NH3 form.
N2 + 3H2
2NH3
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Calculating Percentage Yield, continued
Sample Problem F Solution
mass N2 = 14.0 g N2
molar mass N2 = 28.02 g/mol
mass H2 = 9.0 g H2
molar mass H2 = 2.02 g/mol
theoretical yield of NH3 = ? g NH3
molar mass NH3 = 17.04 g/mol
actual yield of NH3 = 16.1 g NH3
1 mol N2 = 2 mol NH3 and 3 mol H2 = 2 mol NH3
Set up problems that will calculate the mass of NH3 you
would expect to form from each reactant.
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Section 2 Limiting Reactants and
Percentage Yields
Chapter 9
Calculating Percentage Yield, continued
Sample Problem F Solution
Set up problems that will calculate the mass of NH3 you would
expect to form from each reactant.
? g NH3  14.0 g N2 
2 mol NH3 17.04 g NH3
1 mol N2


 17.0 g NH3
28.02 g N2 1 mol N2
1 mol NH3
? g NH3  9.0 g H2 
1 mol H2 2 mol NH3 17.04 g NH3


 51 g NH3
2.02 g H2 3 mol H2
1 mol NH3
The smaller quantity made, 17.0 g NH3, is the theoretical yield.
The limiting reactant is N2.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Calculating Percentage Yield
Sample Problem F Solution, continued
percentage yield 
16.1 g (actual yield)
 100  94.7%
17.0 g (theoretical yield)
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Sample Problem F, practice pg. 317
Determine the limiting reactant and the percentage
yield for each of the following.
1) 14.0 g N2 react with 3.15 g H2 to give an actual
yield of 14.5 g NH3.
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Homework, practice pg.317
Determine the limiting reactant and the percentage yield for each of the
following.
2) In a reaction to make ethyl acetate, 25.5 g CH3COOH
react with 11.5 gC2H5OH to give a yield of 17.6 g
CH3COOC2H5.
CH3COOH + C2H5OH → CH3COOC2H5 + H2O
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3 )16.1 g of bromine are mixed with 8.42 g of
chlorine to give an actual yield of 21.1 g of bromine
monochloride.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Actual Yield and Percentage Yield, continued
Determining Actual Yield
• The actual yield can only be determined experimentally.
• A close estimate of the actual yield can be calculated if
the percentage yield for a reaction is known.
• The percentage yield in a particular reaction is usually
fairly consistent, so it can be used for predicting an
estimated actual yield.
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Calculating Actual Yield
Sample Problem G
How many grams of CH3COOC5H11 should form if
4808 g are theoretically possible and the percentage
yield for the reaction is 80.5%?
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Chapter 9
Section 2 Limiting Reactants and
Percentage Yields
Calculating Actual Yield, continued
Sample Problem G Solution
Use the percentage yield and the theoretical yield to
calculate the actual yield expected.
actual yield
80.5% 
 100
4808 g
actual yield = 4808 g  0.805 =
3.87 103 g CH3COOC5H11
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Sample Problem G , practice pg. 318
1) The percentage yield of NH3 from the
following reaction is 85.0%.What actual yield is
expected from the reaction of 1.00 kg N2 with
225 g H2?
N2 + 3H2 → 2NH3
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Homework, practice pg. 318
2 )If the percentage yield is 92.0%, how many
grams of CH3OH can be made by the reaction of 5.6
× 103 g CO with 1.0 × 103 g H2?
CO + 2H2 → CH3OH
3 )Suppose that the percentage yield of BrCl is
90.0%. How much BrCl can be made by reacting
338 g Br2 with 177 g Cl2?
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HOMEWORK
SectionReview,
pg. 319, Q ( 1- 12)
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6. A chemist reacts 8.85 g of iron with an excess of
hydrogen chloride to form hydrogen gas and iron(II)
chloride. Calculate the theoretical yield and the
percentage yield of hydrogen if 0.27 g H2 are collected.
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7. Use the chemical reaction below to answer
the questions that follow.
P4O10 + H2O → H3PO4
a. Balance the equation.
b. Calculate the theoretical yield if 100.0 g P4O10 react
with 200.0 g H2O.
c. If the actual mass recovered is 126.2 g H3PO4, what
is the percentage yield?
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8. Titanium dioxide is used as a white pigment in
paints. If 3.5 mol TiCl4 reacts with 4.5 mol O2, which
is the limiting reactant? How many moles of each
product are produced? How many moles of the
excess reactant remain?
TiCl4 + O2 →TiO2 + 2Cl2
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9. If 1.85 g Al reacts with an excess of copper(II) sulfate
and the percentage yield of Cu is 56.6%, what mass of
Cu is produced?
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10. Quicklime, CaO, can be prepared by roasting
limestone, CaCO3, according to the chemical
equation below. When 2.00 × 103 g of CaCO3 are
heated, the actual yield of CaO is 1.05 × 103 g.What is
the percentage yield?
CaCO3(s) → CaO(s) + CO2(g)
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11. Magnesium powder reacts with steam to form
magnesium hydroxide and hydrogen gas.
a. Write a balanced equation for this reaction.
b. What is the percentage yield if 10.1 g Mg reacts with
an excess of water and 21.0 g Mg(OH)2 is recovered?
c. If 24 g Mg is used and the percentage yield is 95%,
how many grams of magnesium hydroxide should be
recovered?
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12. Use the chemical reaction below to answer the
questions that follow.
CuO(s) + H2(g) → Cu(s) + H2O(g)
a. What is the limiting reactant when 19.9 g CuO react
with 2.02 g H2?
b. The actual yield of copper was 15.0 g. What is the
percentage yield?
c. How many grams of Cu can be collected if 20.6 g
CuO react with an excess of hydrogen with a yield of
91.0%?
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Chapter 9
Section 3 Stoichiometry and Cars
(
cancelled)
• Brainstorm some reactions that are involved in
Bellringer
operating a car.
• Develop some ideas of how stoichiometry is involved
in these reactions.
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Chapter 9
Section 3 Stoichiometry and Cars
Objectives
• Relate volume calculations in stoichiometry to the
inflation of automobile safety air bags.
• Use the concept of limiting reactants to explain why
fuel-air ratios affect engine performance.
• Compare the efficiency of pollution-control
mechanisms in cars using percentage yield.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Safety Air Bags
• Stoichiometry is important in many aspects of
automobile operation and safety.
An Air Bag Could Save Your Life
• When inflated, air bags slow the motion of a person
so that he or she does not strike the inside of a car
with as much force during a high-speed collision.
• Stoichiometry is used by air-bag designers to ensure
that air bags do not underinflate or overinflate. The
chemicals that inflate the air bag must be present in
just the right proportions.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Safety Air Bags, continued
An Air Bag Could Save Your Life, continued
• Here’s how an air bag works:
• A front-end collision transfers energy to a crash sensor that
causes an igniter to fire.
• The igniter provides energy n to start a very fast reaction
that a mixture of gas called the gas generant.
• The igniter also raises the temperature and pressure so that
the reaction happens very quickly.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Safety Air Bags, continued
Air-Bag Design Depends on Stoichiometric
Precision
• One of the first gas generants used in air bags is a
solid mixture of sodium azide, NaN3, and an oxidizer.
• The gas that inflates the bag is almost pure nitrogen
gas, N2, which is produced in the following reaction.
2NaN3(s) → 2Na(s) + 3N2(g)
• However, this reaction does not inflate the bag
enough, and the sodium metal is dangerous.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Safety Air Bags, continued
Air-Bag Design Depends on Stoichiometric
Precision, continued
• Oxidizers such as ferric oxide, Fe2O3, are included,
which react rapidly with the sodium.
• Energy is released, which heats the gas and causes
the gas to expand and fill the bag.
6Na(s) + Fe2O3(s) → 3Na2O(s) + 2Fe(s) + energy
• One product, sodium oxide, Na2O, is extremely
corrosive.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Safety Air Bags, continued
Air-Bag Design Depends on Stoichiometric
Precision, continued
• Water vapor and CO2 from the air react with sodium
oxide to form less harmful NaHCO3.
Na2O(s) + 2CO2(g) + H2O(g) → 2NaHCO3(s)
• The mass of gas needed depends on the density of
the gas, which depends on temperature.
• To find the amount of gas generant to put into each
system, designers must use stoichiometry.
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Chapter 9
Section 3 Stoichiometry and Cars
Air-Bag Stoichiometry
Sample Problem H
Assume that 65.1 L N2 inflates an air bag to the proper
size. What mass of NaN3 must be used? (density of N2
= 0.92 g/L)
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Chapter 9
Section 3 Stoichiometry and Cars
Air-Bag Stoichiometry, continued
Sample Problem H Solution
Write a balanced chemical equation for the reaction.
2NaN3(s) → 2Na(s) + 3N2(g)
volume of N2 = 65.1 L N2
density of N2 = 0.92 g/L
molar mass of N2 = 28.02 g/mol
mass of reactant = ? g NaN3
molar mass of NaN3 = 65.02 g/mol
2 mol NaN3 = 3 mol N2
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Chapter 9
Section 3 Stoichiometry and Cars
Air-Bag Stoichiometry
Sample Problem H Solution, continued
Start with the volume of N2, and change it to moles using
density and molar mass.
Then use the mole ratio followed by the molar mass of NaN3.
? g NaN3
0.92 g N2
1 mol N2
 65.1 L N2 


1 L N2
28.02 g N2
2 mol NaN3 65.02 g NaN3

 93 g NaN3
3 mol N2
1 mol NaN3
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Engine Efficiency
• The efficiency of a car engine depends on having the
correct stoichiometric ratio of gasoline and oxygen.
• Gasoline contains isooctane, C8H18.
• The other reactant in gasoline combustion is oxygen,
which is about 21% of air by volume.
• The reaction for gasoline combustion is as follows.
2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)
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Chapter 9
Visual Concepts
Energy Conversions in Automobile Engines
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Engine Efficiency, continued
Engine Efficiency Depends on Reactant Proportions
• For efficient combustion, the two reactants must be
mixed in a mole ratio that is close to that of the
balanced chemical equation: 2:25, or 1:12.5.
• The mixture of fuel and oxygen is different when the
car is starting, idling, and running at normal speeds.
• A carburetor or a computer-controlled fuel injector
controls the fuel-oxygen ratio in the engine.
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Chapter 9
Section 3 Stoichiometry and Cars
Fuel-Oxygen Ratio
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Chapter 9
Section 3 Stoichiometry and Cars
Air-Fuel Ratio
Sample Problem I
A cylinder in a car’s engine draws in 0.500 L of air.
How many milliliters of liquid isooctane should be
injected into the cylinder to completely react with the
oxygen present? The density of isooctane is 0.692
g/mL, and the density of oxygen is 1.33 g/L. Air is 21%
oxygen by volume.
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Chapter 9
Section 3 Stoichiometry and Cars
Air-Fuel Ratio, continued
Sample Problem I Solution
Write a balanced equation for the chemical reaction.
2C8H18 + 25O2 → 16CO2 + 18H2O
Use the percentage by volume of O2 in air to find the
volume of O2.
Then set up a volume-volume problem.
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Chapter 9
Section 3 Stoichiometry and Cars
Air-Fuel Ratio, continued
Sample Problem I Solution, continued
? mL C8H18  0.500 L air 
21 L O2 1.33 g O2


100 L air
1 L O2
2 mol C8H18 114.26 g C8H18
1 mol O2



32.00 g O2 25 mol O2
1 mol C8H18
1 mL C8H18
 5.76  10-2 mL C8H18
0.692 g C8H18
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Pollution Control
The table below lists the standards for pollutants in
exhaust set in 1996 by the U.S. Environmental
Protection Agency.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Pollution Control, continued
The Fuel-Air Ratio Influences the Pollutants Formed
• If the fuel-air mixture doesn’t have enough O2, carbon
monoxide can be produced instead of carbon dioxide.
• Also, some unburned fuel (hydrocarbons) comes out
in the exhaust.
• These hydrocarbons are involved in forming smog.
• So the fuel-air ratio is a key factor in determining
how much pollution forms.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Pollution Control, continued
The Fuel-Air Ratio Influences the Pollutants Formed,
continued
• Another factor in auto pollution is the reaction of
nitrogen and oxygen at the high temperatures inside
the engine to form small amounts of highly reactive
nitrogen oxides, including NO and NO2.
N2(g) + O2(g) → 2NO(g)
2NO(g) + O2(g) → 2NO2(g)
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Pollution Control, continued
The Fuel-Air Ratio Influences the Pollutants
Formed, continued
• Nitrogen oxides react with oxygen to form another
harmful chemical, ozone, O3.
NO2(g) + O2(g) → 2NO(g) + O3(g)
• These reactions form what is referred to as
photochemical smog.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Pollution Control, continued
Meeting the Legal Limits Using Stoichiometry
• Automobile makers use stoichiometry to predict when
adjustments will be necessary to keep exhaust
emissions within legal limits.
• Because the units stated in the laws are grams per
kilometer, auto makers must consider how much fuel
the vehicle will burn to move a certain distance.
• Cars with better gas mileage will use less fuel per
kilometer, resulting in lower emissions per kilometer.
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Chapter 9
Section 3 Stoichiometry and Cars
Stoichiometry and Pollution Control, continued
Catalytic Converters Can Help
• Catalytic converters treat exhaust gases before they
are released into the air.
• Platinum, palladium, or rhodium in these converters
act as catalysts.
• They increase the rate of the decomposition of NO and of
NO2 into N2 and O2.
• Catalytic converters also speed the change of CO into
CO2 and the change of unburned hydrocarbons into
CO2 and H2O.
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Chapter 9
Section 3 Stoichiometry and Cars
Calculating Yields: Pollution
Sample Problem J
What mass of ozone, O3, can be produced from 3.50 g
of NO2 contained in a car’s exhaust? The equation is
as follows.
NO2(g) + O2(g)  NO(g) + O3(g)
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Chapter 9
Section 3 Stoichiometry and Cars
Calculating Yields: Pollution, continued
Sample Problem J Solution
volume of air = 0.500 L air
percentage of oxygen in air: 21% by volume
molar mass O2 = 32.00 g/mol
density O2 = 1.33 g/L
volume O2 = ? L
molar mass C8H18 = 114.26 g/mol
density C8H18 = 0.692 g/mL
volume C8H18 = ? mL
2 mol C8H18 = 25 mol O2
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Chapter 9
Section 3 Stoichiometry and Cars
Calculating Yields: Pollution, continued
Sample Problem J Solution, continued
Use the set-up for a mass-mass problem:
21 L O2 1.33 g O2
? mL C8H18  0.500 L air 


100 L air
1 L O2
2 mol C8H18 114.26 g C8H18
1 mol O2



32.00 g O2 25 mol O2
1 mol C8H18
1 mL C8H18
 5.76  10-2 mL C8H18
0.692 g C8H18
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Chapter 9
Standardized Test Preparation
Understanding Concepts
1. Using the mole ratio of the reactants and products
in a chemical reaction, what will you most likely be
able to determine?
A. rate of the reaction
B. energy absorbed or released by the reaction
C. chemical names of the reactants and products
D. mass of a product produced from a known mass
of reactants
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Chapter 9
Standardized Test Preparation
Understanding Concepts
1. Using the mole ratio of the reactants and products
in a chemical reaction, what will you most likely be
able to determine?
A. rate of the reaction
B. energy absorbed or released by the reaction
C. chemical names of the reactants and products
D. mass of a product produced from a known mass
of reactants
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Chapter 9
Standardized Test Preparation
Understanding Concepts
2. Carbon dioxide fire extinguishers were developed to
fight fires where using water would be hazardous.
What effect does the carbon dioxide have on a fire?
F. changes the mole ratio of the reactants
G. decreases the actual yield of the reaction
H. decreases the potential yield of the reaction
I. slows the reaction by limiting the reactant oxygen
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Chapter 9
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Understanding Concepts
2. Carbon dioxide fire extinguishers were developed to
fight fires where using water would be hazardous.
What effect does the carbon dioxide have on a fire?
F. changes the mole ratio of the reactants
G. decreases the actual yield of the reaction
H. decreases the potential yield of the reaction
I. slows the reaction by limiting the reactant oxygen
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Understanding Concepts
3. What is the mole ratio of CO2 to C6H12O6 in the
combustion reaction: C6H12O6 + 6O2
6CO2 + 6H2O?
A. 1:1
B. 1:2
C. 1:6
D. 6:1
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Understanding Concepts
3. What is the mole ratio of CO2 to C6H12O6 in the
combustion reaction: C6H12O6 + 6O2
6CO2 + 6H2O?
A. 1:1
B. 1:2
C. 1:6
D. 6:1
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Understanding Concepts
4. Identify the limiting and excess reactants in the
production of nitric acid when nitrogen dioxide from
combustion of fossil fuels reacts with water vapor in
the air.
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Understanding Concepts
4. Identify the limiting and excess reactants in the
production of nitric acid when nitrogen dioxide from
combustion of fossil fuels reacts with water vapor in
the air.
Answer: The limiting reactant is nitrogen dioxide and the
excess reactant is water.
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Understanding Concepts
5. Write a balanced equation for the conversion of
ozone (O3) to oxygen (O2).
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Understanding Concepts
5. Write a balanced equation for the conversion of
ozone (O3) to oxygen (O2).
Answer: 2O3
3O2
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Understanding Concepts
6. How many grams of oxygen (molar mass = 32) will be
produced by the reaction of 2 moles of O3?
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Understanding Concepts
6. How many grams of oxygen (molar mass = 32) will be
produced by the reaction of 2 moles of O3?
Answer: 96 g O2
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Reading Skills
Read the passage below. Then answer the questions.
Explosives contain substances that, when mixed
together, produce an extremely quick and highly
exothermic reaction. An exothermic reaction is one that
releases energy as heat. The reactants in explosives
should be relatively stable, but should decompose rapidly
when the reaction is initiated. Nitroglycerin, an explosive,
decomposes as shown in the equation:
4C3H5N3O9
6N2 + O2 + 12CO2 + 10H20 + energy
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Reading Skills
7. What is the theoretical yield of nitrogen if 1.0 moles
of nitroglycerin is detonated?
F. 21.0 grams
G. 28.0 grams
H. 42.0 grams
I. 168.0 grams
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Reading Skills
7. What is the theoretical yield of nitrogen if 1.0 moles
of nitroglycerin is detonated?
F. 21.0 grams
G. 28.0 grams
H. 42.0 grams
I. 168.0 grams
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Reading Skills
8. The energy produced by the explosion is heat. How does the
production of large amounts of heat cause the effects observed
in an explosion?
A. The products of the reaction burn.
B. The products of the reaction condense, releasing the heat
energy.
C. Heat makes the gaseous reaction products move and expand
very rapidly.
D. Heat causes the atoms to ionize and the ions that are
produced by this reaction cause the effects of the explosion.
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Reading Skills
8. The energy produced by the explosion is heat. How does the
production of large amounts of heat cause the effects observed
in an explosion?
A. The products of the reaction burn.
B. The products of the reaction condense, releasing the heat
energy.
C. Heat makes the gaseous reaction products move and expand
very rapidly.
D. Heat causes the atoms to ionize and the ions that are
produced by this reaction cause the effects of the explosion.
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Reading Skills
9. Based on the explosive reaction of nitroglycerin, a
typical explosive, how does the stability of the
products of an explosion compare to that of the
reactants?
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Reading Skills
9. Based on the explosive reaction of nitroglycerin, a
typical explosive, how does the stability of the
products of an explosion compare to that of the
reactants?
Answer: In general, the products of an explosion are
much more stable than the reactants.
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Chapter 9
Standardized Test Preparation
Interpreting Graphics
The illustration below shows the parts of an airbag
system on an automobile. Use it to answer questions
10–13.
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Interpreting Graphics
10. What is the purpose of the igniter in this system?
F. pump air into the air bag
G.prevent any reaction until there is a crash
H. provide energy to start a very fast reaction that
produces a gas.
I. provide energy to expand air that is stored in the
bag, inflating it like a hot air balloon
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Interpreting Graphics
10. What is the purpose of the igniter in this system?
F. pump air into the air bag
G.prevent any reaction until there is a crash
H. provide energy to start a very fast reaction that
produces a gas.
I. provide energy to expand air that is stored in the
bag, inflating it like a hot air balloon
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Interpreting Graphics
11. Why does the designer of the air bag need to
understand the stoichiometry of the reaction that
produces the gas?
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Interpreting Graphics
11. Why does the designer of the air bag need to
understand the stoichiometry of the reaction that
produces the gas?
Answer: The stoichiometry of the reaction is needed in
order to calculate the amount of gas produced. If the
wrong amount of reactant is used, the air bag may
over-inflate or under-inflate, making it ineffective.
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Interpreting Graphics
12. Which of the following is a reason why most automobile
manufacturers have replaced NaN3 with other compounds as the
reactants for filling air bags?
A. The sodium produced by the reaction is dangerous.
B. The nitrogen produced by the reaction can be harmful.
C. The materials used to make sodium azide are rare and
expensive.
D. The decomposition of sodium azide is too fast so it fills the
air bags too quickly.
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Interpreting Graphics
12. Which of the following is a reason why most automobile
manufacturers have replaced NaN3 with other compounds as the
reactants for filling air bags?
A. The sodium produced by the reaction is dangerous.
B. The nitrogen produced by the reaction can be harmful.
C. The materials used to make sodium azide are rare and
expensive.
D. The decomposition of sodium azide is too fast so it fills the
air bags too quickly.
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Interpreting Graphics
13. If the reaction that fills the air bag is the
decomposition of sodium azide, represented by the
equation, 2NaN3(s)
2Na(s) + 3N2(g), how many
moles of products are produced by the decomposition
of 3.0 moles of sodium azide?
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Interpreting Graphics
13. If the reaction that fills the air bag is the
decomposition of sodium azide, represented by the
equation, 2NaN3(s)
2Na(s) + 3N2(g), how many
moles of products are produced by the decomposition
of 3.0 moles of sodium azide?
Answer: 7.5 mol
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