Chemistry 100(02) Fall 2013
Instructor: Dr. Upali Siriwardane
e-mail: upali@coes.latech.edu
Office: CTH 311
Phone 257-4941
Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m
Tu,Th,F 8:00 - 10:00 a.m. Or by appointment
Test Dates:
September 30,
2013 (Test 1): Chapter 1 & 2
October 21,
2013 (Test 2): Chapter 3 & 4
November 13, 2013 (Test 3) Chapter 5 & 6
November 14, 2013 (Make-up test) comprehensive:
Chapters 1-6 9:30-10:45:15 AM, CTH 328
CHEM 100 Fall 2013 .
chapter 4 -1 .
Text Book & Resources
REQUIRED :
Textbook: Principles of Chemistry: A Molecular Approach, 2nd
Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase
the Mastering Chemistry
Group Homework, Slides and Exam review guides and sample
exam questions are available online: http://moodle.latech.edu/
and follow the course information links.
OPTIONAL :
Study Guide: Chemistry: A Molecular Approach, 2nd EditionNivaldo J. Tro 2nd Edition
Student Solutions Manual: Chemistry: A Molecular Approach, 2nd
Edition-Nivaldo J. Tro 2nd
CHEM 100 Fall 2013 .
chapter 4 -2 .
Chapter 4. Chemical Quantities and Aqueous Reactions
4.1 Global Warming and the Combustion of Fossil Fuels…………………. 127
4.2 Reaction Stoichiometry: How Much Carbon Dioxide?......................... 128
4.3 Limiting Reactant, Theoretical Yield, and Percent Yield………………. 133
4.4 Solution Concentration and Solution Stoichiometry………………….. 140
4.5 Types of Aqueous Solutions and Solubility…………………………….. 146
4.6 Precipitation Reactions………………………………………………….. 150
4.7 Representing Aqueous Reactions: Molecular, Ionic, and Complete Ionic
Equations…………………………………………………………………........ 153
4.8 Acid–Base and Gas-Evolution Reactions……………………………..... 155
4.9 Oxidation–Reduction Reactions…………………………………………. 162
CHEM 100 Fall 2013 .
chapter 4 -3 .
Chapter 4. Chemical Quantities and Aqueous Reactions
•Global Warming
•Stoichiometry
•Reactions in One reactant in
Limited Supply
•Limiting reactant
•Evaluating Success of Synthesis
•Theoretical Yield
•Actual Yield
•Percent Yield
•Solution Concentration
•Molecular, Ionic, and Complete
Ionic Equations
CHEM 100 Fall 2013 .
•Precipitation Reactions
•Acid–Base and Gas-Evolution
Reactions
•Oxidation–Reduction Reactions
•Stoichiometry in Solution
Reactions
chapter 4 -4 .
What is Global Warming?
CHEM 100 Fall 2013 .
chapter 4 -5 .
Reaction Stoichiometry
• According to the Law of Conversion of Matter:
– Matter is neither created nor destroyed in a chemical reaction.
– A balanced chemical equation illustrates the law of
conversation of matter.
• In a balanced reaction:
– Total mass of reactants = Total mass of products
– A balanced reaction has the same type and quantity of
atoms on both sides of the reaction.
• Stoichiometry is based on the law of conversion of matter.
– Stoichiometry studies the quantitative aspects of chemical
reactions.
CHEM 100 Fall 2013 .
chapter 4 -6 .
Reaction Stoichiometry: What it means
4 Fe(s)
+
 2 Fe2O3(s)
3 O2(g)
This equation means:
4 atoms Fe +
4 moles Fe +
3 molecules O2

2 molecules
Fe2O3
3 moles O2

2 moles Fe2O3
4 moles Fe atoms
and 6 moles O atoms
atoms
23.4 g Fe + 96.0 g O2
415.4 g
of reactants
CHEM 100 Fall 2013 .
=
Fe atoms
and 6 mole O
=
415.4 g Fe2O3
=
415.4 g
of products
chapter 4 -7 .
1) Given the balanced reactions:
2H2(g) + O2(g) = 2 H2O(l),
Write the mole conversion factors:
CHEM 100 Fall 2013 .
chapter 4 -8 .
Strategy Behind Solving STOICHIOMETRY Problems
1. Need a balanced reaction to determine the stoichiometric
relationship between:
a. Reactants and reactants or
b. Reactants and products or
c. Products and products
2. Go to the mole:
a. If mass is given, then divide by molecular mass:
mass (g) /mol. mass (g/mole) = mole
b. If volume and molarity (M) are given, then:
Ma x Va = moles A
3. Use the stoichiometric factor to convert from mole A to mole B to
solve problem.
CHEM 100 Fall 2013 .
chapter 4 -9 .
Limiting Reactant
• For reactions with multiple reactants, it is likely that
one of the reactants will be completely used before
the others.
• When this reactant is used up, the reaction stops and
no more product is made.
• The reactant that limits the amount of product is
called the limiting reactant (limiting reagent).
– The limiting reactant is completely consumed.
• Reactants not completely consumed are called excess
reactants.
• The amount of product that can be made from the
limiting reactant is called the theoretical yield.
10
CHEM 100 Fall 2013 .
chapter 4 -10 .
Limiting Reactant
CHEM 100 Fall 2013 .
chapter 4 -11 .
Analogy in Recipe : Making Cheese Sandwiches
You were given
20 slices bread,
5 slices of cheese,
4 slices of ham If you want to make
sandwiches containing two slices bread and one slice of
cheese and one slice of ham
How many sandwiches you could make?
What is the limiting ingredient?
CHEM 100 Fall 2013 .
chapter 4 -12 .
Using Stoichiometry to Predict
• Theoretical Yield and Limiting Reagent (Reactants)
• Percent Yield
CHEM 100 Fall 2013 .
chapter 4 -13 .
Problem:
The following unbalanced equation is the chemical reaction
associated with photosynthesis.
CO2(g) + H2O(l)  C6H6O6(s) + O2(g)
With adequate water, suppose a plant consumes 37.8 grams of CO2
during the week.
Determine how many grams of glucose (C6H6O6) would be produced
by the plant during this one-week period.
CHEM 100 Fall 2013 .
chapter 4 -14 .
Problem Strategy:
Determine how many grams of glucose (C6H6O6) would be produced
by the plant if 37.8 grams of CO2 is consumed.
1. Need a balanced equation:
6 CO2(g) + 6 H2O(l)  C6H6O6(s) + 6 O2(g)
2. Determine moles of CO2 consumed.
37.8 g CO2 x (1 mol CO2/44.0 g) = 0.859 mol CO2
3. Determine how many moles of C6H6O6 would be produced. Use the
stoichiometric relationship between CO2 and C6H6O6. In this reaction the
relationship is 6 mole CO2 to 1 mole C6H6O6.
0.859 mol CO2 x (1 mol C6H6O6/6 mol CO2) = 0.143 mol C6H6O6
4. Determine the grams of C6H6O6 produced.
0.143 mol C6H6O6 x (1 mol / 180.0 g/1 mol C6H6O6) = 25.8 g C6H6O6
CHEM 100 Fall 2013 .
chapter 4 -15 .
2) My recipe ratio for a bacon double cheeseburger is:
1 bun + 2 hamburger patties + 1 slices of cheese + 2 strips of bacon
If
you start with:
2 bun, 6 patties, 4 slices of cheese, 6 strips of bacon
a) How many bacon double cheeseburgers can you make?
b) Which ingredient is limiting?
c) What ingredients would be left over or in excess?
hamburger bun
hamburger patties
slices of cheese
strips of bacon
1
2
1
2
2
6
4
6
CHEM 100 Fall 2013 .
chapter 4 -16 .
3) Determine if the reaction is stoichiometric, or one reagent
is in excess/one reagent is limiting for the reaction:
3H2(g) + 1N2(g) = 2 NH3(g)
a) If 3 mole H2 and 1.00 mole of N2 are reacted according to
the equation:
Stoichiometric ratio (mole H2/ mole of N2)=
Actual mole ratio (mole H2/ mole of N2) =
b) Is stoichiometric ratio equal, grate or less than actual mole
ratio?
c) Is the reaction stoichiometric, H2 limiting or N2 limiting?
CHEM 100 Fall 2013 .
chapter 4 -17 .
4) Determine if the reaction is stoichiometric, or
one reagent is in excess/one reagent is limiting for
the reaction: reaction:
2H2(g) + 1O2(g) = 2 H2O (l)
a) If 1 mole H2 and 2.00 mole of O2 are reacted
according to the equation:
Stoichiometric ratio (mole H2/ mole of O 2)=
Actual mole ratio (mole H2/ mole of O 2) =
b) Is stoichiometric equal, grate or less than actual
mole ratio?
c) Is the reaction stochiometric, H2 limiting or O 2
limiting?
CHEM 100 Fall 2013 .
chapter 4 -18 .
COMPARE THEORETICAL YIELDS TO DETERMINE THE LIMITING
REACTANT:
• If all 8.10 g O2 were used, then 17.2 g of Al2O3 would be produced.
• If all 5.40 g Al were used, then 10.2 g of Al2O3 would be produced.
10.2 g < 17.2 g
• The limiting reactant is Al.
• Theoretical yield is 10.2 g Al2O3.
To determine the percent yield of the reaction:
(4.50 g/10.2) x 100 = 44.1%
44.1% is the percent yield for this reaction.
CHEM 100 Fall 2013 .
chapter 4 -19 .
Which reactant will remain when the reaction is complete?
• Al was the limiting reactant.
– Therefore, O2 was in excess. But by how much?
– First find how much oxygen gas was required.
– Then find how much oxygen gas is in excess.
CHEM 100 Fall 2013 .
chapter 4 -20 .
How to Determine the Amount of Excess
Reagent Left Over
4 Al + 3 O2
0.200 mol = LR
products
0.253 mol
0.200 mol Al x (3 mol O2/4 mol Al) = 0.15 mol O2
0.150 mol of O2 is required to react with all 0.200 mol of Al.
O2 available – O2 required = excess O2
0.253 mol O2 - 0.150 mol O2 = 0.103 mol O2 left over
= 0.103 mol O2 in excess, or 3.30 grams O2
CHEM 100 Fall 2013 .
chapter 4 -21 .
5) 2 Al (s) + Fe2O3 (s) = 2 Fe(l) + Al2O3
takes place in the thermite mixture when it is ignited by a
magnesium ribbon. A thermite mixture contains a mass
ratio of 1 to 2 for Al and Fe2O3. Which one is the limiting
reagent?
CHEM 100 Fall 2013 .
chapter 4 -22 .
6) Consider the reaction: 2H2(g) + O2(g) = 2 H2O(l)
Equal weights of H2 and O2 are placed in a balloon and then
ignited. Assume reaction goes to completion, which gas is the
excess reagent?
a) How many moles of H2O will be produced by 0.80 mole of O2 and excess
H2, according to the equation?
b) How many moles of H2O will be produced by 25.6g of O2 and excess H2,
according to the equation?
c) How many grams moles of H2O will be produced by 25.6g of O2 and
excess H2, according to the equation?
CHEM 100 Fall 2013 .
chapter 4 -23 .
7) Two moles of Mg and five moles of O2 are placed
in a reaction vessel, and then the Mg is ignited to
produce MgO(F.W.=40.31 g/mole)
a) The balanced chemical reaction:
b) The limiting reactant?
c) Actual mole ratio:
d) Calculated mole ratio:
e) How many moles of MgO are formed?
f) What is the weight of MgO formed?
CHEM 100 Fall 2013 .
chapter 4 -24 .
9) Write the balanced chemical equation for the reaction where
zinc is producing silver in a single displacement reaction with
silver chloride.
CHEM 100 Fall 2013 .
chapter 4 -25 .
Theoretical and Percent Yield
• Theoretical Yield
– Predicting what could/should/would be produced
from a given amount of reactant(s)
• Limiting reagent (reactant)
– The determining reactant
– Using stoichiometric relationship(s) from balanced
reaction
• Percent Yield
– Actual amount produced
– % yield = (actual yield/theoretical yield) x 100
CHEM 100 Fall 2013 .
chapter 4 -26 .
Strategy:
1. Check to see if reaction given is balanced.
2. Determine limiting reactant and theoretical yield.
Solution:
1. Balance reaction: 2 Mg(s) + O2(g)  2 MgO(s)
2. Determine limiting reactant * theoretical yield.
CHEM 100 Fall 2013 .
chapter 4 -27 .
Predicting Theoretical Yield
Problem:
Determine the theoretical yield for the
following reaction:
Mg(s) + O2(g)  MgO(s)
when 42.5 g Mg(s) and 33.8 g O2(g) are
reacted.
CHEM 100 Fall 2013 .
chapter 4 -28 .
10) If 15.00 g of Zn (A.M.= 65.39 g/mole) reacts with silver with
25.00 g AgCl (F.M.= 143.35 g/mole)?
a) Actual moles of Zn= 0.229
b) Actual moles of AgCl= 0.174
c) Stoichiometric ratio (mole Zn/ mole of AgCl)= ½ = 0.5
d) Actual mole ratio (mole Zn/ mole of AgCl)= 1.32
e) What is limiting reactant?
b) Mole of Ag produced?
c) Grams of Ag produced (theoretical yield)?
d) If the actual yield of Ag was found to be 16.1 g Ag,
the percent yield =
CHEM 100 Fall 2013 .
chapter 4 -29 .
8) When 10.0 g pure solid calcium carbonate, CaCO3 is heated
and converted to solid calcium oxide CaO and CO2 gas:
a)
The balanced chemical reaction:
b) How much calcium oxide should be theoretically obtained?
c) If 4.20 g of CaO is actually produced what is the percent yield of
CaO?
CHEM 100 Fall 2013 .
chapter 4 -30 .
Problem: Ammonium nitrate (NH4NO3) decomposes
to N2O(g) and H2O(g).
1. Predict how many grams of H2O will be produced
from the decomposition of 0.454 kg of NH4NO3.
2. If 131 grams of water is produced, what is the %
yield for this reaction?
Strategy:
CHEM 100 Fall 2013 .
1. Write a balanced reaction.
2. Determine the theoretical yield.
3. Calculate the percent (%) yield.
chapter 4 -31 .
Problem Solution: Ammonium nitrate (NH4NO3) decomposes
to N2O(g) and H2O(g).
Balance equation:
NH4NO3(s)  N2O(g) + 2 H2O(g)
1. Predict how many grams of H2O will be produced from the
decomposition of 0.454 kg of NH4NO3.
0.454 kg NH4NO3 x (1000 g/1 kg) x (1 mol NH4NO3/80.04 g)
= 5.68 mol NH4NO3
5.68 mol NH4NO3 x (2 mol H2O/1mol NH4NO3 ) = 11.4 mol H2O
11.4 mol H2O x (18.0 g/1 mol H2O) = 204 grams H2O
204 grams of H2O is the predicted or theoretical yield.
CHEM 100 Fall 2013 .
chapter 4 -32 .
Problem:
Ammonium nitrate (NH4NO3) decomposes
to N2O(g) and H2O(g).
Balance equation:
NH4NO3(s)  N2O(g) + 2 H2O(g)
If 131 grams of water is produced, what is the % yield for this
reaction?
% yield = (131 g/204 g) x 100 = 64.2%
62.4% is the percent yield for the reaction.
CHEM 100 Fall 2013 .
chapter 4 -33 .
Problem:
When 28.6 kg of carbon reacted with 88.2 kg
of TiO2, 42.8 kg of Ti was obtained.
Reaction: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Strategy: 1. Determine limiting reactant.
2. Calculate theoretical yield.
3. Calculate percent yield.
CHEM 100 Fall 2013 .
chapter 4 -34 .
Problem Solution:
Reaction:
When 28.6 kg of carbon reacted with 88.2 kg of
TiO2, 42.8 kg of Ti was obtained.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) It is balanced!
1. Determine limiting reactant.
28.6 kg carbon:
28.6 kg C x (1000 g/1 kg) = 2.86 x 104 g C
88.2 kg TiO2
88.2 kg TiO2 x (1000 g/1 kg) = 8.82 x 104 g TiO2
2.86 x 104 g C x (1 mol C/12.0 g) x (1 mol Ti/2 mol C)
= 1.19 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL
CARBON IS USED.
8.82 x 104 g TiO2 x (1 mol TiO2/79.9 g) x (1 mol Ti/1 mol TiO2)
= 1.10 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL TiO2
IS USED.
CHEM 100 Fall 2013 .
chapter 4 -35 .
Problem Solution:
Reaction:
When 28.6 kg of carbon reacted with 88.2 kg of
TiO2, 42.8 kg of Ti was obtained.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
2. Determining the theoretical yield from limiting reactant.
= 1.19 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL
CARBON IS USED.
= 1.10 x 103 mol Ti
THIS IS THE MAXIMUM MOLE OF Ti
THAT CAN BE PRODUCED IF ALL TiO2
IS USED.
1.10 x 103 < 1.19 x 103; THEREFORE, the limiting reagent is TIO2 and
the theoretical yield in moles of Ti is 1.10 x 103.
CHEM 100 Fall 2013 .
chapter 4 -36 .
Problem Solution:
Reaction:
When 28.6 kg of carbon reacted with 88.2 kg of
TiO2, 42.8 kg of Ti was obtained.
TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
3. Determining percent yield:
The theoretical yield in moles of Ti is 1.10 x 103.
1.10 x 103 mol Ti x (47.9 g/1 mol Ti) = 5.29 x 104 g Ti
42.8 kg is actual yield of Ti.
42.8 kg x (1000 g/1 kg) = 4.28 x 104 g Ti was produced.
(4.28 x 104 g Ti/5.29 x 104 g Ti) x 100 = 80.9%
80.9% is the percent yield for this reaction.
CHEM 100 Fall 2013 .
chapter 4 -37 .
PROBLEM: Given the following chemical reaction:
Al(s) + O2(g)  Al2O3(s)
1. Determine how many grams of Al2O3 can form when
5.40 grams of Al and 8.10 grams of O2 are reacted.
2. If 4.50 grams of Al2O3 was produced, what is the
percent yield for this reaction?
3. Which reactant was in excess and how much (grams)
of this reactant remained after the reaction came to
completion?
CHEM 100 Fall 2013 .
chapter 4 -38 .
Step 1: Balance the reaction and calculate theoretical yield for each
reactant.
Balanced reaction:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
5.40 g Al x (1 mol/27.0 g Al) = 0.200 mol Al
0.200 mol Al x 2 mol Al2O3 = 0.100 mol Al2O3
4 mol Al
stoichiometric factor
0.100 mol Al2O3 x 101.96 g = 10.2 g Al2O3
1 mol Al2O3
This means that if all 5.40 g Al were consumed, then only 10.2 grams of
Al2O3 COULD be produced.
NOW DETERMINE THE THEORITICAL YIELD IF ALL 8.10 grams of O2
WERE USED.
CHEM 100 Fall 2013 .
chapter 4 -39 .
Step 1: Balance the reaction and calculate theoretical yield for each
reactant.
Balanced reaction:
4 Al(s) + 3 O2(g)  2 Al2O3(s)
8.10 g O2 x ( 1 mol/32.00 g) = 0.253 mol O2
0.253 mol O2 x 2 mol Al2O3 = 0.169 mol Al2O3
3 mol O2
stoichiometric factor
0.169 mol Al2O3 x 101.96 g = 17.2 g Al2O3
1 mol Al2O3
This means that if all 8.10 grams of O2 were consumed, then 17.2 grams is
the MOST that COULD be produced.
CHEM 100 Fall 2013 .
chapter 4 -40 .
PROBLEM:
Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn metal, what volume of 2.50 M HCl is
needed to covert the Zn metal completely to Zn2+ ions?
Step 1: Write the balanced equation.
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
Step 2: Calculate amount of Zn.
10.0 g Zn x (1 mol Zn/65.39 g) = 0.153 mol Zn
Step 3:
Use the stoichiometric factor.
0.153 mol Zn x (2 mol HCl/1 mol Zn) = 0.306 mol HCl
Step 4:
Calculate volume of HCl required.
0.306 mol HCl x (1.00 L/2.50 mol) = 0.122 L HCl
CHEM 100 Fall 2013 .
chapter 4 -41 .
Problem:
76.80 grams of apple juice (malic acid) requires 34.56 mL
of 0.664 M NaOH to reach the endpoint in a titration.
What is the (w/w)% of malic acid in this sample?
Reaction:
C4H6O5(aq) + 2 NaOH(aq) --> Na2C4H4O5(aq) + 2 H2O(l)
(malic acid)
Step 1:
Calculate amount of NaOH used.
M x V = (0.664 M)(0.03456 L) = 0.0229 mol NaOH
Step 2:
Calculate amount of malic acid titrated.
0.0229 mol NaOH x (1 mol malic acid/2 mol NaOH)
= 0.0115 mol malic acid
Step 3:
Calculate the grams of malic acid in 0.0115 moles.
0.0115 mol malic acid x (134 g/1 mol malic acid) = 1.54 g
Step 4:
Calculate the weight (w/w)% malic acid in apple sample.
(1.54 g/76.80) x 100 = 2.01 (w/w)%
CHEM 100 Fall 2013 .
chapter 4 -42 .
Chemical Analysis Problem:
An impure sample of the mineral contains Na2SO4.
The mass of mineral sample = 0.123 g
All of the Na2SO4 in the sample is converted to insoluble
BaSO4.
The mass of BaSO4 is 0.177 g.
Determine the mass percent of Na2SO4 in the mineral.
CHEM 100 Fall 2013 .
chapter 4 -43 .
Answer:
Reaction equations:
Na2SO4(aq) + BaCl2(aq)  2 NaCl(aq) + BaSO4(s)
0.177 g BaSO4 x (1 mol/233.4 g) = 7.58 x 10-4 mol BaSO4
7.58 x 10-4 mol BaSO4 x (1 mol Na2SO4/1 mol BaSO4)
= 7.58 x 10-4 mol Na2SO4
7.58 x 10-4 mol Na2SO4 x (142.0 g/1 mol) = 0.108 g Na2SO4
(0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4
CHEM 100 Fall 2013 .
chapter 4 -44 .