Transcript CH 8 blackboardx

7.10 Classifying Chemical
Reactions by What Atoms Do
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Chapter 8
Quantities in Chemical Reactions
8.2 Stoichiometry: Relationships
between Ingredients

The numerical relationship between chemical
quantities in a balanced chemical equation is called
reaction stoichiometry.

We can predict the amounts of products that form in
a chemical reaction based on the amounts of
reactants.

We can predict how much of the reactants are
necessary to form a given amount of product.

We can predict how much of one reactant is required
to completely react with another reactant.
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8.2 Making Pancakes:
Relationships between Ingredients

A recipe gives numerical relationships between the
ingredients and the number of pancakes.
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8.2 Making Pancakes:
Relationships between Ingredients



The recipe shows the numerical relationships between
the pancake ingredients.
If we have 2 eggs—and enough of everything else—we
can make 5 pancakes.
We can write this relationship as a ratio.

2 eggs:5 pancakes
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What if we have 8 eggs? Assuming that we
have enough of everything else, how many
pancakes can we make?
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8.3 Making Molecules:
Mole-to-Mole Conversions


In a balanced chemical equation, we have a “recipe” for how
reactants combine to form products.
The following equation shows how hydrogen and nitrogen
combine to form ammonia (NH3).
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3 H2(g) + N2(g)  2 NH3(g)


The balanced equation shows that 3 H2 molecules react
with 1 N2 molecule to form 2 NH3 molecules.
We can express these relationships as ratios.
3 H2 molecules : 1 N2 molecule : 2 NH3 molecules

Since we do not ordinarily deal with individual molecules,
we can express the same ratios in moles.
3 mol H2 : 1 mol N2 : 2 mol NH3
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3 H2(g) + N2(g)  2 NH3(g)

If we have 3 mol of N2, and more than enough H2, how much
NH3 can we make?
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Example 8.1 Mole-to-Mole Conversions
Sodium chloride, NaCl, forms by this reaction between sodium and chlorine.
2 Na(s) + Cl2(g)2 NaCl(s)
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.
SORT
GIVEN: 3.4 mol Cl2
You are given the number of moles of a reactant (Cl2)
and asked to find the number of moles of product
(NaCl) that will form if the reactant completely reacts.
FIND: mol NaCl
STRATEGIZE
SOLUTION MAP
Draw the solution map beginning with moles of chlorine
and using the stoichiometric conversion factor to
calculate moles of sodium chloride. The conversion
factor comes from the balanced chemical equation.
RELATIONSHIPS USED
1 mol Cl2 : 2 mol NaCl (from balanced chemical
equation)
Example 8.1 Mole-to-Mole Conversions
Continued
SOLVE
SOLUTION
Follow the solution map to solve the problem. There is
enough Cl2 to produce 6.8 mol of NaCl.
CHECK
Check your answer. Are the units correct? Does the
answer make physical sense?
The answer has the correct units, moles. The
answer is reasonable because each mole of Cl2
makes two moles of NaCl.
SKILLBUILDER 8.1 Mole-to-Mole Conversions
Water is formed when hydrogen gas reacts explosively with oxygen gas according to the balanced equation:
O2(g) + 2 H2(g)2 H2O(g)
How many moles of H2O result from the complete reaction of 24.6 mol of O2? Assume that there is more than
enough H2.
Answer: 49.2 mol H20
For More Practice Example 8.8; Problems 15, 16, 17, 18.

8.4 Making Molecules:
Mass-to-Mass Conversions
A chemical equation contains conversion factors between
moles of reactants and moles of products.
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We are often interested in relationships between mass of
reactants and mass of products.
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The general outline for this type of calculation is:
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2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
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What mass of carbon dioxide is emitted by an automobile
per 5.0 x 102 g pure octane used?
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The balanced chemical equation gives us a relationship
between moles of C8H18 and moles of CO2.
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Before using that relationship, we must convert from
grams to moles.
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2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
SOLUTION:
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2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
SOLUTION:
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Example 8.2 Mass-to-Mass Conversions
In photosynthesis, plants convert carbon dioxide and water into glucose (C 6H12O6) according to the 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.
SORT
GIVEN: 58.5 g CO2
You are given the mass of carbon
dioxide and asked to find the mass of
glucose that can form if the carbon
dioxide completely reacts.
FIND: g C6H12O6
STRATEGIZE
SOLUTION MAP
The solution map uses the general
outline
Mass A  Moles A 
Moles B  Mass B
where A is carbon dioxide and B
is glucose.
Example 8.2 Mass-to-Mass Conversions
Continued
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.
RELATIONSHIPS USED
6 mol CO2 : 1 mol C6H12O6 (from balanced chemical equation)
Molar mass CO2 = 44.01 g/mol
Molar mass C6H12O6 = 180.2 g/mol
SOLVE
SOLUTION
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.
CHECK
Are the units correct? Does the
answer make physical sense?
The units, g C6H12O6, are correct. The magnitude of the answer seems reasonable
because it is of the same order of magnitude as the given mass of carbon dioxide.
An answer that is orders of magnitude different would immediately be suspect.
Example 8.2 Mass-to-Mass Conversions
Continued
SKILLBUILDER 8.2 Mass-to-Mass Conversions
Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl,
according to the reaction:
Mg(OH)2(aq) + 2 HCl(aq)  2 H2O(l) + MgCl2(aq)
How much HCl in grams can be neutralized by 5.50 g of Mg(OH)2?
Answer: 6.88 g HCl
For More Practice Example 8.9; Problems 31, 32, 33, 34.
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 simplified reaction.
4 NO2(g) + O2(g) + 2 H2O(l)  4 HNO3(aq)
Assuming that there is more than enough O2 and H2O, how much HNO3 in kilograms forms from 1.5  103 kg of
NO2 pollutant?
SORT
GIVEN: 1.5  103 kg NO2
You are given the mass of nitrogen
dioxide (a reactant) and asked to find the
mass of nitric acid that can form if the
nitrogen dioxide completely reacts.
FIND: kg HNO3
Example 8.3 Mass-to-Mass Conversions
Continued
SOLUTION MAP
STRATEGIZE
The solution map follows the general
format of:
Mass  Moles 
Moles  Mass
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.
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.
RELATIONSHIPS USED
4 mol NO2 : 4 mol HNO3 (from balanced chemical equation)
Molar mass NO2 = 46.01 g/mol
Molar mass HNO3 = 63.02 g/mol
1 kg = 1000 g
Example 8.3 Mass-to-Mass Conversions
Continued
SOLVE
SOLUTION
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.
CHECK
Are the units correct? Does the answer
make physical sense?
The units, kg HNO3 are correct. The magnitude of the answer seems
reasonable because it is of the same order of magnitude as the given
mass of nitrogen dioxide. An answer that is orders of magnitude
different would immediately be suspect.
SKILLBUILDER 8.3 Mass-to-Mass Conversions
Another component of acid rain is sulfuric acid, which forms when SO2, also a pollutant, reacts with oxygen
and rainwater according to the following reaction.
2 SO2(g) + O2(g) + 2 H2O(l)  2 H2SO4(aq)
Assuming that there is more than enough O2 and H2O, how much in kilograms forms from
H2SO4 2.6  103 kg of SO2?
Answer: 4.0  103 kg H2SO4
For More Practice Problems 35, 36, 37, 38.
Administrative Stuff

Silence your electronic devices and make sure your
clickers are ready.

Chapter 7 Homework is due TODAY at 5pm.

Continue learning the polyatomic ions and solubility rules
from chapters 5 and 6.

Exam 2 is 10/14/13 and covers Chapters 6, 7, 8, and 9.

We have 5 classes between now and then including today.
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8.5 More Pancakes: Limiting Reactant,
Theoretical Yield, and Percent Yield
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
Suppose we have 3 cups flour, 10 eggs, and 4 tsp baking
powder.
How many pancakes can we make?
We have enough flour for 15 pancakes, enough eggs for
25 pancakes, and enough baking powder for 40 pancakes.
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8.5 More Pancakes: Limiting Reactant,
Theoretical Yield, and Percent Yield
If this were a chemical reaction, the flour would be the
limiting reactant and 15 pancakes would be the
theoretical yield.
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8.5 More Pancakes: Limiting Reactant,
Theoretical Yield, and Percent Yield
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Suppose we cook our pancakes. We accidentally
burn three of them and one falls on the floor.
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So even though we had enough flour for 15
pancakes, we finished with only 11 pancakes.
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If this were a chemical reaction, the 11 pancakes
would be our actual yield, the amount of
product actually produced by a chemical reaction.
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8.5 More Pancakes: Limiting Reactant,
Theoretical Yield, and Percent Yield

Our percent yield, the percentage of the theoretical yield
that was actually attained, is:
Since four of the pancakes were ruined,
we got only 73% of our theoretical yield.
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Actual Yield and Percent Yield

The actual yield of a chemical reaction must be determined
experimentally and depends on the reaction conditions.

The actual yield is almost always less than 100%.

Some of the product does not form.

Product is lost in the process of recovering it.
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Limiting Reactant, Theoretical Yield,
Actual Yield, and Percent Yield
To summarize:
• Limiting reactant (or limiting reagent)—the
reactant that is completely consumed in a chemical
reaction.
• Theoretical yield—the amount of product that can be
made in a chemical reaction based on the amount of
limiting reactant.
• Actual yield—the amount of product actually produced
by a chemical reaction.
• Percent yield—(actual yield/theoretical yield)*100%.
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Example: Ti(s) + 2 Cl2(g)  TiCl4(s)
Given (moles): 1.8 mol Ti and 3.2 mol Cl2
Find: limiting reactant and theoretical yield
SOLUTION MAP:
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Example: Ti(s) + 2 Cl2(g)  TiCl4(s)
Given (moles): 1.8 mol Ti and 3.2 mol Cl2
Find: limiting reactant and theoretical yield
SOLUTION:
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Limiting Reactant, Theoretical Yield,
Actual Yield, and Percent Yield

In many industrial applications, the more costly reactant
or the reactant that is most difficult to remove from the
product mixture is chosen to be the limiting reactant.

When working in the laboratory, we measure the
amounts of reactants in grams.

To find limiting reactants and theoretical yields from initial
masses, we must add two steps to our calculations.
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Example: Na(s) + Cl2(g)  2 NaCl(s)
Given (grams): 53.2 g Na and 65.8 g Cl2
Find: limiting reactant and theoretical yield
SOLUTION MAP:
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Example: Na(s) + Cl2(g)  2 NaCl(s)
Given (grams): 53.2 g Na and 65.8 g Cl2
Find: limiting reactant and theoretical yield
SOLUTION:
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Example: Na(s) + Cl2(g)  2 NaCl(s)
Given (grams): actual yield 86.4 g NaCl
Find: percent yield

The actual yield is usually less than the theoretical
yield because at least a small amount of product is
lost or does not form during a reaction.
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Example 8.6 Finding Limiting Reactant, Theoretical Yield,
and Percent Yield
Consider the reaction:
Cu2O(s) + C(s)  2 Cu(s) + CO(g)
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.
SORT
You are given the mass of the reactants, carbon and
copper(I) oxide, as well as the mass of copper formed
by the reaction. You are asked to find the limiting
reactant, theoretical yield, and percent yield.
GIVEN: 11.5 g C
114.5 g Cu2O
87.4 g Cu produced
FIND:
limiting reactant
theoretical yield
percent yield
Example 8.6 Finding Limiting Reactant, Theoretical Yield,
and Percent Yield
Continued
STRATEGIZE
SOLUTION MAP
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.
The main conversion factors are the stoichiometric
relationships between moles of each reactant and moles
of copper. The other conversion factors are the molar
masses of copper(I) oxide, carbon, and copper.
RELATIONSHIPS USED
1 mol Cu2O: 2 mol Cu
1 mol C : 2 mol Cu
Molar mass Cu = 63.55 g/mol
Molar mass C = 12.01 g/mol
Molar mass Cu2O = 143.10 g/mol
Example 8.6 Finding Limiting Reactant, Theoretical Yield,
and Percent Yield
Continued
SOLVE
SOLUTION
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.
Since Cu2O makes the least amount of product, Cu2O is
the limiting reactant. The theoretical yield is then 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%.
CHECK
Are the units correct? Does the answer make
physical sense?
The theoretical yield has the right units (g Cu). The
magnitude of the theoretical yield seems reasonable
because it is of the same order of magnitude as the
given masses of C and Cu2O. The theoretical yield is
reasonable because it is less than 100%. Any calculated
theoretical yield above 100% would be suspect.
Example 8.6 Finding Limiting Reactant, Theoretical Yield,
and Percent Yield
Continued
SKILLBUILDER 8.6 Finding Limiting Reactant, Theoretical Yield, and Percent Yield
This reaction is used to obtain iron from iron ore:
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
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.
Answer: Limiting reactant is CO;
theoretical yield = 127 g Fe;
percent yield = 68.8%
For More Practice Example 8.10; Problems 61, 62, 63, 64, 65, 66.
8.7 Enthalpy: A Measure of the Heat
Evolved or Absorbed in a Reaction

Chemical reactions can be exothermic (they emit thermal
energy when they occur).

Chemical reactions can be endothermic (they absorb
thermal energy when they occur).

The amount of thermal energy emitted or absorbed by a
chemical reaction, under conditions of constant pressure
(which are common for most everyday reactions), can be
quantified with a function called enthalpy.
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8.7 Enthalpy: A Measure of the Heat
Evolved or Absorbed in a Reaction

We define the enthalpy of reaction, ΔHrxn, as the amount of
thermal energy (or heat) that flows when a reaction occurs at
constant pressure.
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The sign of ΔHrxn (positive or negative) depends on the
direction in which thermal energy flows when the reaction
occurs.

Energy flowing out of the chemical system is like a withdrawal
and carries a negative sign.

Energy flowing into the system is like a deposit and carries a
positive sign.
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Figure 8.3 Exothermic and
Endothermic reactions

(a) In an exothermic reaction, energy is released into the
surroundings. (b) In an endothermic reaction, energy is
absorbed from the surroundings.
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Sign of ΔHrxn
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When thermal energy flows out of the reaction and
into the surroundings (as in an exothermic reaction),
then ΔHrxn is negative.
The enthalpy of reaction for the combustion of CH4,
the main component in natural gas:
This reaction is exothermic and therefore has a
negative enthalpy of reaction.
The magnitude of ΔHrxn tells us that 802.3 kJ of heat are
emitted when 1 mol CH4 reacts with 2 mol O2.
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Sign of ΔHrxn
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When thermal energy flows into the reaction and out
of the surroundings (as in an endothermic reaction),
then ΔHrxn is positive.
The enthalpy of reaction for the reaction between
nitrogen and oxygen gas to form nitrogen monoxide:
This reaction is endothermic and therefore has a
positive enthalpy of reaction.
The magnitude of ΔHrxn tells us that 182.6 kJ of heat are
absorbed from the surroundings when 1 mol N2 reacts
with 1 mol O2.
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Stoichiometry of ΔHrxn

The amount of heat emitted or absorbed when a
chemical reaction occurs depends on the amounts of
reactants that actually react.

We usually specify ΔHrxn in combination with the
balanced chemical equation for the reaction.

The magnitude of ΔHrxn is for the stoichiometric amounts
of reactants and products for the reaction as written.
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Stoichiometry of ΔHrxn

For example, the balanced equation and ΔHrxn for the
combustion of propane (the fuel used in LP gas) is:

When 1 mole of C3H8 reacts with 5 moles of O2 to
form 3 moles of CO2 and 4 moles of H2O, 2044 kJ of
heat are emitted.
These ratios can be used to construct conversion
factors between amounts of reactants or products and
the quantity of heat exchanged.

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Stoichiometry of ΔHrxn

To find out how much heat is emitted upon the
combustion of a certain mass in grams of propane C3H8,
we can use the following solution map:
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Example 8.7 Stoichiometry
Involving ΔHrxn

An LP gas tank in a home barbecue contains 11.8 x 103 g
of propane (C3H8).

Calculate the heat (in kJ) associated with the complete
combustion of all of the propane in the tank.
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Example: Complete combustion of
11.8 x 103 g of propane (C3H8)
SOLUTION MAP:
RELATIONSHIPS USED:
1 mol C3H8 : –2044 kJ (from balanced equation)
Molar mass C3H8 = 44.11 g/mol
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Example: Complete combustion of
11.8 x 103 g of propane (C3H8)
SOLUTION:
Often in the homework, the absolute
value of Q, |Q|, is requested and words
are used to convey the sign of the heat
absorbed or given off in the reaction.
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Everyday Chemistry
Bunsen Burners

Most Bunsen burners have a mechanism to adjust the
amount of air (and therefore of oxygen) that is mixed with
the methane.

If you light the burner with the air completely closed off, you
get a yellow, smoky flame that is not very hot.

As you increase the amount of air going into the burner, the
flame becomes bluer, less smoky, and hotter.

When you reach the optimum adjustment, the flame has a
sharp, inner blue triangle, gives off no smoke, and is hot
enough to melt glass easily.

Continuing to increase the air beyond this point causes the
flame to become cooler again and may actually extinguish it.
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A Bunsen burner at various stages
of air intake adjustment.
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