Download Chapter 4: Reaction Stoichiometry Reaction Stoichiometry

Chapter 4: Reaction Stoichiometry
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Mole-to-mole ratios and calculations [4.2]
Limiting Reactants and Yields [4.3]
Molarity and stoichiometry [4.4]
Solubility Rules [4.5]
Types of Reactions [4.6, 8, 9]
Writing reactions [4.7]
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Reaction Stoichiometry
• For all reactions, we need to do many
calculations:
– How much of each starting material is needed?
– How much product is made?
– What materials are left over?
• These calculations follow methods we have
already learned: g → mol → mol → g
• A balanced equation is crucial to the success
of problem-solving!
– Coefficients provide mole-to-mole ratio!
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1
Reaction Stoichiometry
Steps for problem-solving:
1) Make sure you have a balanced equation.
2) Calculate molar masses of the compounds
you are interested in.
3) Determine the mole-to-mole ratio from the
coefficients in the balanced equation.
4) Set-up and solve the problem, watching
units to guide the process.
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Reaction Stoichiometry
How much Fe will be produced from 152.6 g of carbon
monoxide and an excess of iron (III) oxide?
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
28.01 g/mol
55.85 g/mol
g CO ! mol CO ! mol Fe ! g Fe
1 mol CO
2 mol Fe
55.85 g Fe
152.6 g CO x
x
x
= 202.8 g Fe
28.01 g CO 3 mol CO
1 mol Fe
Start with the mass
amount given in
problem statement.
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Use the mol-to-mol ratio
from the coefficients in
the balanced equation
Double-check that
your units cancel,
and you have the
correct sig figs!
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2
Reaction Stoichiometry
How many grams of CO2 are produced in the
combustion of 50.0 g of propane, C3H8?
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
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Reaction Stoichiometry
Determine the amount of oxygen produced when
0.549 g of KClO3 decomposes.
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
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3
Limiting Reactants
In all of the previous examples, the amount of only one starting
material is given. What if you know about both?
Example:
In the summer, I love fresh tomato sandwiches. I like 3 slices of
tomatoes per sandwich on two slices of bread. If I have 20 slices
of bread and 24 slices of tomatoes, how many sandwiches can I
make? What do I have left over, if anything?
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Limiting Reactants - Demo
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
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0.6 g
Mg
1.2 g
Mg
2.4 g
Mg
100 mL
1.0 M HCl
100 mL
1.0 M HCl
100 mL
1.0 M HCl
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Limiting Reactants
Steps for problem-solving:
1) Make sure you have a balanced equation.
2) Calculate the molar masses of all compounds that
you are interested in.
3) From each starting material, determine the amount
of the desired product that can be formed.
4) Compare the two product amounts; the lower
amount is the amount that will be formed, and the
reactant is the limiting reactant (also called limiting
reagent).
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Limiting Reactants
Methanol (CH3OH) can be burned as an alternative to fossil fuels.
For the following reaction, determine the amount of CO2 that is
formed when 130.0 g of CH3OH reacts with 150.0 g of O2.
2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)
32.04 g/mol 32.00 g/mol 44.01 g/mol
CH 3OH: 130.0 g CH 3OH x
O 2 : 150.0 g O 2 x
1 mol CH 3OH
32.04 g CH 3OH
1 mol O 2
32.00 g O 2
x
2 mol CO 2
3 mol O 2
x
x
2 mol CO 2
2 mol CH 3OH
44.01 g CO 2
1 mol CO 2
x
44.01 g CO 2
1 mol CO 2
= 178.6 g CO 2
= 137.5 g CO 2
Less CO2 is produced from the oxygen,
so oxygen is the limiting reactant and
the theoretical yield is 137.5 g of CO2.
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5
Limiting Reactants
The compound IF5 can be prepared by the reaction of SF4 with I2O5. The
only other product obtained is SO2.
5 SF4 + 2 I2O5 → 4 IF5 + 5 SO2
What is the maximum number of grams of IF5 that can be obtained from
40.0 g of SF4 and 40.0 g of I2O5?
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Limiting Reactants
Consider the following unbalanced reaction:
TiO2 +
C→
Ti +
CO
When 28.6 g of C is allowed to react with 88.2 g of TiO2, how many grams
of Ti can be produced? Which reactant is the limiting reactant? Which
reactant is in excess? How much of that reactant is leftover?
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Limiting Reactants
Consider the following unbalanced reaction:
1 TiO2 + 2 C → 1 Ti + 2 CO
When 28.6 g of C is allowed to react with 88.2 g of TiO2, how many grams
of Ti can be produced? Which reactant is the limiting reactant? Which
reactant is in excess? How much of that reactant is leftover?
1 mol C
1 mol Ti 47.88 g Ti
x
x
= 57.0 g Ti
12.011 g C 2 mol C
1 mol Ti
1 mol TiO 2
1 mol Ti
47.88 g Ti
TiO 2 : 88.2 g TiO 2 x
x
x
= 52.9 g Ti
79.87 g TiO 2 1 mol TiO 2
1 mol Ti
C: 28.6 g C x
TiO 2 is the limiting reagent, 52.9 g Ti can be produced from the reaction.
C is in excess, to determine the amount leftover:
57.0 g Tifrom C - 52.9 g Tifrom TiO = 4.1 g excess Tifrom C
2
1 mol Ti
2 mol C 12.011 g C
4.1 g Ti x
x
x
= 2.1 g C in excess
47.88 g Ti 1 mol Ti
1 mol C
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Percent Yield
• When we determine the amount of a product,
we are calculating a theoretical yield for the
reaction.
• Unfortunately, no reaction is perfect!
Percent Yield =
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Actual Yield
x 100%
Theoretical Yield
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Percent Yield
Example: For the IF5 reaction, suppose only 47.68 g of IF5 were
isolated. What is the percent yield for the reaction?
Example: For the methanol combustion reaction, suppose only
75.0 g of carbon dioxide is formed. What is percent yield for the
reaction?
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Solutions - Definitions
• Solute
• Solvent
• Solution
• Electrolyte
– Strong electrolytes
– Weak electrolytes
• Nonelectrolyte
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8
Molarity and Solutions
Many reactions are done in the solution phase, where
the reactants are dissolves in a solvent.
Molarity (M) =
moles of solute
L of solution
Problem
A 1.192 g sample of oxalic acid, H2C2O4, is placed in water to reach a total
of 100.0 mL. What is the molar concentration of the resulting solution?
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Molarity Problems
1.000 g of sulfuric acid is required for a reaction. How many mL of
a 0.200 M of sulfuric acid is needed to obtain this?
How do you prepare 500.0 mL of 0.25 M HCl solution from
concentrated HCl, which is 12.4 M?
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Atoms
Mass
Atoms
Moles A
Moles B
Volume
Mass
Volume
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Molarity and Stoichiometry
An aluminum can will dissolve in an aqueous solution of strong base,
according to the following reaction:
2 Al(s) + 2 KOH(aq) + 6 H2O(l)  2 KAl(OH)4(aq) + 3 H2(g)
2.05 g of Al are placed in a beaker with 185 mL of 1.35 M KOH. What is
the limiting reagent? How much (in g) KAl(OH)4 is produced?
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Writing and Predicting Equations
• We will focus on aqueous reactions, or
reactions where water is the solvent.
– Biological relevance
– Environmental relevance
• What causes a reaction to “go” or proceed?
– A reaction needs a “driving force”!
• A solid forms, called a precipitate (ppt)
• A gas forms, ↑
• Water forms
• Electrons are exchanged
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Writing and Predicting Equations
Reactions can be classified/grouped in two different ways:
Chemical
• Precipitation
Pattern
•
– 2 ionic compounds react to form 2 new
ionic compounds (or water)
– A solid forms
• Acid/Base
– Water forms
– Some gas-forming
• Reduction/Oxidation, or
“Redox”
– Electrons exchange
– Elements ⇔ compounds
– Some gas-forming
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Double Displacement
•
Single Displacement
– 1 ionic compound + 1 element yield a
new ionic compound and a new element
•
Combination
•
Decomposition
– 2 or more reactants yield one product
– One reactant yields 2 or more products
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Writing Reactions
• For reactions to proceed in water, the
reactants should be soluble - they form a
homogeneous solution.
– Reactions proceed faster if this is the case!
• How do we determine this?
– Type of compound
• Ionic - can be soluble or insoluble, based on
rules
• Molecular - often insoluble
– Electrolyte behavior
• Does the solution conduct electricity?
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Ionic Compounds - Solubility
Definitions (for ionic compounds mixing in H2O):
Soluble -
K
The molecule breaks apart and individual
ions “roam” independently.
SO4 K
H2O
K+1
SO4 -2
K+1
K2SO4 → 2 K+ + SO4-2
Insoluble - The molecule stays together- ions stay joined.
Ag
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SO4 Ag
H2O
Ag SO4 Ag
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Solubility Rules
Ion(s)
Li+, Na+, K+, NH4+
C2H3O2-, NO3Cl-, Br-, ISO42-
Rule
Group IA and ammonium salts are
all soluble.
Acetates and nitrates are soluble.
Most halides are soluble.
Most sulfates are soluble.
CO32-
Most carbonates are insoluble.
3-
Most phosphates are insoluble.
PO4
2-
S
Most sulfides are insoluble.
-
OH
Most hydroxides are insoluble.
Exceptions
none
none
Ag+, Hg22+, Pb2+
CaSO4, SrSO4, BaSO4, Ag2SO4,
Hg2SO4, PbSO4
Group IA and ammonium
carbonates
Group IA and ammonium
phosphates
Group IA and ammonium sulfides
and CaS, SrS, BaS
Group IA and ammonium
hydroxides, Ca(OH)2, Sr(OH)2,
Ba(OH)2
A similar table may be found in your text (Table 4.1).
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Predicting Solubility
• General Rules:
– Look at cation first - they are the overriding predictor of solubility!
• All compounds containing group I (including H+) and NH4 + ions are
soluble - no exception!
– If the cation does not make it soluble, then look at the anion.
Ex: Determine if the following compounds are soluble
or insoluble. If soluble, split into ions.
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(NH4)2CO3
CuSO4
AgBr
LiOH
BaSO4
Cobalt (III) Sulfide
Mercury (I) chloride
Sodium phosphate
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Writing Reactions - Double Displacement
Most precipitation and acid/base reactions are
double displacement reactions. How do we
write these reactions?
1. Split the reactants into individual ions.
2. Swap the cations to form new compounds these are your products.
3. Write the new compounds with balanced
formulas.
4. Determine the phases of the new products - are
the new compounds soluble or insoluble?
5. Balance the overall equation.
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Writing Reactions - Double Displacement
Write a balanced equation with the following starting materials:
NaNO3(aq) + CaCl2(aq)
H2SO4(aq) + LiOH(aq)
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Writing Reactions - Double Displacement
Write a balanced equation with the following starting materials:
Copper (II) nitrate and ammonium sulfide
Hydrochloric acid and silver hydroxide
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Redox Reactions - Definitions
Most redox reactions are single displacement reactions. How do
we recognize a redox reaction?
Oxidation:
Reduction:
Oxidizing agent:
Reducing agent:
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Assigning Oxidation States
• In free elements, each atom has an oxidation state of zero.
• For single-atom ions, the oxidation state equals the charge of the
atom. **Predict off periodic table!!
• Hydrogen is either +1 or –1, depending on the atom it is bonded to.
• Oxygen is typically –2, unless bonded to itself (e.g. H2O2)
• The sum of the oxidation states must equal the charge of the molecule
or ion.
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Assigning Oxidation States
Problem: Determine the oxidation states for each atom in the following
compounds:
CH3OH
NaBH4
Perchlorate ion
FeBr3
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Single Displacement Reactions
How do we write these SD reactions?
1. Split the ionic compound into individual ions.
2. Take the elemental reactant and turn it into a cation of
appropriate charge.
3. Swap the cations to form a new compound - this will be
one of your products.
4. Take the remaining cation and convert it to an element this will be the other product.
5. Write the new compounds with balanced formulas.
6. Determine the phases of the new products - are the new
compounds soluble or insoluble?
7. Balance the overall equation.
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Writing Reactions - Single Displacement
Write a balanced equation with the following starting materials:
Ca(s) + HCl(aq)
Cu(s) + Ag2SO4(aq)
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Writing Reactions - Single Displacement
Write a balanced equation with the following starting materials:
Na(s) + H2O(l)
Lead and hydroiodic acid to make a lead (II) compound
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Net Ionic Equations
We often “translate” chemical equations into net ionic equations
(or reactions). These reactions strip out the ions that do not
participate in the chemical process described by the reaction.
Example:
Molecular Equation
CaCl2(aq) + Li2 SO4(aq) → CaSO4(s) + 2 LiCl(aq)
Ionic Equation
Ca+2(aq) + 2 Cl-(aq) + 2 Li+(aq) + SO4-2(aq) → CaSO4(s) + 2 Li+(aq) + 2 Cl-(aq)
Net Ionic Equation
Ca+2(aq) + SO4-2(aq) → CaSO4(s)
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Net Ionic Equations
Do the following steps to generate a net ionic equation:
1) You must begin with a balanced molecular equation, with all phase
labels listed.
2) For each and only (aq) species, break it up into individual ions (keep
polyatomic ion units together).
3) Do not break up any (s), (l), or (g). This generates the ionic equation.
4) Inspect the ionic equation - for each species that appears in the
identical form on both sides of the equation, cross those out.
Reconcile any species that appear in differing amounts on both sides.
5)
Rewrite only the species that were not crossed out in a new
reaction - the net ionic equation!
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Net Ionic Equations
Write net ionic equations for the following balanced equations.
2 AgNO3(aq) + MgCl2(aq) → 2 AgCl(s) + Mg(NO3)2(aq)
2 Ag+(aq) + 2 NO3-(aq) + Mg+2(aq) + 2 Cl-(aq) → 2 AgCl(s) + Mg+2(aq) + 2 NO3-(aq)
2 Ag+(aq) + 2 Cl-(aq) → 2 AgCl(s)
H2SO4(aq) + Zn(s) → H2(g) + ZnSO4(aq)
2 NaOH(aq) + Pb(C2H3O2)2(aq) → Pb(OH)2(s) + 2 NaC2H3O2(aq)
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Some Special Reactions
Gas-Forming Reactions fall under either redox or acid/base
reactions.
Redox: Metal + acid → Hydrogen gas + ionic compound
H2SO4(aq) + Zn(s) → H2(g) + ZnSO4(aq)
2 HClO4(aq) + Pt(s) → H2(g) + Pt(ClO4)2(aq)
Acid/Base: Reactions that generate H2CO3 decompose further to
CO2 and H2O
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2CO3(aq) → H2O(l) + CO2(g)
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
K2CO3(s) + 2 HBr(aq) → 2 KBr(aq) + H2O(l) + CO2(g)
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Putting it all together
Perform the following:
1) Write a complete and balanced chemical equation that describes the
reaction between aluminum metal and sulfuric acid.
2) Determine how much 0.125 M sulfuric acid is needed (in mL) to react
with 22.5 g of aluminum.
3) Determine how much of the elemental product is liberated in that
reaction as well.
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