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EXAMPLE 8.1
Mole-to-Mole Conversions
Sodium chloride, NaCl, forms by the following reaction between sodium and chlorine.
How many moles of NaCl result from the complete reaction of 3.4 mol of Cl 2 ? Assume that there is more than
enough Na.
You are given the number of moles of a reactant (Cl2)
and asked to find how many moles of product (NaCl)
will result if the reactant completely reacts. The
conversion factor comes from the balanced chemical
equation.
The solution map begins with moles of chlorine and
uses the stoichiometric conversion factor to obtain
moles of sodium chloride
Solution Map:
Follow the solution map to solve the problem. There
is enough Cl2 to produce 6.8 mol of NaCl.
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Mole-to-Mole Conversions
EXAMPLE 8.1
Continued
SKILLBUILDER 8.1
Mole-to-Mole Conversions
Water is formed when hydrogen gas reacts explosively with oxygen gas according to the following balanced
equation.
How many moles of H2O result from the complete reaction of 24.6 mol of O2? Assume that there is more than
Enough H2.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 8.8; Problems 17, 18, 19, 20.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.2
Mass-to-Mass Conversions
In photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H12O6) according to the following
reaction.
How many grams of glucose can be synthesized from 58.5 g of CO2 ? Assume that there is more than enough water
present to react with all of the CO2 .
Set up the problem in the normal way.
The main conversion factor is the
stoichiometric relationship between moles of
carbon dioxide and moles of glucose. This
conversion factor comes from the balanced
equation. The other conversion factors are
simply the molar masses of carbon dioxide
and glucose.
The solution map uses the general outline
Solution Map:
where A is carbon dioxide and B is glucose.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.2
Mass-to-Mass Conversions
Continued
Follow the solution map to solve the
problem. Begin with grams of carbon dioxide
and multiply by the appropriate factors to
arrive at grams of glucose.
SKILLBUILDER 8.2
Solution:
Mass-to-Mass Conversions
Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl,
according to the following reaction.
How much HCl in grams can be neutralized by 5.50 g of Mg(OH)2?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 8.9; Problems 33, 34, 35, 36.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.3
Mass-to-Mass Conversions
One of the components of acid rain is nitric acid, which forms when NO 2 a pollutant, reacts with oxygen and rainwater according to the following reaction.
Assuming that there is more than enough O2 and H2O, how much HNO3 in kilograms forms from 1.5  103 kg of
NO2 pollutant?
Set up the problem in the normal way.
The main conversion factor is the
stoichiometric relationship between moles of
nitrogen dioxide and moles of nitric acid.
This conversion factor comes from the
balanced equation. The other conversion
factors are simply the molar masses of
nitrogen dioxide and nitric acid and the
relationship between kilograms and grams.
The solution map follows the general format
of:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Solution Map:
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.3
Mass-to-Mass Conversions
Continued
However, since the original quantity of NO2
is given in kilograms, you must first convert
to grams. Since the final quantity is
requested in kilograms, you must convert
back to kilograms at the end.
Solution Map:
Follow the solution map to solve the
problem. Begin with kilograms of nitrogen
dioxide and multiply by the appropriate
conversion factors to arrive at kilograms of
nitric acid.
Solution:
SKILLBUILDER 8.3
Mass-to-Mass Conversions
Another component of acid rain is sulfuric acid, which forms when SO 2 also a pollutant, reacts with oxygen and
rainwater according to the following reaction.
Assuming that there is plenty of O2 and H2O how much H2SO4 in kilograms forms from 2.6 × 103 kg of SO2?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 37, 38, 39, 40.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.4
Limiting Reactant and Theoretical Yield from Initial Moles of
Reactants
Consider the following reaction.
If we begin with 0.552 mol of aluminum and 0.887 mol of chlorine, what is the limiting reactant and theoretical
yield of AlCl3 in moles?
Set up the problem in the normal way.
The conversion factors are the stoichiometric
relationships (from the balanced equation)
between the moles of each reactant and the
moles of product.
The solution map shows how to get from
moles of each reactant to moles of AlCl3. The
reactant that makes the least amount of AlCl3
is the limiting reactant..
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Given:
0.552 mol Al
0.887 mol Cl2
Find:
limiting reactant
theoretical yield
Solution Map:
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.4
Limiting Reactant and Theoretical Yield from Initial Moles of
Reactants
Continued
Solution:
Since the 0.552 mol of Al makes the least amount of AlCl 3 is the limiting reactant. The theoretical yield is 0.552 mol
of AlCl3.
SKILLBUILDER 8.4
Limiting Reactant and Theoretical Yield from Initial Moles of
Reactants
Consider the following reaction.
If you begin with 4.8 mol of sodium and 2.6 mol of fluorine, what is the limiting reactant and theoretical yield of
NaF in moles?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 45, 46, 47, 48, 49, 50, 51, 52.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.5
Finding Limiting Reactant and Theoretical Yield
Ammonia, NH3, can be synthesized by the following reaction.
What is the maximum amount of ammonia in grams that can be synthesized from 45.8 g of NO and 12.4 g of H 2?
Although this problem does not specifically
ask for the limiting reactant, it must be found
to determine the theoretical yield, which is
the maximum amount of ammonia that can
be synthesized. Begin by setting up the
problem in the normal way.
Given: 545.8 g NO, 12.4 g H2
Find: maximum amount of NH3 (theoretical yield)
Conversion Factors:
The main conversion factors are the
stoichiometric relationship between moles of
each reactant and moles of ammonia. The
other conversion factors are simply the molar
masses of nitrogen monoxide, hydrogen gas,
and ammonia.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.5
Finding Limiting Reactant and Theoretical Yield
Continued
Find the limiting reactant by calculating how
much product can be made from each
reactant. The reactant that makes the least
amount of product is the limiting reactant.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Solution Map:
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.5
Finding Limiting Reactant and Theoretical Yield
Continued
Follow the solution map, beginning with the
actual amount of each reactant given, to
calculate the amount of product that can be
made from each reactant.
Solution:
There is enough NO to make 26.0 g of NH3 and enough H2 to make 41.8 g of HN3. Therefore, NO is the limiting
reactant, and the maximum amount of ammonia that can possibly be made is 26.0 g, the theoretical yield.
SKILLBUILDER 8.5
Finding Limiting Reactant and Theoretical Yield
Ammonia can also be synthesized by the following reaction.
What is the maximum amount of ammonia in grams that can be synthesized from 25.2 g of N 2 and 8.42 g of H2?
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.5
Finding Limiting Reactant and Theoretical Yield
Continued
SKILLBUILDER PLUS
What is the maximum amount of ammonia in kilograms that can be synthesized from 5.22 kg of H 2 and 31.5 kg of
N2?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 55, 56, 57, 58.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.6
Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Consider the following reaction.
When 11.5 g of C are allowed to react with 114.5 g of Cu2O, 87.4 g of Cu, are obtained. Find the limiting reactant,
theoretical yield, and percent yield.
Set up the problem in the normal
way.
Given:
11.5 g C
114.5 g Cu2O
87.4 g Cu produced
Find:
limiting reactant
theoretical yield
percent yield
The main conversion factors are the
stoichiometric relationships between moles
of each reactant and moles of copper. The
other conversion factors are simply the molar
masses of copper(I) oxide, carbon, and
copper
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.6
Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
The solution map shows how to find the
mass of Cu formed by the initial masses of
Cu2O and C. The reactant that makes the
least amount of product is the limiting
reactant and determines the theoretical yield.
Solution Map:
Follow the solution map, beginning with the
actual amount of each reactant given, to
calculate the amount of product that can be
made from each reactant.
Solution:
Since Cu2O makes the least amount of
product, Cu2O is the limiting reactant. The
theoretical yield is simply the amount of
product made by the limiting reactant. The
percent yield is the actual yield (87.4 g Cu)
divided by the theoretical yield (101.7 g Cu)
multiplied by 100%.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.6
Finding Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
SKILLBUILDER 8.6
Finding Limiting Reactant, Theoretical Yield, and Percent Yield
The following reaction is used to obtain iron from iron ore:
The reaction of 185 g of Fe2O3 with 95.3 g of CO produces 87.4 g of Fe. Find the limiting reactant, theoretical
yield, and percent yield.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 8.10; Problems 61, 62, 63, 64, 65, 66.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.7
Stoichiometry Involving ΔH
An LP gas tank in a home barbecue contains 11.8 × 103 g of propane (C3H8). Calculate the heat (in kJ) associated
with the complete combustion of all of the propane in the tank.
You are given the mass of propane and asked
to find the heat evolved (in kJ) in its
combustion.
Given: 11.8 × 103 g C3H8
Find: kJ
Conversion Factors:
Start with g C3H8 and then use the molar
mass of C3H8 to find the number of moles.
Next, use the stoichiometric relationship
between mol C3H8 and kJ to find the heat
evolved.
Solution Map:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.7
Stoichiometry Involving ΔH
Continued
Follow the solution map to solve the
problem. Begin with 11.8 × 103 g C3H8 and
multiply by the appropriate conversion
factors to arrive at kJ. The answer is
negative, as it should be for heat evolved by
the reaction.
SKILLBUILDER 8.7
Solution:
Stoichiometry Involving ΔH
Ammonia reacts with oxygen according to the following equation:
Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH 3.
SKILLBUILDER PLUS
What mass of butane in grams is necessary to produce 1.5 × 103 kJ of heat? What mass of CO2 is produced?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 8.11; Problems 69, 70, 71, 72, 73, 74.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.8
Mole-to-Mole Conversions
How many moles of sodium oxide can be synthesized from 4.8 mol of sodium? Assume that there is more than
enough oxygen present. The balanced equation is:
Given: 4.8 mol Na
Find: mol Na2O
Conversion Factor:
Solution Map:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.9
Mass-to-Mass Conversions
How many grams of sodium oxide can be synthesized from 17.4 g of sodium? Assume that there is more than
enough oxygen present. The balanced equation is:
Given: 17.4 g Na
Find: g Na2O
Conversion Factors:
Molar mass Na = 22.99 g/mol
Molar mass Na2O = 61.98 g/mol
Solution Map:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.9
Mass-to-Mass Conversions
Continued
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.10
Limiting Reactant, Theoretical Yield, and Percent Yield
10.4 g of Fe reacts with 11.8 g of S to produce 14.2 g of Fe2S3. Find the limiting reactant, theoretical yield, and
percent yield for this reaction. The balanced chemical equation is:
Given: 10.4 g Fe
11.8 g S
14.2 g Fe2S3
Find:
limiting reactant
theoretical yield
percent yield
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.10
Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
Solution Map:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.10
Limiting Reactant, Theoretical Yield, and Percent Yield
Continued
Solution:
The limiting reactant is Fe.
The theoretical yield is 19.4 g of Fe2S3 .
The percent yield is 73.2%.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 8.11
Stoichiometry Involving ΔH
Calculate the heat evolved (in kJ) upon complete combustion of 25.0 g of CH 4.
Given: 25 g CH4
Find: kJ
Conversion Factors:
Solution Map:
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.