Transcript (a) (b) - Haverford Alchemy

Worked Example 5.1 Balancing Chemical Reactions
Use words to explain the following equation for the reaction used in extracting lead metal from its ores. Show that
the equation is balanced.
Solution
The equation can be read as, “Solid lead (II) sulfide plus gaseous oxygen yields solid lead (II) oxide plus gaseous
sulfur dioxide.” To show that the equation is balanced, count the atoms of each element on each side of the arrow:
On the left:
2 Pb
2S
On the right:
2 Pb
2S
The numbers of atoms of each element are the same in the reactants and products, so the equation is balanced.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.2 Balancing Chemical Equations
Write a balanced chemical equation for the Haber process, an important industrial reaction in which elemental
nitrogen and hydrogen combine to form ammonia.
Solution
STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products.
By examination, we see that only two elements, N and H, need to be balanced. Both these elements exist in
nature as diatomic gases, as indicated on the reactant side of the unbalanced equation.
STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Remember that the
subscript 2 in and
indicates that these are diatomic molecules (that is, 2 N atoms or 2 H atoms per
molecule). Since there are 2 nitrogen atoms on the left, we must add a coefficient of 2 in front of the
on the
right side of the equation to balance the equation with respect to N:
Now we see that there are 2 H atoms on the left, but 6 H atoms on the right. We can balance the
equation with respect to hydrogen by adding a coefficient of 3 in front of the (g) on the left side:
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
STEP 3: Check the equation to make sure the numbers and kinds of atoms on both
sides of the equation are the same.
On the left:
(1 × 2) N = 2 N
(3 × 2) H = 6 H
On the right:
(2 × 1) N = 2 N
(2 × 3) H = 6 H
STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, the coefficients
already represent the lowest whole-number values.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.3 Balancing Chemical Equations
Natural gas (methane,
for the reaction.
) burns in oxygen to yield water and carbon dioxide
. Write a balanced equation
Solution
STEP 1: Write the unbalanced equation, using correct formulas for all substances:
(Unbalanced)
STEP 2: Since carbon appears in one formula on each side of the arrow, let us begin with that element. In fact, there
is only 1 carbon atom in each formula, so the equation is already balanced for that element. Next, note that there are
4 hydrogen atoms on the left (in
) and only 2 on the right (in
). Placing a coefficient of 2 before
gives the same number of hydrogen atoms on both sides:
(Balanced for C and H)
Finally, look at the number of oxygen atoms. There are 2 on the left (in
each
). If we place a 2 before the
) but 4 on the right (2 in
and 1 in
, the number of oxygen atoms will be the same on both sides, but the
numbers of other elements will not change:
(Balanced for C, H, and O)
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
STEP 3: Check to be sure the numbers of atoms on both sides are the same.
On the left:
1C
On the right:
1C
4H
STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, the answer is
already correct.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.4 Balancing Chemical Equations
Sodium chlorate
decomposes when heated to yield sodium chloride and oxygen, a reaction used to
provide oxygen for the emergency breathing masks in airliners. Write a balanced equation for this reaction.
Solution
STEP 1: The unbalanced equation is:
STEP 2: Both the Na and the Cl are already balanced, with only one
atom of each on the left and right sides of the equation. There are
3 O atoms on the left, but only 2 on the right. The O atoms can be
balanced by placing a coefficient of 1½ in front of
on the right side
of the equation:
The oxygen in emergency breathing masks comes
from heating sodium chlorate.
STEP 3: Checking to make sure the same number of atoms of each type occurs on both sides of the
equation, we see 1 atom of Na and Cl on both sides, and 3 O atoms on both sides.
STEP 4: In this case, obtaining all coefficients in their smallest whole-number values requires that we
multiply all coefficients by 2 to obtain:
Checking gives
On the left:
On the right:
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.5 Classifying Chemical Reactions
Classify the following as a precipitation, an acid–base neutralization, or a redox reaction.
(a)
(b)
(c)
Analysis One way to identify the class of reaction is to examine the products that form and match them with
the descriptions for the types of reactions provided in this section. By a process of elimination, we
can readily identify the appropriate reaction classification.
Solution
(a) The products of this reaction are water and an ionic compound, or salt
with the description of an acid–base neutralization reaction.
. This is consistent
(b) This reaction involves two aqueous reactants,
and NaCl, which combine to form a
solid product,
. This is consistent with a precipitation reaction.
(c) The products of this reaction are a solid, Ag (s), and an aqueous ionic compound,
.
This does not match the description of a neutralization reaction, which would form water and an
ionic compound. One of the products is a solid, but the reactants are not both aqueous compound;
one of the reactants is also a solid (Cu). Therefore, this reaction would not be classified as a
precipitation reaction. By the process of elimination, then, it must be a redox reaction.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.6
Chemical Reactions: Solubility Rules
Will a precipitation reaction occur when aqueous solutions of
Solution
and
are mixed?
Identify the two potential products, and predict the solubility of each using the guidelines in the
text. In this instance,
and
might give CdS and
. Since the guidelines predict
that CdS is insoluble, a precipitation reaction will occur:
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.7
Chemical Reactions: Acid–Base Neutralization
Write an equation for the neutralization reaction of aqueous
Solution
and aqueous
The reaction of
with
involves the combination of a proton
from the base to yield water and a salt
.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
.
from the acid with
© 2013 Pearson Education, Inc.
Worked Example 5.8 Chemical Reactions: Redox Reactions
For the following reactions, indicate which atom is oxidized and which is reduced, based on the definitions provided
in this section. Identify the oxidizing and reducing agents.
(a)
(b)
Analysis The definitions for oxidation include a loss of electrons, an increase in charge, and a gain of oxygen
atoms; reduction is defined as a gain of electrons, a decrease in charge, and a loss of oxygen atoms.
Solution
(a) In this reaction, the charge on the Cu atom increases from 0 to 2+. This corresponds to a loss of
2 electrons. The Cu is therefore oxidized and acts as the reducing agent. Conversely, the
ion
undergoes a decrease in charge from 2+ to 0, corresponding to a gain of 2 electrons for the
ion. The
is reduced, and acts as the oxidizing agent.
(b) In this case, the gain or loss of oxygen atoms is the easiest way to identify which atoms
are oxidized and reduced. The Mg atom is gaining oxygen to form MgO; therefore, the Mg is being
oxidized and acts as the reducing agent. The C atom in
is losing oxygen. Therefore, the C
atom in
is being reduced, and so
acts as the oxidizing agent.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.9 Chemical Reactions: Identifying Oxidizing/Reducing Agents
For the respiration and metallurgy examples discussed previously, identify the atoms being oxidized and reduced,
and label the oxidizing and reducing agents.
Analysis
Solution
Again, using the definitions of oxidation and reduction provided in this section, we can determine
which atom (s) are gaining/losing electrons or gaining/losing oxygen atoms.
Respiration:
Because the charge associated with the individual atoms is not evident, we will use the definition of
oxidation/reduction as the gaining/losing of oxygen atoms. In this reaction, there is only one
reactant besides oxygen
, so we must determine which atom in the compound is
changing. The ratio of carbon to oxygen in
is 1:1, while the ratio in
is 1:2. Therefore,
the C atoms are gaining oxygen and are oxidized; the
is the reducing agent and
is the
oxidizing agent. Note that the ratio of hydrogen to oxygen in
and in
is 2:1. The H
atoms are neither oxidized nor reduced.
Metallurgy:
The
is losing oxygen to form Fe (s); it is being reduced and acts as the oxidizing agent. In
contrast, the CO is gaining oxygen to form
; it is being oxidized and acts as the reducing agent.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.10 Chemical Reactions: Identifying Redox Reactions
For the following reactions, identify the atom (s) being oxidized and reduced:
(a)
(b)
Analysis Again, there is no obvious increase or decrease in charge to indicate a gain or loss of electrons.
Also, the reactions do not involve a gain or loss of oxygen. We can, however, evaluate the reactions
in terms of the typical behavior of metals and nonmetals in reactions.
Solution
(a) In this case, we have the reaction of a metal (Al) with a nonmetal ( ). Because metals tend to
lose electrons and nonmetals tend to gain electrons, we can assume that the Al atom is oxidized
(loses electrons) and the
is reduced (gains electrons).
(b) The carbon atom is the less electronegative element (farther to the left) and is less likely to gain
an electron. The more electronegative element (Cl) will tend to gain electrons (be reduced).
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.11 Redox Reactions: Oxidation Numbers
What is the oxidation number of the titanium atom in
(Section 3.10).
? Name the compound using a Roman numeral
Solution
Chlorine, a reactive nonmetal, is more electronegative than titanium and has an oxidation number of –1. Because
there are 4 chlorine atoms in
, the oxidation number of titanium must be +4. The compound is named
titanium (IV) chloride. Note that the Roman numeral IV in the name of this molecular compound refers to the
oxidation number +4 rather than to a true ionic charge.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.12 Redox Reactions: Identifying Redox Reactions
Use oxidation numbers to show that the production of iron metal from its ore (
) by reaction with charcoal (C)
is a redox reaction. Which reactant has been oxidized, and which has been reduced? Which reactant is the oxidizing
agent, and which is the reducing agent?
Solution
The idea is to assign oxidation numbers to both reactants and products and see if there has been a change. In the
production of iron from
, the oxidation number of Fe changes from +3 to 0, and the oxidation number of C
changes from 0 to +4. Iron has thus been reduced (decrease in oxidation number), and carbon has been oxidized
(increase in oxidation number). Oxygen is neither oxidized nor reduced because its oxidation number does not
change. Carbon is the reducing agent, and
is the oxidizing agent.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
Worked Example 5.13 Chemical Reactions: Net Ionic Reactions
Write balanced net ionic equations for the following reactions:
(a)
(b)
(c)
Solution
(a) The solubility guidelines discussed in Section 5.4 predict that a precipitate of insoluble AgCl
forms when aqueous solutions of
and
are mixed. Writing all the ions separately gives an
ionic equation, and eliminating spectator ions
and
gives the net ionic equation.
Ionic equation:
Net ionic equation:
The coefficients can all be divided by 2 to give:
Net ionic equation:
A check shows that the equation is balanced for atoms and charge (zero on each side).
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.
(b) Allowing the acid HCl to react with the base
leads to a neutralization reaction. Writing the ions
separately, and remembering to write a complete formula for water, gives an ionic equation. Then eliminating the
spectator ions and dividing the coefficients by 2 gives the net ionic equation.
Ionic equation:
Net ionic equation:
A check shows that atoms and charges are the same on both sides of the equation.
Fundamentals of General, Organic, and Biological Chemistry, 7e
John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson
© 2013 Pearson Education, Inc.