Transcript Chem 1201 - LSU Department of Chemistry

Watkins
Chapter 3
CHAPTER 3
Chemical Reaction Calculations
• CR calculations always involve a
balanced chemical reaction equation.
• CR calculations always involve
mole  mole conversions
• CR calculations often involve
mass  mole conversions.
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Chapter 3
Chemical Reaction Calculations
A Chemical Reaction Equation
states a chemical fact
(in Chemish)
N2(g) + 3H2(g) → 2NH3(g)
Reactants
Products
•Nitrogen gas and hydrogen gas
(in English) •react under the appropriate conditions of T and P
•to produce ammonia gas.
The chemical formula of each reactant
and product must be correct!
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Chapter 3
Coefficients
Chemical Reaction Equation
1N2(g) + 3H2(g) → 2NH3(g)
Stoichiometric
Coefficients
Conservation of Matter
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Chapter 3
States
Chemical Reaction Equation
N2(g) + 3H2(g) → 2NH3(g)
Physical
States
g, l, s, aq
An important part of the reaction equation.
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Chapter 3
1N2(g) + 3H2(g) → 2NH3(g)
N
H
H
H
N
H
H
H
Conservation of Matter
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Before reaction:
1 molecule of
nitrogen gas (2 N
atoms) & 3
molecules of
hydrogen gas (6 H
atoms).
After reaction:
2 molecules of
ammonia gas (2
N atoms & 6 H
atoms)
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Chapter 3
Coefficients
Chemical Reaction Equation
1N2(g) + 3H2(g) → 2NH3(g)
Molecules of reactants produce Molecules of product
Balancing: atoms left = atoms right
Conservation of Matter
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Chapter 3
1N2(g) + 3H2(g) → 2NH3(g)
2N2(g) + 6H2(g) → 4NH3(g)
N
N
N
H
H
H
H
H
H
H
H
H
H
H
H
1/ (doz)N (g)
6
2
H
N
N
H
H
H
→
H
N
N
H
H
H
H
H
N
H
H
+ 1/2(doz)H2(g) → 1/3(doz)NH3(g)
12N2(g) + 36H2(g) → 24NH3(g)
→
1(doz)N2(g) + 3(doz)H2(g) → 2(doz)NH3(g)
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Chapter 3
BALANCING EQUATIONS
1 2H4(g) + __O
2
__C
3 2(g) → __CO
2 2O(l)
2(g) + __H
2
6
2
3
3/
1
3
1
__Al(s)
+ __HCl(aq)
→ __AlCl
(s)
+
__H
2
3
2(g)
2
3
2
4
3/
2
1
__CH
O(l)
+
__O
1
2 2(g) → __CO
4
2(g) + __H2O(l)
Smallest whole number coefficients;
not necessary, but sometimes asked for in problems.
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ATOMIC MASS
Chapter 3
Periodic Table
No C atom weighs 12.011 amu!!!
6
Atomic Number
C
Atomic Symbol
12.011
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Atomic Weight (amu)
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Chapter 3
Atomic Mass Scale
No C atom weighs 12.011 amu!!!
• Any sample of pure carbon on this planet is a
mixture of two isotopes: 12C and 13C
• In a sample containing 10,000 C atoms:
•9,890 atoms are 12C (12 amu exactly)(12.0 amu)
•110 atoms are 13C (13.003325 amu)
• The average atomic mass of the sample is
isotopic masses
average atomic mass
atomic "weight"
(9,890)(12.000000) + (110)(13.003325)
10,000
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10
= 12.011
amu
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Chapter 3
Atomic Mass Scale
No C atom weighs 12.011 amu!!!
In a sample containing 10,000 C atoms:
9,890 atoms are 12C (12 amu exactly)
110 atoms are 13C (13.003325 amu)
The fraction of each naturally occuring isotope is
0.9890 12C
0.0110 13C
Natural Abundance is this fraction expressed as %
98.90% 12C isotopic abundances are listed in tables of
1.10% 13C isotopes: Handbook of Chemistry & Physics.
Artificial isotopes (like 14C) have 0% natural
abundance.
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Chapter 3
FORMULA WEIGHTS
C
F.W. (or A.W.) = 12.01 amu
NaCl
F.W. = 22.99 + 35.45 = 58.44 amu
H2O
F.W. (or M.W.) = 2(1.01) + 16.00 = 18.02 amu
H2O2 (HO)
E.F.W. = 1.01 + 16.00 = 17.01 amu
F.W. (or M.W.) = 2(17.02) = 34.02 amu
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Chapter 3
Number Names
pair
quartet
dozen
gross
mole (mol)
(the "Chemical Dozen")
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Chapter 3
Numbers of Atoms
1 pair of 12C atoms = 2 12C atoms
1 quartet of 12C atoms = 4 12C atoms
1 dozen 12C atoms = 12 12C atoms
1 gross of 12C atoms = 144 12C atoms
1 mol of 12C atoms = 6.02×1023 12C atoms
(Avogadro’s number NA )
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Chapter 3
Atomic Mass
1 12C atom weighs 12 amu exactly
6.02E23 amu = 1 g
12 amu ×
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1g
6.02E23 amu
15
= 1.99E-23 g
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Chapter 3
Molar Mass
1 12C atom = 1.99E-23 g
1 pair of 12C atoms = 3.98E-23 g
1 quartet of 12C atoms = 7.96E-23 g
1 dozen 12C atoms = 2.39E-22 g
1 gross of 12C atoms = 2.87E-21 g
1 mol of 12C atoms = 12 g exactly
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Chapter 3
Molar Mass
Mol: the number of C-12 atoms in exactly
12 grams of C-12.
I have counted Avogadro’s number:
I weighed out exactly 12 g of C-12
12.000000000000000000000000
I counted all the atoms in the pile of C-12
atoms using a Tiger Stadium Fan Clicker:
602 214 179 331 069 112 495 297
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Chapter 3
Molar Mass
Mol: the number of C-12 atoms in exactly
12 grams of C-12.
This number has been approximated by NIST:
the most recent value (www.nist.gov) is
(6.02214179 ± 0.00000030) × 1023
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Chapter 3
MOLAR MASS
Periodic Table
6
Atomic Number
C
Atomic Symbol
12.011
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Atomic Weight (amu)
Average Molar Mass (g)
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Chapter 3
(AVERAGE) MOLAR MASSES
12 g
1 mol of C atoms = 12.011 g
1 mol of O atoms = 15.999 g 16 g
1 mol of U atoms = 238.03 g 238 g
1 mol of H2O molecules = 18.01 g 18 g
For homework, use two decimal places.
For exams, usually use zero decimal places.
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Chapter 3
Molecular Interpretation
Chemical Reaction Equation
1N2(g) + 3H2(g) → 2NH3(g)
1 molecule + 3 molecules → 2 molecules
1 pair + 3 pair → 2 pair
1 dozen + 3 dozen → 2 dozen
1 mol + 3 mol → 2 mol
Any coefficients in the ratio 1:3:2 are OK!
½ mol + 1½ mol → 1 mol
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Chapter 3
Mass Interpretation
Chemical Reaction Equation
1N2(g) + 3H2(g) → 2NH3(g)
28.01 amu + 6.05 amu = 34.06 amu
28.01 g
+ 6.05 g
= 34.06 g
Molecules of reactants produces Molecules of product
Moles
of reactants produces Moles
of product
Reactant mass (28.01 + 6.05) = Product mass (34.06)
Conservation of Matter
Mass Balance
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Chapter 3
Elemental Composition
BY WEIGHT
Element Weight
× 100
%Element =
Formula Weight
e.g. glucose, C6H12O6: F.W. = 180.16 amu
%C
%H
%O
6(12.01)
× 100
180.16
12(1.01)
× 100
180.16
6(16.00)
× 100
180.16
40.00%
6.72%
53.28%
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Chapter 3
Elemental Composition
BY WEIGHT
glucose: 40.00%C, 6.72%H, 53.28%O
glucose: CH2O E.F.W. = 30.03 amu
%C
%H
%O
12.01
x 100
30.03
2(1.01)
x 100
30.03
16.00
x 100
30.03
40.00%
6.72%
53.28%
Empirical composition is identical to elemental
composition.
Empirical composition is found from experiments
(empirically) by analytical chemists.
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Chapter 3
Empirical Formulas
from Composition
The Process
1. Convert % composition to grams:
Glucose: 40.00%C, 6.72%H, 53.28%O
=> 40.00 g C, 6.72 g H, 53.28 g O
(in 100 g of glucose)
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Chapter 3
Empirical Formulas
from Composition
Glucose: 40.00%C, 6.72%H, 53.28%O
2. Convert grams to moles:
mass (g) / molar mass (g/mol)
40.00
40.00 g C →
12.01
6.72 g H → 6.72
1.01
53.28 g O → 53.28
16.00
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= 3.33 mol C
= 6.66 mol H
= 3.33 mol O
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Chapter 3
Empirical Formulas
from Composition
3. Calculate Mole Ratios:
mol C : mol H : mol O
3.33 : 6.66 : 3.33
C3.33H6.66O3.33
3.33 : 6.66 : 3.33
3.33 3.33 3.33
divide each number
by the smallest
1 : 2
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{
C1H2O1 = CH2O
: 1
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Chapter 3
Empirical Formulas
from Composition
Summary of The Process
1. Convert % composition to grams of each
element;
2. Convert grams of each element to moles of
each element;
3. Calculate smallest-whole-number mole ratio
(empirical formula).
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Chapter 3
Empirical Composition
What experiment does the analytical chemist perform to
determine the composition and empirical formula of a
substance?
She burns it!
"Burning" is a chemical reaction between a "fuel" and
oxygen in the air.
Any reaction with oxygen = combustion
Combustion converts each element in the fuel (except
oxygen) into its oxide.
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Chapter 3
Combustion Analysis
Objective: to find the empirical formula of a fuel
If the fuel is “organic” (contains only C, H, O):
fuel + oxygen → carbon dioxide + water
Every Atom
Oz and O2 are
CxHyOz + O2 → CO2 + H2O mixed up in
both products
Every Atom
The “C-H” Process
Capture and weigh the CO2
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Capture and weigh the H2O
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Chapter 3
Combustion Analysis
Combustion of 0.255 grams of an organic
fuel produces 0.561 grams of carbon
dioxide and 0.306 grams of water.
Step 1: weigh the fuel and the products
weight of fuel (0.255 g)
weight of CO2 (0.561 g)
weight of H2O (0.306 g)
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Chapter 3
Combustion Analysis
Step 2: convert to grams of C and H:
0.561 g CO2 ×
1 mol CO2
44.01 g CO2 1 mol CO2
molar
mass
0.306 g H2O ×
×
1 mol C
×
12.01 g C
1 mol C
molar
ratio
1 mol H2O
molar
mass
2 mol H
×
×
18.02 g H O 1 mol H O
2
= 0.153 g C
2
1.008 g H
1 mol H
= 0.034 g H
Use at least 4 significant figures for molar masses!
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Chapter 3
Combustion Analysis
Step 3: grams of O by difference:
• The mass of the fuel = the mass of all of its atoms
• g fuel = g C + g H + g O
• g O = g fuel - (g C + g H)
• g O = 0.255 g fuel - (0.153 g C + 0.034 g H) = 0.068 g O
(in some organic fuels, there is no oxygen!)
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Chapter 3
Combustion Analysis
Step 4: convert to moles of C, H, and O:
0.153 g C
×
0.034 g H ×
0.068 g O ×
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1 mol C
12.01 g C
1 mol H
1.008 g H
1 mol O
16.00 g O
35
= 0.0128 mol C
= 0.0340 mol H
= 0.0043 mol O
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Chapter 3
Combustion Analysis
Step 5: ratio of moles => Empirical Formula
mol C : mol H : mol O
0.0128 :
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0.0128
0.0043
0.0340 :
0.0043
:
0.0340
0.0043
:
0.0043
0.0043
2.98
:
7.91
:
1
~3
:
~8
:
C3H8O
1
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Chapter 3
Combustion Analysis
Step 5: ratio of moles => Empirical Formula
mol C : mol H : mol O
0.2124 : 0.3721 : 0.1065
1.99
:
3.49 :
1
~2
:
~ 7/2 :
1
4
:
7 :
2
C4H7O2
Watch for rational fractions!
(especially halves and thirds)
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Chapter 3
Combustion Analysis
Combustion analysis yields only the
empirical formula.
To find a molecular formula, more
information is required - usually Molecular
Weight:
n×empirical formula weight = molecular weight
(empirical formula)n = molecular formula
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Chapter 3
Combustion Analysis
Summary
Step 1: weigh the fuel, CO2 and H2O
Step 2: convert to grams of C and H
Step 3: find grams of O by difference
Step 4: convert to moles of C, H, and O
Step 5: smallest whole number molar ratio
=> Empirical Formula
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Chapter 3
General Stoichiometry
What is a stoichiometry problem?
You are given some data for a specific chemical
reaction; e.g., the initial amounts of some of the
reactants and products;
Question: calculate the amounts of one or more of
the reactants and/or products;
usually one or more products;
usually after the reaction goes to completion.
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Chapter 3
General Stoichiometry
Strategy (The Process)
Write and balance the reaction equation;
Gather the data (make a list);
The question(s): what calculations are required
Do the calculations required;
(Note: there is always more than one way to
solve a stoichiometry problem);
Report the answer(s).
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Chapter 3
Sample Problem 1
2.78 moles of hydrogen gas react completely with
excess oxygen gas. How many moles of liquid
water will be produced?
2H2(g) + O2(g) → 2H2O(l)
If the answer involves just one product (or reactant),
dimensional analysis provides a quick solution.
Homepage > Extra > Dimensional Analysis
2 mol H2O
? mol H2O = 2.78 mol H2
= 2.78 mol H2O
2 mol H2
2.78 moles of water will be produced.
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Chapter 3
Sample Problem 1
2.78 moles of hydrogen gas react completely with
excess oxygen gas. How many moles of liquid
water will be produced?
2H2(g) + O2(g) → 2H2O(l)
In this case, the molar ratio provides an even
quicker solution:
The molar ratio of hydrogen to water is 2:2, so
mol H2:mol H2O = 2:2 = 1:1 = 2.78:2.78
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Chapter 3
Sample Problem 2
Most stoichiometry problems involve masses of
reactants and products (not moles).
But chemical reaction equations are always in
moles, so mole  mass conversions are required.
What mass of CO2 is produced when 3.00 g of
ethanol (C2H6O) are burned?
C2H6O + 3O2 → 2CO2 + 3H2O
44 g CO2
2 mol CO2
1 mol C2H6O
? g CO2 = 3.00 g C2H6O ×
×
×
46 g C2H6O 1 mol C2H6O 1 mol CO2
= 5.74 g CO2
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
→
4NO + 6H2O
This problem is more complicated.
There are four answers requiring four separate
calculations.
Also, a separate calculation is required to determine
the limiting reagent.
First, what is a limiting reagent?
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
→
4NO + 6H2O
When a reaction goes to completion (“reacts completely”),
one of the reactants (the limiting reagent) must be used
up first; that is, the reaction just “runs out of fuel”!
If this were like Sample Problem 1 (“1.50 g of ammonia
reacts with excess oxygen”), the answer would be
obvious: some oxygen would be left, the ammonia
would be used up first, and the final mass of ammonia
would be 0 g!
Limiting Reagent Problems: when the amounts of two or
more reagents are given explicitely.
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Chapter 3
Limiting Reagent Problems
If the amounts of two or more reactants are given,
we can't tell which one is used up first
Objective: determine which reactant is used up
first in the reaction.
• LR = Limiting Reagent (Fmol = 0)
• XS = Reagent in Excess (Fmol > 0)
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
→
4NO + 6H2O
So which reactant runs out first, ammonia or oxygen?
One way to find out: compare molar ratios.
moles NH3
moles O2 = 4/5 = 0.8
moles NH3 1.50/17.0
Given molar ratio: moles O = 1.85/32.0 = 1.53
2
Theoretical molar ratio:
There is clearly too much NH3 (excess) and not enough O2
(limiting). Thus, the remaining calculations, to
determine the amount of NH3 left and the amounts of
NO and H2O produced, must be based on 1.85 g O2.
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Chapter 3
Stoichiometry
Quantitative chemical reaction problems can be
simple (few calculations) or complex (many
calculations).
For simple problems, use dimensional analysis –
it’s quick and easy.
For complex problems, keeping track of the data,
the calculations and the results can be very
challenging!
One way to organize complex chemical reaction
problems is the Reaction Table Method.
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Chapter 3
Reaction Table
Chemical Reaction Equation
aA + bB → cC + dD
Molecules of reactants produces Molecules of product
Moles of reactant molecules produce
Moles of product molecules
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Chapter 3
General Stoichiometry
Homepage > Extra > Reaction Table
Objective: to solve a stoichiometry problem
for a balanced chemical reaction.
A Reaction Table organizes the data used in the
stoichiometry calculations
A Reaction Table keeps track of what happens
before, during, and after a reaction.
A Reaction Table shows you exactly what
stoichiometry calculations to make.
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Chapter 3
Reaction Table
4NH3 + 5O2
→
4NO + 6H2O
Initial moles
Change in moles
Final moles
•
•
•
•
•
A spreadsheet for any stoichiometry problem.
One column for each reactant and product.
"Initial" means before the reaction starts
"Change" is what happens during the reaction
"Final" means after the reaction is complete
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Chapter 3
Reaction Table
4NH3 + 5O2
→
4NO + 6H2O
I
C
F
• The first row contains the moles of each reactant
and product before the reaction begins.
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Chapter 3
Reaction Table
4NH3 + 5O2
→
4NO + 6H2O
I
C
F
• The second row shows how these amounts
change during the reaction.
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Chapter 3
Reaction Table
4NH3 + 5O2
→
4NO + 6H2O
I
C
F
• The third row contains the moles of all reactants
and products after the reaction is complete
• Final = Initial + Change
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Chapter 3
Reaction Table
4NH3 + 5O2
→
4NO + 6H2O
I
C
- 4x
-5x
+ 4x
+ 6x
F
The Change row is always the same:
x in every box (the change is unknown)
multiply by the stoichiometric coefficients
minus for reactants (because they decrease)
plus for products (because they increase)
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Chapter 3
Sample Problem 1
2.78 moles of H2 react with excess O2. How many
moles of H2O will be produced?
2H2 + O2 → 2H2O
I
C
F
Step 1: Write the reaction table
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Chapter 3
Sample Problem 1
2.78 moles of H2 react with excess O2. How many
moles of H2O will be produced?
2H2 + O2 → 2H2O
I
2.78
XS
0
C
F
no water has been
produced before the
reaction begins!
XS oxygen means there will be plenty of oxygen
left after the reaction stops, so ...
hydrogen is the limiting reagent!
Step 2: Fill in the initial data
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Chapter 3
Sample Problem 1
2.78 moles of H2 react with excess O2. How many
moles of H2O will be produced?
2H2 + O2 → 2H2O
I
2.78
XS
0
C
-2x
-x
+2x
F
The change row
is filled in the
same way for
every problem.
x is a single unknown, so it will have the
same value in every box.
Step 3: Fill in the change row
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Chapter 3
Sample Problem 1
2.78 moles of H2 react with excess O2. How many
moles of H2O will be produced?
2H2 + O2 → 2H2O
I
2.78
XS
0
C
-2x
-x
F 2.78-2x XS
+2x
2x
Row 3 is always
the sum of the
first two rows.
Some oxygen will remain after the reaction,
but we don't care how much - it's not part of
the problem!
Step 4: F = I + C
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Chapter 3
Sample Problem 1
2.78 moles of H2 react with excess O2. How many
moles of H2O will be produced?
2H2 + O2 → 2H2O
I
2.78
XS
0
C
-2x
-x
F 2.78-2x XS
+2x
2x
How do you solve
for x?
LR: after the reaction, all of the hydrogen has been used up!
F H2 = final moles of H2 = 2.78-2x = 0 x = 2.78/2 = 1.39
Step 5: solve for x (that is, find the value of x)
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Chapter 3
Sample Problem 1
2.78 moles of H2 react with excess O2. How many
moles of H2O will be produced?
2H2 + O2 → 2H2O
I
2.78
XS
0
C
-2x
-x
F 2.78-2x XS
=0
+2x
2x =
2.78
x = 2.78/2 = 1.39
Step 6: substitute the value of x in the third row to get the answer
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Chapter 3
REACTION TABLE
aA + bB → cC + dD
I
C
F
Most stoichiometry problems use masses of
reactants and products instead of moles. These
problems are sometimes called "weight-weight"
problems.
We can add some rows to the three "mole rows"
to include the new data, but these mole rows are
still the most important rows.
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Chapter 3
REACTION TABLE
aA + bB → cC + dD
Imass
MM
I
mA
mB
mC
mD
C
F
Fmass
The first row contains the initial mass, in grams,
of each reactant and product.
mC and mD are usually assumed to be 0. Why?
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Chapter 3
REACTION TABLE
aA + bB → cC + dD
mB
Imass mA
MM MMA MMB
I
C
F
mC
MMC
mD
MMD
Do not multiply the
molar mass by the
stoichiometric
coefficient!!!
Fmass
The second row contains the Molar Mass
(molecular weight), in grams/mol, of each
reactant and product.
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Watkins
Chapter 3
REACTION TABLE
aA + bB → cC + dD
mB
Imass mA
MM MMA MMB
I
mC
MMC
mD
MMD
n=m/MM
C
F
Fmass
Divide the first row numbers by the second row
numbers to calculate the third row numbers.
“n” is the generic symbol for moles.
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Chapter 3
REACTION TABLE
aA + bB → cC + dD
mB
Imass mA
MM MMA MMB
I
C
mC
MMC
mD
MMD
n=m/MM
-ax, etc.
I+C
F
Fmass
The three "mole rows" can be filled in, then solve
for x and substitute the value of x to find the final
moles for each reactant and product, exactly as in
the previous problem.
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Watkins
Chapter 3
REACTION TABLE
aA + bB → cC + dD
mB
Imass mA
MM MMA MMB
I
C
mC
MMC
mD
MMD
n=m/MM
-ax, etc.
I+C
F
m=nMM
Fmass
The final mass of each reactant and product is just the
final moles (row 5) times the molar mass (row 2):
g
F × MM = mol × mol
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Chapter 3
Sample Problem 2
What mass of CO2 is produced when 3.00 g of
ethanol (C2H6O) are burned?
C2H6O + 3O2 → 2CO2 + 3H2O
Imass
3.00
XS
0
0
46
32
44
18
I
0.065
XS
0
0
C
-x
-3x
+2x
+3x
F
0.065-x
XS-3x
2x
3x
MM
Fmass
5.72
Fmol(C2H6O) = 0 = 0.065-x
Fmol(CO2) = 2x = 0.130 mol
Chemistry 1421
x = 0.065
Fmass(CO2) = (0.130 mol)(44 g/mol)
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Brown & Lemay
Watkins
Chapter 3
Sample Problem 2
What mass of CO2 is produced when 3.00 g of
ethanol (C2H6O) are burned?
C2H6O + 3O2 → 2CO2 + 3H2O
Imass
3.00
XS
0
0
46
32
44
18
I
0.065
XS
0
0
C
-x
-3x
+2x
+3x
F
0.065-x
XS-3x
2x
3x
MM
Fmass
5.72
Key Ideas:
1. Once the ethanol is used up, reaction stops;
2. There is plenty of oxygen left.
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Limiting Reagent
XS Reagent
Brown & Lemay
Watkins
Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
Im
→
4NO + 6H2O
1.50
17
1.85
32
0
30
0
18
I
0.0881
0.0578
0
0
C
-4x
-5x
+4x
+6x
4x
6x
MM
F 0.0881-4x 0.0578-5x
Fm
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
Im
→
4NO + 6H2O
1.50
17
1.85
32
0
30
0
18
I
0.0881
0.0578
0
0
C
-4x
-5x
+4x
+6x
4x
6x
MM
F 0.0881-4x 0.0578-5x
Fm
Which reactant is used up first?
Strategy: try both reactants as LR
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
Im
→
4NO + 6H2O
1.50
17
1.85
32
0
30
0
18
I
0.0881
0.0578
0
0
C
-4x
-5x
+4x
+6x
4x
6x
MM
F 0.0881-4x 0.0578-5x
Fm
• If LR = NH3, 0.0881-4x = 0 => x = 0.0220
• If LR = O2, 0.0578-5x = 0 => x = 0.0116
• Always choose the smallest x (WHY???)
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
Im
4NO + 6H2O
1.50
17
1.85
32
0
30
0
18
I
0.0881
0.0578
0
0
C
-4x
-5x
+4x
+6x
MM
x = 0.0116
→
0.0419
0
0.0578-5x
F 0.0881-4x
0.712
0
Fm
XS
0.0462
4x
1.388
0.0694
6x
1.250
3.35 g
3.35 g
LR
Note that Total Initial Mass = Total Final Mass
Mass balance is a GOOD CHECK!
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Chapter 3
Sample Problem 3
What mass of all reactants and products are left after the
reaction of 1.50 g of NH3 with 1.85 g O2 is complete?
4NH3 + 5O2
Im
MM
I
1.50
17
1.85
32
0.0881
0.0578
C
→
4NO + 6H2O
Another method for
finding the LR: compare
the theoretical and
actual ratio of moles
F
Too muchFm
of this
XS
Not enough of this
LR
Chemistry 1421
mol NH3
mol O2
75
Theory
4
= 0.8
5
Actual
0.0881
= 1.52
0.0578
Brown & Lemay
Watkins
Chapter 3
Sample Problem 4
A chemical plant uses electrical energy to
decompose brine solutions. If the plant produces
1.4E6 kg of Cl2 daily, estimate the quantities of H2
and NaOH produced.
2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + H2(g) + Cl2(g)
Dimensional Analysis: the question equations are
? g NaOH = 1.4E9 g Cl2
? g H2 = 1.4E9 g Cl2
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Chapter 3
Sample Problem 4
A chemical plant uses electrical energy to decompose aqueous solutions of NaCl. If the
plant produces 1.4E6 kg of Cl2 daily, estimate the quantities of H2 and NaOH produced.
2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + H2(g) + Cl2 (g)
Im
?
?
0
0
0
58.44
18.02
40.00
2.02
70.92
I
?
?
0
0
0
C
-2x
-2x
+2x
+x
+x
F
?
?
?
?
2x
1.58E9
MM
Fm
x
x
3.98E7 1.40E9
FMass(Cl2) = (Fmol)(MM) => 1.40E9 g = (x)(70.92 g/mol)
x = 1.97E7 mol
Fmass(NaOH) = (2×1.97E7 mol)(40.00 g/mol) = 1.58E9 g
Fmass(H2) = (1.97E7 mol)(2.02 g/mol) = 3.98E7 g
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