chapter 5 – chemical reactions and equationschemweb/chem1000... · chapter 5 – chemical...

51

Chapter 5 – Chemical Reactions and Equations

5.1 (a) single-displacement reaction; (b) anhydrous; (c) molecular equation; (d) decomposition reaction;

(e) balanced equation; (f) reactant; (g) spectator ion; (h) combustion reaction; (i) precipitate

5.2 (a) combination reaction; (b) activity series; (c) chemical equation; (d) product; (e) ionic equation;

(f) neutralization reaction; (g) precipitation reaction; (h) net ionic equation; (i) double-displacement

reaction

5.3 Substances that are formed as a result of the reaction are the products. This reaction produces solid

aluminum oxide, Al2O3(s). You should pay attention to the substance formed and the state of the material

if it is given. The reactants are the substances used to form the products. The reactants are aluminum

metal, Al(s), and oxygen gas, O2(g) (i.e. “Aluminum metal reacts with oxygen gas”).

5.4 Substances that are formed as a result of the reaction are the products. Nitrogen gas and solid iodine, N2(g)

and I2(s), and energy are the reaction products (i.e. “Nitrogen gas and solid iodine form in an explosive

reaction”). You should pay attention to the substance formed and the state of the material if it is given.

The reactants are the substances used to form the products. The reactant is nitrogen triiodide, NI3 (i.e.

“when nitrogen triiodide is detonated”).

5.5 Image A represents the reactants because it contains both fluorine and xenon. Images B-D show product

molecules, but only image C represents the products. In a chemical reaction, mass must be conserved. In

image A there are two xenon atoms and eight fluorine atoms. In order for mass to be conserved, the same

number of each type of atom must be present in the products.

5.6 Make sure you look carefully at the images to distinguish between the chemical formulas for ozone (three

red spheres) and oxygen (two red spheres). Image C represents the reactants because it contains both ozone

and carbon monoxide. The products are represented by image D. Images A and B represent mixtures of

reactants and products.

5.7 Mass must be conserved in a chemical reaction. In order for mass to be conserved, the same number of

each type of atom must be present in both the products and reactants. The image is not accurate because

mass is not conserved. To fix the image, you first need to count the number of each type of atom. Here’s a

summary of the atoms present in the reactants and products:

Atom Reactants Products

H 10 12

N 4 4

The numbers of hydrogen atoms do not match in the reactants and products. If you add one hydrogen

molecule to the reactant image, the numbers will balance and the molecular-level diagram will be accurate.

5.8 Mass must be conserved in a chemical reaction. In order for mass to be conserved, the same number of

each type of atom must be present in both the products and reactants. The image is not accurate because

mass is not conserved. To fix the image, you first need to count the number of each type of atom. Here’s a

summary of the atoms present in the reactants and products:


H 10 10

O 6 5

52

The numbers of oxygen atoms do not match in the reactants and products. If you add one hydrogen

molecule to the reactant image and one molecule of water (H2O) to the product image, the numbers will

balance and the molecular-level diagram will be accurate.


H 12 12

O 6 6

5.9 The law of conservation of mass is obeyed when the same number of each type of atom is present in both

the products and reactants. In the product image you could have 5 molecules and one unreacted atom.

5.10 The law of conservation of mass is obeyed when the same number of each type of atom is present in both

the products and reactants. In the product image you could have 1 unreacted molecule and 5 unreacted

atoms. Two molecules of product should be shown.

5.11 One oxygen molecule (two oxygen atoms) and two sulfur dioxide molecules must react for every two

sulfur trioxide molecules (SO3) produced. Since there are four sulfur dioxide molecules, both oxygen

molecules will react. This means that none of the reactants remain after the reaction, and we should show

four sulfur trioxide molecules as the reaction products.

53

5.12 One oxygen molecule (two oxygen atoms) and two NO molecules must react for every two NO2 produced.

Since there are six NO molecules, all of the oxygen molecules will react. We should show six NO2

molecules as the reaction products.

5.13 Three signs that a chemical reaction is occurring are evident: (1) a brown gas is forming, (2) bubbles are

being produced, and (3) the solution is changing color (assuming that the liquid was not green to start

with).

5.14 Two signs that a chemical reaction is occurring are evident: (1) the color of the mixture is different than the

color of either solution and (2) a new substance (the yellow solid) is forming. You can tell a solid is

formed because the solution becomes opaque (i.e. you can’t see through it).

5.15 We are starting and ending with CO2, so no new substances are formed. A chemical reaction has not taken

place. What we are observing is a change in the physical state of CO2.

5.16 There are several signs that a chemical reaction has taken place: the formation of “a blue solution” indicates

that a color change occurred. In addition, formation of a “black deposit” indicates the production of a new

substance.

5.17 There is a rearrangement of the atoms into new substances. Since new substances are formed, a chemical

reaction has taken place.

5.18 In the image on the left, the molecules are composed of one red and two white spheres, just as they are in

the image on the right. When a chemical reaction takes place new substances are formed. A chemical

reaction is not occurring here, because no new substance is formed.

5.19 The solution shown on the left has the spheres placed closer together, while in the solution shown on the

right the spheres have moved farther apart. Since no new chemical substances have formed, a chemical

reaction has not taken place. This is what you would expect to see in dilution.

5.20 A rearrangement of the atoms in a substance is an indication that a chemical reaction has taken place.

Since the molecules have different shapes in the reactant and product images, we conclude that a chemical

reaction has occurred.

5.21 A chemical equation is a chemist’s shorthand way of showing what happens during a chemical reaction. It

identifies the formulas of the reactants and products and demonstrates how mass is conserved during the

reaction. You can think of a chemical equation as a chemist’s recipe for making new substances. The

reactants are the ingredients (i.e. melted chocolate or solid chocolate chips) and the products are what are

produced by the reaction (i.e. fudge or cookies). Like a recipe, the chemical equation indicates the states of

the materials and the quantities of substances that are used (the reactants) and produced (the products).

5.22 A chemical reaction is any process that produces a new substance. A chemical equation is the shorthand

that chemists use to describe that reaction. The chemical equation includes the quantities and physical

states of the substances used (the reactants) and produced (the products) in the ideal reaction.

54

5.23 To decide if a reaction has taken place, we have to determine whether a new substance has been formed.

(a) This equation represents a chemical reaction. Carbonic acid, H2CO3(aq), has been formed from the

reaction of carbon dioxide (CO2) and water (H2O).

(b) This equation represents a physical change. The chemical compositions of solid water (ice) and liquid

water are the same, so a chemical reaction did not occur.

(c) This equation represents a chemical reaction. Even though the same atoms are present in the reactant

and product, the order in which they are connected is different so the substances are different

compounds.

5.24 (a) This equation represents a chemical reaction. New substances, TiO2(s) and HCl(aq), are formed from

TiCl4 and H2O.

(b) This equation represents a chemical reaction. Solid Li2O forms from the elements Li and O2.

(c) The process of dissolving a salt in water is usually considered a physical change. Salts are composed

of ions (for example, KBr(s) is actually a collection of K+ and Br

ions). When a salt dissolves, the

ions are separated from each other by water molecules. However, the structure of the ions does not

change.

5.25 Balancing a chemical equation demonstrates how mass is conserved during a reaction. A balanced

equation is a quantitative tool for determining the amounts of reactant(s) used and of product(s) produced.

5.26 Chemical equations are descriptions of particular chemical reactions. Changing subscripts in the formulas

of reactant or product species changes the identities of the substances involved in the reaction the equation

is describing.

5.27 In balancing the equation, we add coefficients so that the number of each type of atom is the same on the

reactant and product sides of the equation.

(a) To start with, we write the skeletal equation for the reaction:

NaH(s) + H2O(l) H2(g) + NaOH(aq)

When we count the number of each type of atom in the reactants and products, we find that this

equation is already balanced.

Atoms in Reactants Atoms in Products

1 Na 3 H 1 O 1 Na 3 H 1 O

(b) We start by writing the skeletal equation:

Al(s) + Cl2(g) AlCl3(s) Unbalanced

We count the number of each type of atom on each side of the unbalanced equation to determine where

to start balancing the equation.


1 Al 2 Cl 1 Al 3 Cl

The chlorines are not balanced. To balance, we need to find a common factor. The smallest factor

would be 6. We use coefficients of 3 and 2 on Cl2(g) and AlCl3(s), respectively

Al(s) + 3Cl2(g) 2AlCl3(s)


1 Al 6 Cl 2 Al 6 Cl

Now we balance the Al on the reactant side, using a coefficient of 2.

2Al(s) + 3Cl2(g) 2AlCl3(s)

55


2 Al 6 Cl 2 Al 6 Cl

Since the number of each type of atom on the reactant and product sides of the equation is the same,

the equation is balanced.

2Al(s) + 3Cl2(g) 2AlCl3(s) Balanced

5.28 In balancing the equation, we add coefficients so that the number of each type of atom is the same on the

reactant and product sides of the equation.

(a) We start by writing the skeletal equation:

Na(s) + O2(g) Na2O(s) Unbalanced




1 Na 2 O 2 Na 1 O

You can start by balancing either the oxygen or the sodium. If you start with oxygen, you need a

coefficient of 2 in front of the Na2O.

Na(s) + O2(g) 2Na2O(s) Unbalanced


1 Na 2 O 4 Na 2 O

Since there are four sodium atoms on the right, we use a coefficient of 4 with Na:

4Na(s) + O2(g) 2Na2O(s)


4 Na 2 O 4 Na 2 O



4Na(s) + O2(g) 2Na2O(s) Balanced


Cu(NO3)2(s) CuO(s) + NO2(g) + O2(g) Unbalanced




1 Cu 2 N 6 O 1 Cu 1 N 5 O

Since copper is already balanced, we need to start with either the nitrogen or the oxygen. It is usually

best to start by balancing the atoms that appear in the smallest number of compounds. In this case that

is nitrogen. We balance the nitrogen by placing a coefficient of 2 in front of NO2.

Cu(NO3)2(s) CuO(s) + 2NO2(g) + O2(g)

When you count atoms, don’t forget that you have also increased the number of oxygen atoms in the

product at the same time you increased the number of nitrogen atoms.


1 Cu 2 N 6 O 1 Cu 2 N 7 O

56

Since there are not enough oxygen atoms on the reactant side (and there is only one reactant) we need

to add a coefficient of 2 in front of the Cu(NO3)2(s). We will also have to increase the coefficient for

NO2 to keep the number of nitrogen atoms balanced.

2 Cu(NO3)2(s) CuO(s) + 4 NO2(g) + O2(g)


2 Cu 4 N 12 O 1 Cu 4 N 11 O

The last step is to balance number of copper and oxygen atoms:

2 Cu(NO3)2(s) 2 CuO(s) + 4 NO2(g) + O2(g)


2 Cu 4 N 12 O 2 Cu 4 N 12 O



2 Cu(NO3)2(s) 2 CuO(s) + 4 NO2(g) + O2(g) Balanced

5.29 There are two nitrogen molecules (N2) and six chlorine molecules (Cl2) in the reactant image, and 4 NCl3

molecules in the product image. The large spacing between the molecules allows us to assume that the

reactants and products are in the gas state. The balanced chemical equation is:

2N2(g) + 6Cl2(g) 4NCl3(g)

The coefficients of a balanced equation should be written as the smallest possible whole numbers. Since

each of the coefficients is divisible by 2, we reduce them by dividing each by 2.

N2(g) + 3Cl2(g) 2NCl3(g) Balanced

5.30 There are three methane molecules (CH4) and six oxygen molecules (O2) in the reactant image and three

carbon dioxide molecules (CO2) and six water (H2O) molecules in the product image. The large spacing

between the molecules allows us to assume the reactants and products are in the gas state. The balanced

chemical equation is:

3CH4(g) + 6O2(g) 3CO2(g) + 6H2O(g)

The coefficients of a balanced equation should be written as the smallest possible whole numbers. Since

each of the coefficients is divisible by 3, we reduce them by dividing each by 3.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Balanced

5.31 Magnesium is a solid and it burns in the presence of oxygen, O2(g), from the air. The product is also a

solid (indicated by the statement “ash-like substance”). Image B matches the reactants most closely

because it shows the correct states of materials in the reactants. Also, if we assume that magnesium oxide

is formed, the chemical formula for the product would be MgO(s). This also matches image B. The

skeletal equation for this reaction is:

Mg(s) + O2(g) MgO(s) Unbalanced


1 Mg 2 O 1 Mg 1 O

We add a coefficient of 2 to MgO to balance the oxygen:

Mg(s) + O2(g) 2MgO(s)


1 Mg 2 O 2 Mg 2 O

57

Then we add a coefficient of 2 to Mg to complete the balancing process:

2Mg(s) + O2(g) 2MgO(s)


2 Mg 2 O 2 Mg 2 O

2Mg(s) + O2(g) 2MgO(s) Balanced

5.32 Image A matches this description most closely. The water molecules (shown in the background of the

reactant and product images) are closely spaced and randomly oriented. This is what we would expect for a

substance in the liquid phase. Metallic sodium should be a solid with a well defined structure. The

products include aqueous sodium hydroxide, so we expect to find Na+ and OH

ions surrounded by water

molecules. Finally, we see several H2 molecules in the solution. The skeletal equation for the reaction is:

Na(s) + H2O(l) NaOH(aq) + H2(g) Unbalanced


1 Na 2 H 1 O 1 Na 3 H 1 O

It is easiest to balance this reaction using a fractional coefficient (1/2) for the hydrogen molecule.

Na(s) + H2O(l) NaOH(aq) + ½H2(g)


1 Na 2 H 1 O 1 Na 2 H 1 O

This technique is best applied when everything except atoms of one element (i.e. H2, O2, S8, etc.) is

balanced. By using a fractional coefficient, we can change the amount of one type of atom without

affecting the others. We remove the fractional coefficient in this case by multiplying all the coefficients by

2 (the denominator of the fraction) to remove the fraction.

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) Balanced

5.33 According to the balanced chemical equation for this reaction, for each molecule of H2 and I2 that are

consumed in the reaction, two HI molecules are produced. Since three H2 and three I2 molecules are

shown, we must show six molecules of HI in the product image of the molecular-level diagram.

5.34 According to the balanced chemical equation for this reaction, for every two molecules of NO and CO that

are consumed in the reaction, one molecule of N2 and two molecules of CO2 are produced. Since four NO

and four CO molecules are shown in the left image, we must show two molecules of N2 and four molecules

of CO2 in the product image.

58

5.35 Our molecular-level diagram should include two molecules of NO2 as the reactants and one molecule of

N2O4 as the products. The law of conservation of mass must be satisfied for the diagram to be accurate. In

each image there are two nitrogen atoms and four oxygen atoms.

5.36 Our molecular-level diagram should include two molecules of F2 and two molecules of H2O as the

reactants and four HF molecules and two O2 molecules as the products. The law of conservation of mass

must be obeyed for the diagram to be accurate. In each image there are four fluorine atoms, four hydrogen

atoms, and two oxygen atoms.

5.37 (a) We start by writing the skeletal equation:

Al(s) + Cl2(g) AlCl3(s) Unbalanced

Start by taking an inventory of atoms present on each side to help determine where to start balancing

the equation.


1 Al 2 Cl 1 Al 3 Cl

Only the chlorines are not balanced. To balance, we need to find a common factor for 2 and 3. The

smallest factor would be 6. We use coefficients of 3 and 2 on Cl2(g) and AlCl3(s), respectively

Al(s) + 3Cl2(g) 2AlCl3(s)


1 Al 6 Cl 2 Al 6 Cl

We recheck our inventory and find that we need to balance the Al on the reactant side using a

coefficient of 2.

2Al(s) + 3Cl2(g) 2AlCl3(s)


2 Al 6 Cl 2 Al 6 Cl



2Al(s) + 3Cl2(g) 2AlCl3(s) Balanced


59

Pb(NO3)2(aq) + K2CrO4(aq) PbCrO4(s) + KNO3(aq) Unbalanced

In the inventory it is perferable to keep polyatomic ions together as groups. This works when the

polyatomic ions are not changed in the reaction:

Reactants Products

1 Pb 2 NO3 2 K 1 CrO4 1 Pb 1 NO3 1 K 1 CrO4

Start by balancing nitrates or potassium ions. Using a two in front of KNO3 balances both potassiums

and nitrates!

Pb(NO3)2(aq) + K2CrO4(aq) PbCrO4(s) + 2KNO3(aq)

Reactants Products

1 Pb 2 NO3 2 K 1 CrO4 1 Pb 1 NO3 2 K 1 CrO4

(c) We start by writing the skeletal equation:

Li(s) + H2O(l) LiOH(aq) + H2(g) Unbalanced


1 Li 2 H 1 O 1 Li 3 H 1 O

Notice that the hydrogens are the only atoms not balanced, but on the product side, the hydrogens

appear in two different substances. If we change the number of H2 and H2O molecules, the reaction

will continue to be unbalanced. The solution is to use a coefficient of 2 on the H2O and see where it

might lead you.

Li(s) + 2H2O(l) LiOH(aq) + H2(g) Unbalanced


1 Li 4 H 2 O 1 Li 3 H 1 O

Next balance the oxygens:

Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) Unbalanced


1 Li 4 H 2 O 2 Li 4 H 2 O

The last step is to balance the lithium atoms and check.

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) Balanced


2 Li 4 H 2 O 2 Li 4 H 2 O

(d) We start by writing the skeletal equation:

C6H14(g) + O2(g) CO2(g) + H2O(g) Unbalanced


6C 14H 2O 1C 2H 3O

510

In the combustion reactions of hydrocarbons it is usually easiest to begin by balancing the carbon and

leave the balancing of oxygen for the last step. This follows the general concepts of balancing the

atoms that appear in the smallest number of substances first (you could have also started with

hydrogen) and balancing the substances appearing as pure elements last.

C6H14(g) + O2(g) 6CO2(g) + H2O(g) Unbalanced


6C 14H 2O 6C 2H 13O

C6H14(g) + O2(g) 6CO2(g) + 7H2O(g) Unbalanced


6C 14H 2O 6C 14H 19O

To balance the oxygen, we use a coefficient of 19/2 for O2:

C6H14(g) + 19/2 O2(g) 6CO2(g) + 7H2O(g)

While this equation is balanced, we need to remove the fraction by multiplying the entire equation by a

factor of two.

2C6H14(g) + 19 O2(g) 12CO2(g) + 14H2O(g) Balanced


12C 28H 38O 12C 28H 38O

5.38 (a) You are given the skeletal equation:

Mg(s) + O2(g) MgO(s) Unbalanced

Start by taking an inventory of atoms present on each side to help determine where to start balancing

the equation.


1 Mg 2 O 1 Mg 1 O

Since the oxygens are not balanced, we use a coefficient of 2 in front of MgO

Mg(s) + O2(g) 2MgO(s) Unbalanced


1 Mg 2 O 2 Mg 2 O

Coefficient of 2 on Mg balances the equation:

2Mg(s) + O2(g) 2MgO(s) Balanced


2 Mg 2 O 2 Mg 2 O

511

(b) Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + 2Ag(s) Balanced

Zn(s) + AgNO3(aq) Zn(NO3)2(aq) + Ag(s) Unbalanced

In the inventory it is perferable to keep polyatomic ions together as groups. This works when the

polyatomic ions are not changed in the reaction:

Reactants Products

1 Zn 1 Ag 1 NO3 1 Zn 1 Ag 2 NO3

Since the nitrates are not balanced, we start by using a coefficient of 2 in front of AgNO3

Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + Ag(s)

Reactants Products

1 Zn 2 Ag 2 NO3 1 Zn 1 Ag 2 NO3

A coefficient in front of the Ag in the product finish the balancing process.

Zn(s) + 2AgNO3(aq) Zn(NO3)2(aq) + 2Ag(s)

Reactants Products

1 Zn 2 Ag 2 NO3 1 Zn 2 Ag 2 NO3

(c) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) Balanced

C2H2(g) + O2(g) CO2(g) + H2O(g) unbalanced


2 C 2 H 2 O 1 C 2 H 3 O

In the combustion reactions of hydrocarbons it is usually easiest to begin by balancing the carbon and

leave the balancing of oxygen for the last step. This follows the general concepts of balancing the

atoms that appear in the smallest number of substances first (you could have also started with

hydrogen) and balancing the substances appearing as pure elements last. Start with a coefficient of 2

in front of CO2.

C2H2(g) + O2(g) 2CO2(g) + H2O(g) unbalanced


2 C 2 H 2 O 2 C 2 H 5 O

To balance the oxygen, we use a coefficient of 5/2 for

C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(g) unbalanced


2 C 2 H 5 O 2 C 2 H 5 O

While this equation is balanced, we need to remove the fraction by multiplying the entire equation by a

factor of two.

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)


4 C 4 H 10 O 4 C 4 H 10 O

(d) C12H22O11(s) + 8KClO3(l) 12CO2(g) + 11H2O(g) + 8KCl(s) Balanced

C12H22O11(s) + KClO3(l) CO2(g) + H2O(g) + KCl(s) Unbalanced

Reactants Products

12C 22H 14O 1K 1Cl 1C 2H 3O 1K 1Cl

Start by balaning products with the reactant C12H22O11. As with hydrocarbons, balance C and H first:

C12H22O11(s) + KClO3(l) 12CO2(g) + 11H2O(g) + KCl(s)

Reactants Products

12C 22H 14O 1K 1Cl 12C 22H 35O 1K 1Cl

There are 35 oxygen atoms in the products and 11 in the reactants. Ignore the oxygens in KClO3

because it is unbalanced. This means that KClO3 needs to provide 24 oxgyen atoms. A coefficient of

eight will provide those oxygens.

512

C12H22O11(s) + 8KClO3(l) 12CO2(g) + 11H2O(g) + KCl(s)

Reactants Products

12C 22H 35O 8K 8Cl 12C 22H 35O 1K 1Cl

A coefficient of eight in front of KCl balances the equation:

C12H22O11(s) + 8KClO3(l) 12CO2(g) + 11H2O(g) + 8KCl(s)

Reactants Products

12C 22H 35O 8K 8Cl 12C 22H 35O 8K 8Cl

5.39 (a) We start by writing the skeletal equation:

CuCl2(aq) + AgNO3(aq) Cu(NO3)2(aq) + AgCl(s) Unbalanced

When balancing chemical equations that contain polyatomic ions which do not undergo chemical

change during the reaction, treat the ions as whole units in the same way that you would treat

individual atoms.

Reactants Products

1 Cu 2 Cl 1 Ag 1 NO3 1 Cu 1 Cl 1 Ag 1 NO3

You can balance this equation by starting with either Cl or NO3–.

CuCl2(aq) + AgNO3(aq) Cu(NO3)2(aq) + 2AgCl(s)

Reactants Products

1 Cu 2 Cl 1 Ag 1 NO3– 1 Cu 2 Cl 2 Ag 2 NO3

–

CuCl2(aq) + 2AgNO3(aq) Cu(NO3)2(aq) + 2AgCl(s) Balanced

Reactants Products

1 Cu 2 Cl 2 Ag 2 NO3– 1 Cu 2 Cl 2 Ag 2 NO3

–

Since the number of each type of atom on both sides of the equation is the same, the equation is

balanced.


S8(s) + O2(g) SO2(g) Unbalanced

Reactants Products

8 S 2 O 1 S 2 O

Begin by balancing S, because O already appears to be balanced.

S8(s) + O2(g) 8SO2(g)

Reactants Products

8 S 2 O 8 S 16 O

Add a coefficient of 8 for O2 to balance the equation.

S8(s) + 8O2(g) 8SO2(g) Balanced

Reactants Products

8 S 16 O 8 S 16 O


balanced.

513

(c) We start by writing the skeletal equation:


Reactants Products

3 C 8 H 2 O 1 C 2 H 3 O

In combustion reactions of hydrocarbons (hydrocarbon + molecular oxygen carbon dioxide +

water), we begin by balancing the carbon. This follows the general concept of balancing the atoms

that appear in the smallest number of substances first (you could have also started with hydrogen). We

add a coefficient of 3 for CO2.

C3H8(g) + O2(g) 3CO2(g) + H2O(g)

Reactants Products

3 C 8 H 2 O 3 C 2 H 7 O

Continue balancing all the other types of atoms in C3H8 before proceeding to balance the oxygen. We

add a coefficient of 4 for H2O.

C3H8(g) + O2(g) 3CO2(g) + 4H2O(g)

Reactants Products

3 C 8 H 2 O 3 C 8 H 10 O

Finally, add a coefficient of 5 for O2.

C3H8(l) + 5O2(g) 3CO2(g) + 4H2O(g) Balanced

Reactants Products

3 C 8 H 10 O 3 C 8 H 10 O


balanced.

5.40 (a) Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l) Balanced

Al2O3(s) + HCl(aq) AlCl3(aq) + H2O(l) Unbalanced

Reactants Products

2 Al 3 O

1 H 1 Cl

1 Al 1 O

2 H 1 Cl

We begin by balancing the Al with coefficient of 2 for AlCl3.

Al2O3(s) + HCl(aq) 2AlCl3(aq) + H2O(l)

Reactants Products

2 Al 3 O

1 H 1 Cl

2 Al 1 O

2 H 6 Cl

Continue balancing all the types of atoms in Al2O3 before proceeding to balance the other elements.

Al2O3(s) + HCl(aq) 2AlCl3(aq) + 3H2O(l)

Reactants Products

2 Al 3 O

1 H 1 Cl

2 Al 3 O

6 H 6 Cl

Balance the H and Cl by adding a coefficient of 6 for HCl.

514

Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l) Balanced

Reactants Products

2 Al 3 O

6 H 6 Cl

2 Al 3 O

6 H 6 Cl


balanced.

(b) PCl3(l) + 3AgF(s) PF3(g) + 3AgCl(s) Balanced

PCl3(l) + AgF(s) PF3(g) + AgCl(s) Unbalanced

Reactants Products

1 P 3 Cl

1 Ag 1 F

1 P 1 Cl

1 Ag 3 F

Begin by balancing either Cl or F.

PCl3(l) + AgF(s) PF3(g) + 3AgCl(s)

Reactants Products

1 P 3 Cl

1 Ag 1 F

1 P 3 Cl

3 Ag 3 F

PCl3(l) + 3AgF(s) PF3(g) + 3AgCl(s) Balanced

Reactants Products

1 P 3 Cl

3 Ag 3 F

1 P 3 Cl

3 Ag 3 F


balanced.

(c) 2NO2(g) 2NO(g) + O2(g) Balanced

NO2(g) NO(g) + O2(g) Unbalanced

Reactants Products

1 N 2 O 1 N 3 O

Since nitrogen is balanced, we have to start with O. Since we only need half of the oxygen provided

by O2, we use a coefficient of 1/2 for O2 to balance the equation.

NO2(g) NO(g) + 1/2O2(g)

Reactants Products

1 N 2 O 1 N 2 O

Since the number of each type of atom on both sides of the equation is the same, theoretically, the

reaction is balanced. However, the coefficients of a balanced equation need to be whole numbers so we

multiply all the coefficients by 2 to remove the fraction.

2NO2(g) 2NO(g) + O2(g) Balanced

Reactants Products

2 N 4 O 2 N 4 O

5.41 Copper and silver, Cu and Ag, are metals, so they, like virtually all the metals, occur in the solid state.

Silver nitrate is an ionic compound. The formulas for silver ion and nitrate ion are Ag+ and NO3

–,

respectively (see Figures 3.12 and 3.17 if you need help with the ion formulas). The formula for an

aqueous solution of silver nitrate is AgNO3(aq). From the ion name, copper(II), we know that the formula

515

for copper ion is Cu2+

. The formula for an aqueous solution of copper(II) nitrate is Cu(NO3)2(aq). The

unbalanced chemical equation is:

Cu (s) + AgNO3(aq) Cu(NO3)2(aq) + Ag(s) Unbalanced

When balancing chemical equations that contain polyatomic ions which do not undergo chemical change

during the reaction, treat the ions as whole units in the same way that you would treat individual atoms.

Reactants Products

1 Cu 1 Ag 1 NO3– 1 Cu 1 Ag 2 NO3

–

Balance NO3– by placing a coefficient of 2 in front of AgNO3.

Cu (s) + 2AgNO3(aq) Cu(NO3)2(aq) + Ag(s)

Reactants Products

1 Cu 2 Ag 2 NO3– 1 Cu 1 Ag 2 NO3

–

The equation is balanced when we add a coefficient of 2 for Ag(s).

Cu (s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) Balanced

5.42 The formulas for barium and chloride ions are Ba2+

and Cl, respectively. The formula for aqueous barium

chloride is BaCl2(aq), and the formulas for sulfuric acid and hydrochloric acid are H2SO4(aq) and HCl(aq),

respectively. Since the charges on barium and sulfate ions are of the same magnitude (2+ and 2-,

respectively), the formula of barium sulfate is BaSO4(s) (Table 5.3). See Figures 3.12 and 3.17 and the

nomenclature rules for acids in Chapter 3 if you need more help with these formulas. The unbalanced

equation is:

BaCl2(aq) + H2SO4(aq) BaSO4(s) + HCl(aq) Unbalanced

When balancing chemical equations that contain polyatomic ions which do not undergo chemical change

during the reaction, treat the ions as whole units in the same way that you would treat individual atoms.

Reactants Products

1 Ba 2 Cl 2 H 1 SO42–

1 Ba 1 Cl 1 H 1 SO42–

Begin by balancing Cl.

BaCl2(aq) + H2SO4(aq) BaSO4(s) + 2HCl(aq) Balanced

Reactants Products

1 Ba 2 Cl 2 H 1 SO42–

1 Ba 2 Cl 2 H 1 SO42–

Since the number of each type of atom is the same on both sides of the equation, the equation is balanced.

5.43 Table 5.1 lists the characteristics of the reactants and products for the different classifications of reactions.

Decomposition: Reactants: 1 compound; Products: 2 elements or smaller compounds

Combination: Reactants: 2 elements or compounds; Products: 1 compound

Single-displacement: Reactants: 1 element and 1 compound; Products: 1 element and 1 compound

Double-displacement: Reactants: 2 compounds; Products: 2 compounds

5.44 Table 5.1 lists the characteristics of the reactants and products for the different classifications of reactions.

There are many different examples that you could have picked (if you need ideas, look through the sections of

this chapter). As you select example reactions, make sure they match the descriptions given in Table 5.1.

Decomposition: Reactants: 1compound; Products: 2 elements or smaller compounds

H2CO3(aq) CO2(g) + H2O(l)

Combination: Reactants: 2 elements or compounds; Products: 1 compound

2H2(g) + O2(g) 2H2O(l)

516

Single-displacement: Reactants: 1 element and 1 compound; Products: 1 element and 1 compound

Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq)

Double-displacement: Reactants: 2 compounds; Products: 2 compounds

CuCl2(aq) + 2AgNO3(aq) Cu(NO3)2(aq) + 2AgCl(s)

5.45 For help classifying reactions, refer to Table 5.1. We can classify two of the types based on the number of

products and reactants in the chemical equation. In combination reactions, 2 substances combine to form 1

new substance; in decomposition reactions, 1 compound forms several new substances. Equations

indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If one of

the reactants and one of the products is an element, the reaction is single-displacement.

(a) combination

(b) single-displacement

(c) decomposition




indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If one of

the reactants and one of the products is an element, the reaction is single-displacement.

(a) decomposition

(b) double-displacement

(c) combination

5.47 From the description we learn that there are two reactants: sodium chloride and lead(II) nitrate. These are

both compounds. There are also two products: lead(II) chloride and sodium nitrate. Since these are also

compounds, the reaction is double-displacement (2 compounds as reactants and 2 compounds as products).

5.48 The reactants are solid sulfur and oxygen gas. These are both elements. Since there is only one product,

sulfur dioxide, which is a compound, this is a combination reaction (2 elements or compounds as reactants

and 1compound as the product).

5.49 (a) There are two different substances in the reactant image and only one substance in the product image.

This is a combination reaction.

(b) The dark red and gray spheres of solid represent two different elements. The gray spheres are

displacing the dark red spheres. One compound and 1 element are producing another compound and

element. This is a single-displacement reaction.

5.50 (a) The image on the left represents some type of ionic compound. On the right, the different types of

atoms have recombined into elements. This is a decomposition reaction.

(b) The image on the left represents an aqueous solution of two different ionic compounds. The image on

the right indicates the presence of two new compounds, one of which is water soluble and one that is

not. This is a double displacement reaction.



new substance. In decomposition reactions, 1 compound forms several new substances. Equations

indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If 1 of the

reactants and 1 of the products is an element, the reaction is single-displacement.

(a) double-displacement: CaCl2(aq) + Na2SO4(aq) CaSO4(s) + 2NaCl(aq)

(b) single-displacement: Ba(s) + 2HCl(aq) BaCl2(aq) + H2(g)

(c) combination: N2(g) + 3H2(g) 2NH3(g)

(d) This reaction does not easily fit the classification scheme given in Table 5.1; however, it is most like a

single-displacement reaction. CO is displacing iron from FeO:

FeO(s) + CO(g) heat Fe(s) + CO2(g)

517

(e) combination: CaO(s) + H2O(l) Ca(OH)2(aq)

(f) double-displacement: Na2CrO4(aq) + Pb(NO3)2(aq) PbCrO4(s) + 2NaNO3(aq)

(g) single-displacement: 2KI(aq) + Cl2(g) 2KCl(aq) + I2(aq)

(h) decomposition: 2NaHCO3(s) heat Na2CO3(s) + CO2(g) + H2O(g)

5.52 For help classifying reactions, refer to Table 5.1. We can classify two of the types on the number of



indicating 2 reactants and 2 products represent either single- or double-displacement reactions. If 1 of the

reactants and 1 of the products is an element, the reaction is single-displacement.

(a) combination: GaH3 + N(CH3)3 (CH3)3NGaH3

(b) single-displacement: Ca(s) +2H2O(l) Ca(OH)2(aq) + H2(g)

(c) single-displacement: N2(g) + CaC2(s) 2C(s) + CaNCN(s)

(d) combination: N2(g) + 3Mg(s) Mg3N2(s)

(e) decomposition: NH4Cl(s) NH3(g) + HCl(g)

(f) combination: CaO(s) + SO3(g) heat CaSO4(s)

(g) decomposition: PCl5(g) heat PCl3(g) + Cl2(g)

(h) double-displacement: Ca3N2(s) + 6H2O(l) heat 3Ca(OH)2(aq) + 2NH3(g)

5.53 Carbonate-containing compounds (except Group IA (1)), when heated, produce the metal oxide and carbon

dioxide gas (Table 5.2). In this case, we know that nickel has a 2+ charge to balance the charge of the 2

carbonate ion. The oxide will be NiO. The balanced chemical equation is NiCO3(s) NiO(s) + CO2(g).

5.54 When chloride-containing compounds are heated they decompose, forming chlorine gas and the elemental

metal (Table 5.2). In this case, platinum(IV) chloride, PtCl4, decomposes to produce platinum metal, Pt,

and chlorine gas, Cl2. The balanced chemical equation is PtCl4(s) Pt(s) + 2Cl2(g). Note: Most ionic

compounds (such as PtCl4) are solids, and molecular chlorine is a gas.

5.55 See Table 5.2.

(a) CaCO3(s) heat CaO(s) + CO2(g)

Metal carbonates (except Group IA (1)) decompose to produce carbon dioxide and the metal oxide.

We determine the formula of the metal oxide from the charge on the metal and the 2 charge of the

oxide ion (i.e. Ca2+

and O2

produce CaO).

(b) CuSO45H2O(s) heat CuSO4(s) + 5H2O(g)

Hydrates decompose by separation of the water from the ionic compound.

5.56 See Table 5.2.

(a) Cu(OH)2(s) heat CuO(s) + H2O(g)

Hydroxides decompose to the metal oxide with the loss of water (as a vapor or gas). Even at the high

temperatures needed to decompose many of these compounds, ionic compounds remain in the solid

state.

(b) 2NaN3(s)heat 2Na(s) + 3N2(g)

Azides often decompose to produce nitrogen gas.

5.57 See Table 5.3

2Mg(s) + O2(g) 2MgO(s)

Combination reactions result in the formation of 1 product from 2 reactants. If the reactants are a metal and

a nonmetal, we can reliably predict that the metal will form a cation and the nonmetal will form an anion.

518

When oxygen reacts with elements, the oxides of those elements are formed. With metals, the products are

the metal oxides. Magnesium is a Group IIA (2) metal. This means that it will form an ion with a 2+

charge. The formula for oxide ion is O2

. The formula for the metal oxide product in this reaction is MgO.

5.58 See Table 5.3

2Na(s) + Cl2(g) 2NaCl(s)

Combination reactions result in the formation of 1 product from 2 reactants. If the reactants are a metal and

a nonmetal, we can reliably predict that the metal will form a cation and the nonmetal will form an anion.

When molecular chlorine, Cl2, reacts with metals, the chlorides of the metals are formed. Sodium is a

Group IA (1) metal so it will form a 1+ ion. The formula for chloride ion is Cl. The formula for the metal

chloride is NaCl (salt).

5.59 Combination reactions result in the formation of 1 product from 2 reactants (see Table 5.3). If the reactants

are a metal and a nonmetal, you can reliably predict that the metal will form a cation and the nonmetal will

form an anion. In other situations, simply try to combine the two reactants to form a product you

recognize.

(a) 3Ca(s) + N2(g) Ca3N2(s)

Calcium and nitrogen (metal and nonmetal) form the Ca2+

and N3

ions based on their positions on the

periodic table. We predict that the product they form is Ca3N2.

(b) 2K(s) + Br2(l) 2KBr(s)

Metals and nonmetals often react to form metal salts.

(c) 4Al(s) +3O2(g) 2Al2O3(s)

Aluminum and oxygen (metal and nonmetal) form Al3+

and O2

ions, based on their positions on the

periodic table. We predict that the product they form is Al2O3.

5.60 Combination reactions result in the formation of 1 product from 2 reactants (see Table 5.3). Metals and

nonmetals often react to form metal salts. Each of these products is the metal salt formed the two reactants.

(a) 4Na(s) + O2(g) 2Na2O(s)

(b) 2Al(s) + 3Cl2(g) 2AlCl3(s)

(c) Ca(s) + F2(g) CaF2(s)

5.61 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a

metal displaces the metal ion, or a nonmetal displaces the nonmetal ion. To determine whether a reaction

occurs, we compare the activities of the metals involved (Figure 5.21). If the metal is more active than the

metal whose ion appears in the compound, a reaction occurs. In these cases, the formula for the ionic

product is determined by the metal ion that is produced. If you can’t predict the charge (i.e. it is not listed

in Figure 3.12), assume that it has a charge of 2+. For example if Fe displaces Cu, Fe will form a 2+ ion.

In this type of single-displacement reaction, the formula for the anion remains unchanged.

(a) Zn is higher on the activity series than Ag so a reaction occurs. Zn forms a 2+ ion, and Ag+ is

displaced, forming silver metal, Ag(s).

Zn(s) + 2AgNO3(aq) 2Ag(s) + Zn(NO3)2(aq)

(b) Na is higher on the activity series than Fe so a reaction occurs. Na forms a 1+ ion, and Fe2+

is

displaced, forming iron metal, Fe(s).

2Na(s) + FeCl2(s) 2NaCl(s) + Fe(s)

5. 62 In a single-displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a



metal whose ion appears in the compound, a reaction occurs. In these cases, the formula for the ionic

519

product is determined by the metal ion that is produced. If you can’t predict the charge (i.e. it is not listed

in Figure 3.12), assume that it has a charge of 2+. For example if Fe displaces Cu, Fe will form a 2+ ion.

In this type of single-displacement reaction, the formula for the anion remains unchanged.

(a) Al (s) higher on the activity series than Cu so a reaction occurs. Al forms a 3+ ion, and Cu2+

is

displaced as copper metal, Cu(s).

2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)

(b) Cs is higher in activity than H2 so a reaction occurs. In reactions of active metals with water, hydrogen

gas is formed. Cs forms a 1+ ion, and H+ is displaced as H2(g). When metals displace H

+ from H2O

(or HOH), the product is a metal hydroxide.

2Cs(s) + 2H2O(l) 2CsOH(aq) + H2(g)


metal displaces the metal ion or a nonmetal displaces the nonmetal ion. The formulas for the products are

determined by the element or ions that are produced (i.e. metal or diatomic molecules, etc.). When Zn

displaces Sn, Zn will form a 2+ ion (Figure 3.12). In this single-displacement reaction, the formula for the

anion remains unchanged. Since the formula for chloride ion is Cl, the product is ZnCl2. The balanced

equation is:

Zn(s) + SnCl2(aq) ZnCl2(aq) + Sn(s)


metal displaces the metal ion or a nonmetal displaces the nonmetal ion. The formulas for the products are

determined by the element or ions that are produced (i.e. metal or diatomic molecules, etc.). When Mg

displaces Cu, Mg will form a 2+ ion. In this single-displacement reaction, the formula for the anion

remains unchanged.

Mg(s) + CuSO4(aq) Cu(s) + MgSO4(aq)

5.65 In reactions of metals with hydrochloric acid solution or water, one product will always be hydrogen gas

(H2). To determine whether a reaction takes place, compare the activity of the metal (not the metal ion

found in the product) with the activity of hydrogen gas (Figure 5.21). If the metal is higher in the activity

series than hydrogen gas, a reaction will take place. Some metals are more active than others. Those that

are more active react under less extreme conditions.

(a) Ca is more active than H2. Because it is so high on the activity series, it will displace hydrogen from

water and form an acidic solution. As noted on the activity series shown in Figure 5.21, if a substance

reacts with cold water, it also reacts with steam and acid.

Ca reacts with water: Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)

Ca reacts with HCl: Ca(s) + 2HCl(aq) CaCl2(aq) + H2(g)

(b) Fe is more active than H2. This means that Fe can displace H2 under the proper conditions. As is noted

in Figure 5.21, Fe is not active enough to displace hydrogen in cold water, but can do so from either

hydrochloric acid solution or from steam.

Fe reacts with steam: Fe(s) + 2 H2O(g) Fe(OH)2(s) + H2(g)

Fe reacts with HCl: Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g)

(c) No reaction. Copper is below H2 on the activity series. There are no reaction conditions which favor a

reaction of copper with water or HCl solutions.

5.66 In reactions of metals with hydrochloric acid solutions or water, one product will always be hydrogen gas

(H2). To determine whether a reaction takes place, compare the activity of the metal (not the metal ion

found in the product) with the activity of hydrogen gas (Figure 5.21). If the metal is higher in the activity

series than hydrogen gas, a reaction will take place. Some metals are more active than others. Those that

are more active react under less extreme conditions.

520

(a) Cr is more active than H2. This means that it can displace hydrogen under the right conditions.

However, Cr is not active enough to displace hydrogen in cold water. Either hydrochloric acid

solution or steam is required to cause a reaction. Chromium forms a 2+ ion in the activity series.

Cr reacts with HCl: Cr(s) + 2HCl(aq) CrCl2(aq) + H2(g)

Cr reacts with steam: Cr(s) + 2H2O(g) Cr(OH)2(s) + H2(g)

(b) No reaction. Bismuth is below H2 on the activity series. There are no reaction conditions which favor

a reaction of bismuth with water or HCl solutions.

(c) K is more active than H2. Because it is so high on the activity series, K will displace hydrogen from

water and from hydrochloric acid solution. As noted on the activity series, if a substance reacts with

cold water, it also reacts with steam and acid.

K reacts with water: 2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

K reacts with HCl: 2K(s) + 2HCl(aq) 2KCl(aq) + H2(g)




metal whose ion appears in the compound, a reaction occurs.

(a) Magnesium is more active than aluminum. The reaction occurs.

(b) Zinc is less active than magnesium. No reaction occurs.

(c) Copper is less active than lead. No reaction occurs.

(d) Nickel is more active than silver. The reaction occurs.

5.68 In a single displacement reaction, 1 element displaces its ionic counterpart from a compound. That is, a



metal whose ion appears in the compound, a reaction occurs.

(a) Nickel is less active than iron. No reaction occurs.

(b) Aluminum is more active than nickel. The reaction occurs.

(c) Iron is more active than copper. The reaction occurs.

(d) Tin is less active than aluminum. No reaction occurs.

5.69 To determine solubility, refer to the solubility rules given in Table 5.4. Except for compounds containing

NH4+ and/or Group IA (1) cations (including Na

+ and K

+), the solubility rules focus on the anions in ionic

compounds. It is worth memorizing that Group IA (1), nitrates, acetates, and ammonium ion compounds

are soluble, because we encounter them so frequently when we study chemistry.

(a) CuCl2: The anion is Cl; the compound is soluble.

(b) AgNO3: The anion is NO3–; the compound is soluble.

(c) PbCl2: The anion is Cl; most chlorides are soluble, but PbCl2 is an exception. It is insoluble.

(d) Cu(OH)2: The anion is OH; the compound is insoluble.

5.70 To determine solubility, refer to the solubility rules given in Table 5.4. Except for compounds containing

NH4+ and/or Group IA (1) cations (including Na

+ and K

+), the solubility rules focus on the anions in ionic

compounds. It is worth memorizing that Group IA (1), nitrates, acetates, and ammonium ion compounds

are soluble, because we encounter them so frequently when we study chemistry.

(a) Cr2S3: The anion is S2

; the compound is insoluble.

(b) Ca(OH)2: The anion is OH; the compound is soluble. The solubility rules indicate that Ca(OH)2 is

somewhat soluble. This means that a small amount of Ca(OH)2 will dissolve in water.

(c) BaSO4: The anion is SO42–

; the compound is insoluble. While most sulfates are soluble in water, there

are exceptions, such as BaSO4.

(d) (NH4)2CO3: The cation is NH4+; the compound is soluble. All ammonium compounds are soluble.

521

5.71 For each reaction we: 1) determine the ion formulas, 2) predict the products based on ionic charges, and 3)

determine the states of the products. If one or both products is insoluble, is a gas, or is a molecule (for

example, water), a reaction takes place. Finally, we balance the resulting equation.

(a) Reactants: K2CO3(aq) and BaCl2(aq)

Ions: K+, CO3

2–, Ba

2+, Cl

Products: BaCO3 and KCl

Solubility (from Table 5.4): BaCO3(s) and KCl(aq)

Balanced equation: K2CO3(aq) + BaCl2(aq) BaCO3(s) + 2KCl(aq)

(b) Reactants: CaS(aq) and Hg(NO3)2(aq)

Ions: Ca2+

, S2

, Hg2+

, NO3–

Products: Ca(NO3)2 and HgS

Solubility (from Table 5.4): Ca(NO3)2(aq) and HgS(s)

Balanced equation: CaS(aq) + Hg(NO3)2(aq) Ca(NO3)2(aq) + HgS(s)

(c) Reactants: Pb(NO3)2(aq) and K2SO4(aq)

Ions: Pb2+

, NO3–, K

+, SO4

2–

Products: PbSO4 and KNO3

Solubility (from Table 5.4): PbSO4(s) and KNO3(aq)

Balanced equation: Pb(NO3)2(aq) + K2SO4(aq) PbSO4(s) + 2KNO3(aq)

5.72 For each reaction we: 1) determine the ion formulas, 2) predict the products based on ionic charges,

and 3) determine the states of the products. If one or both products is insoluble, is a gas, or is a

molecule (for example, water), a reaction takes place. Finally, we balance the resulting equation.

(a) Reactants: MgSO4(aq) and BaCl2(aq)

Ions: Mg2+

, SO42–

, Ba2+

, Cl

Products: MgCl2 and BaSO4

Solubility (from Table 5.4): MgCl2(aq) and BaSO4(s)

Balanced equation: MgSO4(aq) + BaCl2(aq) MgCl2(aq) and BaSO4(s)

(b) Reactants: K2SO4(aq) and MgCl2(aq)

Ions: K+, SO4

2–, Mg

2+, Cl

Products: KCl and MgSO4

Solubility (from Table 5.4): KCl(aq) and MgSO4(aq)

Balanced equation: No reaction occurs because both products are ionic and soluble in water.

(c) Reactants: MgCl2(aq) and Pb(NO3)2(aq)

Ions: Mg2+

, Cl, Pb

2+, NO3

–

Products: Mg(NO3)2 and PbCl2

Solubility (from Table 5.4): Mg(NO3)2(aq) and PbCl2(s)

Balanced equation: MgCl2(aq) + Pb(NO3)2(aq) Mg(NO3)2(aq) + PbCl2(s)

5.73 For each reaction, 1) determine the ion formulas, 2) predict the products based on ion charges, and 3)


example, water), a reaction takes place. Balance the resulting equations.

(a) Reactants: BaCO3(s) and H2SO4(aq)

Ions: Ba2+

, CO32–

, H+, SO4

2–

Products: BaSO4 and H2CO3

According to the solubility rules (Table 5.4), barium sulfate is insoluble in water. Carbonic acid

decomposes to carbon dioxide and water (as described in Table 5.2). Since a precipitate is formed and

carbonic acid decomposes to form water and carbon dioxide, we predict that the following reaction

takes place:

BaCO3(s) + H2SO4(aq) BaSO4(s) + H2CO3(aq) carbonic acid decomposes

522

BaCO3(s) + H2SO4(aq) BaSO4(s) + H2O(l) + CO2(g) Balanced

(b) Reactants: CuCl2(aq) and AgNO3(aq)

Ions: Cu2+

(based on charge of chloride), Cl, Ag

+, NO3

–

Products: Cu(NO3)2 and AgCl

According to the solubility rules (Table 5.4), nitrates are soluble in water. While most chlorides are

soluble in water, AgCl is an exception. Since a precipitate is formed, we can predict that the following

reaction takes place:

CuCl2(aq) + 2AgNO3(aq) Cu(NO3)2(aq) + 2AgCl(s) Balanced



example, water), a reaction takes place. Balance the resulting equations.

(a) Reactants: NaOH(aq) and CuCl2(aq)

Ions: Na+, OH

, Cu

2+ (based on the charge of chloride ion), Cl

Products: NaCl and Cu(OH)2

According to the solubility rules (Table 5.4), most chlorides are soluble in water, so NaCl will be

dissolved. Most hydroxides are insoluble in water, so Cu(OH)2 will form a precipitate (that is, a solid).

Because the precipitate forms, we predict that the following reaction takes place:

2NaOH(aq) + CuCl2(aq) 2NaCl(aq) + Cu(OH)2(s) Balanced

(b) Reactants: H2SO4(aq) and KOH(aq)

Ions: H+, SO4

2, K

+, OH

Products: H2O and K2SO4

According to the solubility rules (Table 5.4), sulfates are generally insoluble in water; however, K2SO4

is an exception. Because water is a stable molecule, we can predict that the following reaction takes

place:

H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) Balanced

5.75 To identify the precipitate, we first determine the products of the reaction, and then check the solubility of

the products using the solubility rules (Table 5.4). The precipitate will be the compound that is insoluble in

water. We deduce the reactants from the description given in the problem.

Reactants: Na2SO4 and Pb(NO3)2

Ions: Na+, SO4

2–, Pb

2+, NO3

–

Products: PbSO4 and NaNO3

Solubility (from Table 5.4): PbSO4(s), NaNO3(aq)

Lead(II) sulfate is the white solid that forms when the two solutions are mixed.

5.76 To identify the precipitate, we first determine the products of the reaction, and then check the solubility of

the products using the solubility rules (Table 5.4). The precipitate will be the compound that is insoluble in

water. We deduce the reactants from the description given in the problem. In this particular problem, we

encounter the mercury(I) ion. Mercury(I) ions (Hg22+

) represent one of a very few diatomic metal ions (see

Figure 3.25). Notice that the ion has a 2+ charge because it is diatomic. The average charge of mercury in

this ion is 1+.

Reactants: Hg2(NO3)2 and CaI2

Ions: Hg22+

, NO3–, Ca

2+, I

Products: Hg2I2 and Ca(NO3)2

Solubility (from Table 5.3): Hg2I2(s), Ca(NO3)2(aq)

Mercury(I) iodide is the yellow solid that forms when the two solutions are mixed.

5.77 From the problem description we can write:

CaCl2(aq) + K2CO3(aq)

523

In double-displacement reactions, the cations switch places to form new products (i.e. the metal ions

displace one another). To avoid common errors, and to be able to spot mistakes easily, it is helpful if we

write the complete formulas for each ion before we determine the products.

Reactants: CaCl2(aq) and K2CO3(aq)

Ions: Ca2+

, Cl, K

+, CO3

2

Considering the charges of the ions, we can write the chemical formulas for the products. The products

will be calcium carbonate, CaCO3, and potassium chloride, KCl. Using the solubility rules (Table 5.4) we

can determine whether these products are soluble or insoluble in water (i.e. whether we should write (aq),

or (s), respectively). From the solubility rules, we determine that carbonates are insoluble (i.e. CaCO3(s))

and chlorides are generally soluble (i.e. KCl(aq)).

CaCl2(aq) + K2CO3(aq) CaCO3(s) + 2KCl(aq) Balanced

5.78 From the problem description we can write:

(NH4)2CrO4(aq) + Pb(NO3)2(aq)

In double-displacement reactions, the cations switch places to form new products (i.e. the metal ions

displace one another). To avoid common errors, and to be able to spot mistakes easily, it is helpful if we

write the complete formulas for each ion before we determine the products.

Reactants: (NH4)2CrO4(aq) and Pb(NO3)2(aq)

Ions: NH4+, CrO4

2, Pb

2+, NO3

–

Considering the charges of the ions, we can write the chemical formulas for the products. The products

will be lead(II) chromate, PbCrO4, and ammonium nitrate, NH4NO3. Using the solubility rules (Table 5.4)

we can determine whether these products are soluble or insoluble in water (i.e. whether we should write

(aq) or (s), respectively). From the solubility rules we determine that chromates are generally insoluble in

water and compounds containing ammonium and/or nitrate ions are generally soluble in water.

(NH4)2CrO4(aq) + Pb(NO3)2(aq) PbCrO4(s) + 2NH4NO3(aq) Balanced

5.79 Chemical reactions occur for a “reason.” For many reactions, the reason is the nature of the products that

form. This “driving force” of a reaction can often be identified by formation of a precipitate, formation of a

gas (water-insoluble gas), or formation of molecular compounds, such as water. Water-insoluble gases will

leave a solution, providing an additional driving force for the reaction. The driving force for each reaction

is:

(a) precipitation of HgS(s)

(b) formation of the insoluble gas H2S(g)

(c) formation of the molecular compound H2O(l), and precipitation of BaSO4(s)

5.80 Chemical reactions occur for a “reason.” For many reactions, the reason is the nature of the products that

form. This “driving force” of a reaction can often be identified by formation of a precipitate, formation of a

gas (water-insoluble gas), or formation of molecular compounds, such as water. Water-insoluble gases will

leave a solution, providing an addition driving force for the reaction. The driving force for each reaction is:

(a) formation of the molecular compound H2O(l)

(b) formation of a precipitate, CaSO4(s), a molecular compound, H2O(l), and a gas, CO2(g)

(c) formation of a precipitate, BaSO4(s)

5.81 Neutralization reactions involve the reaction of an acid with a base. The products are a salt (i.e. ionic

compound) and water. Just as with precipitation reactions we: 1) determine the ionic formulas, 2) predict

the products based on ion charges, and 3) determine the states of the products. Acid-base reactions are

usually driven by the formation of water, a molecular compound. We must use the solubility rules to

determine whether or not the salt will dissolve in water.

(a) Reactants: H2S(aq) and Cu(OH)2(s)

Ions: H+, S

2, Cu

2+, OH

Products: H2O(l) and CuS(s) (determined from solubility rules)

Balanced equation: H2S(aq) + Cu(OH)2(s) CuS(s) + 2H2O(l)

524

(b) CH4 is not an acid, so no reaction is expected.

(c) Reactants: KHSO4(aq) and KOH(aq)

Ions: K+, H

+, SO4

2, OH

Products: H2O(l) and K2SO4. Note that the H+ reacts with OH

, leaving K

+ and SO4

2 to form the salt,

which is water-soluble.

Balanced equation: KHSO4(aq) + KOH(aq) K2SO4(aq) + H2O(l)

5.82 Neutralization reactions involve the reaction of an acid with a base. The products are a salt (i.e. ionic

compound) and water. Just as with precipitation reactions we: 1) determine the ion formulas, 2) predict the

products based on ion charges, and 3) determine the states of the products. Acid-base reactions are usually

driven by the formation of water (a molecular compound). We must use the solubility rules to determine

whether or not the salt will dissolve in water.

(a) H2(g) is not an acid, so no reaction is expected.

(b) Reactants: H3PO4(aq) and NaOH(aq)

Ions: H+, PO4

3-, Na

+, OH

Products: Na3PO4(aq) and H2O(l)

Balanced equation: H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l)

(c) Reactants: H2SO4(aq) and Fe(OH)2(aq)

Ions: H+, SO4

2–, Fe

2+, OH

Products: FeSO4(aq) and H2O(l)

Balanced equation: H2SO4(aq) + Fe(OH)2(aq) FeSO4(aq) + 2H2O(l)

5.83 Oxygen, O2(g) is always one of the reactants of a combustion reaction.

5.84 Carbon dioxide and water are the usual products of hydrocarbon combustion.

5.85 The combustion reaction products are the oxides of those elements. For metals, we can predict the oxide

formula based on the common charges of the metal ions and the charge of oxide ion, O2

. Metals that form

several different cations can form several different oxides.

(a) Cesium only forms a 1+ ion, so the oxide that forms is Cs2O(s).

(b) Lead forms either a 2+ or 4+ ion (Figure 3.24), so the oxides that lead can form are PbO(s) and

PbO2(s).

(c) Aluminum only forms a 3+ ion, so the formula of aluminum oxide is Al2O3(s).

(d) Combustion of hydrogen produces H2O(g).

(e) Carbon produces two different oxides, CO(g) and CO2(g).

5.86 The combustion reaction products are the oxides of those elements. For metals, we can predict the oxide

formula based on the common charges of the metal ions and the charge of oxide ion, O2

. Metals that form

several different cations (see Table 3.4) can form several different oxides.

(a) Magnesium only forms a 2+ ion, so the oxide that forms is MgO(s).

(b) Iron forms either a 2+ or 3+ ion, so the oxides that iron can form are FeO(s) and Fe2O3(s).

(c) Lithium only forms a 1+ ion, so the formula of lithium oxide is Li2O(s).

(d) Combustion of S8 produces SO2(g).

(e) Nitrogen produces many different oxides. Some examples are NO and NO2.

5.87 The combustion reaction products of compounds are the oxides of the elements which make up the

compound. In many cases, there are multiple products.

(a) CH4 (and all hydrocarbons) combusts to form oxides of carbon and hydrogen. The products are CO(g)

or CO2(g) and H2O(g).

(b) The combustion of CO results in formation of CO2(g).

(c) The combustion of aluminum metal results in formation of solid aluminum oxide solid, Al2O3(s).

(d) CH3OH (and all compounds containing C, H, and O) combusts to form CO(g) or CO2(g) and H2O(g).

525

5.88 The combustion reaction products of compounds are the oxides of the elements which make up the

compound. In many cases, there are multiple products.

(a) The combustion of sodium metal results in the formation of sodium oxide, Na2O(s).

(b) The combustion of elemental phosphorus results in the formation of diphosphorus pentoxide solid,

P2O5(s).

(c) C3H8 (and all hydrocarbons) combusts to form oxides of carbon and hydrogen. The products are

CO(g) or CO2(g) and H2O(g).

(d) HCO2H (and all compounds containing C, H, and O) combust to form the oxides of carbon and

hydrogen. The products are CO(g) or CO2(g) and H2O(g).

5.89 An electrolyte produces ions when it dissolves in water. A nonelectrolyte does not produce ions when it

dissolves in water.

5.90 Dissolve the solid, then determine whether or not the resulting liquid conducts electricity with an apparatus

such as that depicted in Figure 3.3. If the solid is not soluble in water, melt it and then see if it conducts

electricity.

5.91 Aqueous solutions of soluble ionic compounds, acids, or bases are electrolytes because the substances

produce ions when they dissolve.

(a) NaOH(aq): base; electrolyte

(b) HCl(aq): acid; electrolyte

(c) C12H22O11: not a soluble ionic compound, acid, or base; nonelectrolyte

5.92 Aqueous solutions of soluble ionic compounds, acids, or bases are electrolytes because the substances

produce ions when they dissolve.

(a) CH3CH2OH: not a soluble ionic compound, acid, or base; nonelectrolyte

(b) H2O(l): not a soluble ionic compound, acid, or base; nonelectrolyte

(c) NaCl(aq): soluble ionic compound: electrolyte

5.93 The substance is a nonelectrolyte because it does not appear to have dissociated in the solution. In order to

be an electrolyte, ions must form in the solution.

5.94 Since we started with a single substance and now it has formed multiple species, the substance is probably

an electrolyte that has dissociated into its ions.

5.95 Soluble ionic compounds, acids, and bases produce ions when they dissolve in water.

(a) C6H12O6: not a soluble ionic compound, acid, or base; does not form ions in solution

(b) CH4: not a soluble ionic compound, acid, or base; does not form ions in solution

(c) NaCl: soluble ionic compound; forms Na+ and Cl

ions in solution; Na

+(aq) and Cl

(aq)

5.96 Soluble ionic compounds, acids, or bases produce ions when they dissolve in water.

(a) I2: not a soluble ionic compound, acid, or base; does not form ions in solution

(b) KI: soluble ionic compound; forms K+ and I

ions in solution

(a) CH3OH: not a soluble ionic compound, acid, or base; does not form ions in solution

5.97 Molecular equations: We write the chemical formulas for all substances in their complete form (for

example, O2(g), NaCl(aq), Ag(s)).

Ionic Equations: We write the formulas of all soluble salts, strong acids and bases as individual, solvated

ions (for example, NaCl(aq) becomes Na+(aq) and Cl

(aq)).

Net ionic equations: We remove all spectator ions from the ionic equation.

5.98 It is necessary to identify substances as electrolytes or nonelectrolytes because electrolytes are written as

separated ions in a net ionic equation. This allows us to identify and eliminate the spectator ions.

526

5.99 Spectator ions are those ions that do not actually participate in the chemical change that occurs during the

reaction. These are the ions that appear both as reactants and as products in an ionic equation.

5.100 We find spectator ions only in ionic equations. We do not write compounds in ionic form in molecular

equations. When we write net ionic equations, we do not include any spectator ions.

5.101 (a) From the solubility rules (Table 5.4) we find that all the substances are soluble ionic compounds

except Ag2SO4 and AgCl, which are insoluble in water.

2NaCl(aq) + Ag2SO4(s) Na2SO4(aq) + 2AgCl(s) Molecular equation

Then we write the formulas of all soluble ionic compounds as ions, and eliminate the spectator ions.

2Na+(aq) + 2Cl

(aq) + Ag2SO4(s) 2Na

+(aq) + SO4

2+(aq) + 2AgCl(s) Ionic equation

2Na+(aq) + 2Cl

(aq) + Ag2SO4(s) 2Na

+(aq) + SO4

2+(aq) + 2AgCl(s)

This leaves the balanced net ionic equation.

2Cl(aq) + Ag2SO4(s) 2AgCl(s) +SO4

2–(aq) Net ionic equation

(b) From the solubility rules we know that Cu(OH)2 is insoluble, while the other ionic compounds are

soluble. Water is a liquid.

Cu(OH)2(s) + 2HCl(aq) CuCl2(aq) + 2H2O(l) Molecular equation

Next we write the ionic equation by writing all the aqueous ionic compounds and strong acids (Table

3.10) in ionic form.

Cu(OH)2(s) + 2H+(aq) + 2Cl

(aq) Cu

2+ + 2Cl

(aq) + 2H2O(l)

Finally, we write the net ionic equation by identifying and eliminating the spectator ions.

Cu(OH)2(s) + 2H+(aq) + 2Cl

(aq) Cu

2+ + 2Cl

(aq) + 2H2O(l)

Cu(OH)2(s) + 2H+(aq) Cu

2+(aq) + 2H2O(l) Net ionic equation

(c) From the solubility rules we know that only BaCl2 is soluble, while the other ionic compounds are

insoluble.

BaCl2(aq) + Ag2SO4(s) BaSO4(s) + 2AgCl(s) Molecular equation

Only BaCl2 forms ions in solution. Since none of the products form ions, there are no spectator ions.

Ba2+

(aq) + 2Cl(aq) + Ag2SO4(s) BaSO4(s) + 2AgCl(s) Net ionic equation

5.102 (a) From the solubility rules (Table 5.4) we find that all the substances are soluble ionic compounds

except Cr(OH)3, which is insoluble.

Cr2(SO4)3(aq) + 6KOH(aq) 2Cr(OH)3(s) + 3K2SO4(aq) Molecular equation

Next we write the ionic equation by separating all the aqueous ionic compounds into their respective

ions.

2Cr3+

(aq) + 3SO42–

(aq) + 6K+(aq) + 6OH

(aq) 2Cr(OH)3(s) + 6K

+(aq) + 3SO4

2–(aq)

Finally, we write the net ionic equation by identifying and eliminating the spectator ions, and reducing

the coefficients as much as possible.

2Cr3+

(aq) + 3SO42–

(aq) + 6K+(aq) + 6OH

(aq) 2Cr(OH)3(s) + 6K

+(aq) + 3SO4

2–(aq)

2Cr3+

(aq) + 6OH(aq) 2Cr(OH)3(s)

Cr3+

(aq) + 3OH(aq) Cr(OH)3(s) Net ionic equation

(b) From the solubility rules we find that all the substances are soluble ionic compounds except PbCl2(s)

and PbCrO4, which are insoluble.

527

K2CrO4(aq) + PbCl2(s) 2KCl(aq) + PbCrO4(s) Molecular equation

Then we write the formulas of all soluble ionic compounds as ions, and eliminate the spectator ions.

2K+(aq) + CrO4

2(aq) + PbCl2(s) 2K

+(aq) + 2Cl

(aq) + PbCrO4(s)

2K+(aq) + CrO4

2(aq) + PbCl2(s) 2K

+(aq) + 2Cl

(aq) + PbCrO4(s)

CrO42

(aq) + PbCl2(s) 2Cl(aq) + PbCrO4(s) Net ionic equation

(c) From the solubility rules we find that Na2SO3 and Na2SO4 are soluble ionic compounds. Sulfuric acid,

H2SO4, is a strong acid and SO2 is a gas.

Na2SO3(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l) + SO2(g) Molecular equation

Next we write the ionic equation by writing all the aqueous ionic compounds and strong acids as

separate ions. Although sulfuric acid is a strong acid, only the first hydrogen atom ionizes completely.

2Na+(aq) + SO3

2(aq) + 2H

+(aq) + SO4

(aq) 2Na

+ + SO4

2–(aq) + H2O(l) + SO2(g)


2Na+(aq) + SO3

2(aq) + 2H

+(aq) + SO4

2(aq) 2Na

+ + SO4

2–(aq) + H2O(l) + SO2(g)

SO32

(aq) + 2H+(aq) H2O(l) + SO2(g) Net ionic equation

5.103 From the solution descriptions, we know that there are four ions present in the solution before any

precipitation takes place: Na+, CO3

2, Ca

2+, and Br

.

(a) The only possible precipitate is CaCO3 (see Table 5.4). The other combinations represent soluble

compounds.

(b) Na2CO3(aq) + CaBr2(aq) CaCO3(s) + 2NaBr(aq)

(c) 2Na+(aq) + CO3

2(aq) + Ca

2+(aq) + 2Br

(aq) CaCO3(s) + 2Na

+(aq) + 2Br

(aq)

(d) Na+(aq) and Br

(aq)

(e) CO32

(aq) + Ca2+

(aq) CaCO3(s)

5.104 From the solution descriptions, we know that there are four ions present in the solution before any

precipitation takes place: K+, CrO4

2, Ba

2+, and Cl

.

(a) Although most chromates are insoluble, potassium chromate is an exception. The only possible

precipitate is barium chromate, BaCrO4 (see Table 5.4).

(b) K2CrO4(aq) + BaCl2(aq) BaCrO4(s) + 2KCl(aq)

(c) 2K+(aq) + CrO4

2(aq) + Ba

2+(aq) + 2Cl

(aq) BaCrO4(s) + 2K

+(aq) + 2Cl

(aq)

(d) K+(aq) and Cl

(aq)

(e) CrO42

(aq) + Ba2+

(aq) BaCrO4(s)

5.105 To write the net ionic equation for this reaction, we start by writing the molecular equation. From the

description in the question we have:

CaCl2(aq) + 2AgNO3(aq) Ca(NO3)2(aq) + 2AgCl(s) Molecular equation

Next we write the ionic equation by separating the formulas of all the water-soluble ionic compounds into

their respective ions. All the ions written in the equation below are solvated by water.

Ca2+

(aq) + 2Cl(aq) + 2Ag

+(aq) + 2NO3

(aq) Ca

2+(aq) + 2NO3

(aq) + 2AgCl(s)

Next we eliminate the spectator ions.

Ca2+

(aq) + 2Cl(aq) + 2Ag

+(aq) + 2NO3

(aq) Ca

2+(aq) + 2NO3

(aq) + 2AgCl(s)

2Cl(aq) + 2Ag

+(aq) 2AgCl(s)

Because each of the coefficients is divisible by two, we simplify the net ionic equation to:

Ag+(aq) + Cl

(aq) AgCl(s) Net ionic equation

528

A quick way to check the answer is to see if the total charge on the reactant and product sides of the

equation is the same. In this case, the total charge is zero for both the reactants (Ag+ and Cl

) and the

products.

5.106 To write the net ionic equation for this reaction, we start by writing the molecular equation. From the

description in the question we have:

Cr2(SO4)3(aq) + 6NaOH(aq) 2Cr(OH)3(s) + 3Na2SO4(aq) Molecular equation

Next we write the ionic equation by separating the formulas of all water-soluble compounds into their

respective ions. All the ions written in the equation below are solvated by water.

2Cr3+

(aq) + 3SO42

(aq) + 6Na+(aq) + 6OH

(aq) 2Cr(OH)3(s) + 6Na

+(aq) + 3SO4

2(aq)

Next we eliminate the spectator ions.

2Cr3+

(aq) + 3SO42

(aq) + 6Na+(aq) + 6OH

(aq) 2Cr(OH)3(s) + 6Na

+(aq) + 3SO4

2(aq)

2Cr3+

(aq) + 6OH(aq) 2Cr(OH)3(s)

Because each of the coefficients in this equation is divisible by 2, we simplify the net ionic equation to:

Cr3+

(aq) + 3OH(aq) Cr(OH)3(s) Net ionic equation


equation is the same. In this case, the total charge is zero for both the reactants (the Cr3+

is balanced by the

three negative charges of OH) and the products.

5.107 The reactants are sodium metal (you can tell from its well ordered structure) and H2O(l). The products are

Na+, OH

, and H2. Na

+ and OH

are ions dissolved in water. Molecular hydrogen is a gas. The balanced

ionic equation is:

2Na(s) + 2H2O(l) 2Na+(aq) + 2OH

(aq) + H2(g)

There are no spectator ions, so the balanced net ionic equation and the balanced ionic equations are

identical.

2Na(s) + 2H2O(l) 2Na+(aq) + 2OH

(aq) + H2(g) Net ionic equation

5.108 Copper and aqueous silver nitrate participate in a single displacement reaction because copper is more

active than silver (Figure 5.21). We know that copper nitrate dissolves in water from the diagram, because

the ions are separated.

Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq) Molecular equation

Next we write the ionic equation by showing all the soluble ionic compounds in their ionic forms.

Cu(s) + 2Ag+(aq) + 2NO3

(aq) 2Ag(s) + Cu

2+(aq) + 2NO3

(aq)

Eliminating spectator ions we have:

Cu(s) + 2Ag+(aq) + 2NO3

(aq) 2Ag(s) + Cu

2+(aq) + 2NO3

(aq)

Cu(s) + 2Ag+(aq) 2Ag(s) + Cu

2+(aq) Net ionic equation


equation is the same. In this case, the total charge is 2+ on both the reactant and product sides.

5.109 (a) First we write the molecular equation and determine the solubility of ionic reactants and products

(Table 5.4).

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) Molecular equation

Next we write the ionic equation by writing all the aqueous ionic compounds in ionic form.

Cu(s) + 2Ag+(aq) + 2NO3

(aq) Cu

2+(aq) + 2NO3

(aq) + 2Ag(s)

529

Finally, we eliminate the spectator ions and write the net ionic equation.

Cu(s) + 2Ag+(aq) + 2NO3

(aq) Cu

2+(aq) + 2NO3

(aq) + 2Ag(s)

Cu(s) + 2Ag+(aq) Cu

2+(aq) + 2Ag(s) Net ionic equation

(b) First we write the molecular equation and determine the solubility of ionic reactants and products

(Table 5.4).

FeO(aq) + 2HCl(aq) FeCl2(aq) + H2O(l) Molecular equation

Next we write the ionic equation by writing all the aqueous ionic compounds and strong acids

(Table 3.10) in ionic form.

Fe2+

(aq) + O2

(aq) + 2H+(aq) + Cl

(aq) Fe

2+(aq) + 2Cl

(aq) + H2O(l)


Fe2+

(aq) + O2

(aq) + 2H+(aq) + 2Cl

(aq) Fe

2+(aq) + 2Cl

-(aq) + H2O(l)

O2

(aq) + 2H+(aq) H2O(l) Net ionic equation

5.110 (a) First we write the molecular equation and determine the solubility of ionic reactants and products

(Table 5.4).

Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g) Molecular equation


Ca(s) + 2H2O(l) Ca2+

(aq) + 2OH(aq) + H2(g)

There are no spectator ions, so the balanced net ionic equation and the balanced ionic equations are

identical.

Ca(s) + 2H2O(l) Ca2+

(aq) + 2OH(aq) + H2(g) Net ionic equation

(b) First we write the molecular equation and determine the solubility of ionic reactants and products

(Table 5.4).

BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq) Molecular equation


Ba2+

(aq) + 2Cl(aq) + 2Na

+(aq) + SO4

2 (aq) BaSO4(s) + 2Na

+(aq) + 2Cl

(aq) Ionic equation


Ba2+


+(aq) + SO4

2–(aq) BaSO4(s) + 2Na

+(aq) + 2Cl

-(aq)

Ba2+

(aq) + SO42–

(aq) BaSO4(s) Net ionic equation


determine the states for the products. If one or both products is insoluble, is a gas, or is a molecule (for

example, water), a reaction takes place.

(a) Reactants: Sr(NO3)2(aq) and H2SO4(aq)

Ions: Sr2+

, NO3–, H

+, SO4

2–

Products: SrSO4 and HNO3

Solubility (Table 5.4, Table 3.10): Sr(NO3)2(aq), H2SO4(aq), SrSO4(s), HNO3(aq)

Sr(NO3)2(aq) + H2SO4(aq) SrSO4(s) + 2HNO3(aq) Molecular equation

Next we write the ionic equation by separating all the aqueous ionic compounds and strong acids into

their respective ions.

Sr2+

+ 2NO3–(aq) + 2H

+(aq)

+ SO4

2(aq) SrSO4(s) + 2H

+(aq) + 2NO3

–(aq)

Finally, we eliminate the spectator ions:

530

Sr2+

+ 2NO3–(aq) + 2H

+(aq) + SO4

2(aq) SrSO4(s) + 2H

+(aq) + 2NO3

–(aq)

Sr2+

(aq) + SO42

(aq) SrSO4(s) Net ionic equation

(b) Reactants: Zn(NO3)2 and Na2SO4

Ions: Zn2+

, NO3–, Na

+, SO4

2–

Products: ZnSO4 and NaNO3

Solubility: Both products are aqueous ionic compounds.

The products do not precipitate. In addition, no molecular products are formed. No reaction takes

place.

(c) Reactants: CuSO4(aq) and BaS(s)

Ions: Cu2+

, SO42

, Ba2+

, S2

Products: CuS and BaSO4

Solubility: CuSO4(aq), BaS(s), CuS(s), BaSO4(s)

CuSO4(aq) + BaS(s) CuS(s) + BaSO4(s) Molecular equation


ions.

Cu2+

(aq) + SO42–

(aq) + BaS(s) CuS(s) + BaSO4(s) Ionic & net ionic equation

Since only one reactant is soluble, there are no spectator ions, so the ionic and net ionic equations are

the same.

(d) Reactants: NaHCO3and HCH3CO2

Ions: Na+, HCO3

, H

+, and CH3CO2

(see note below)

Products: NaCH3CO2 and H2CO3 (decomposes to H2O(l) and CO2(g)

Solubility: NaHCO3(aq), HCH3CO2(aq), NaCH3CO2(aq): Acids are generally soluble. Other

solubilities are obtained from Table 5.4.

A reaction takes place because of the formation of water and carbon dioxide.

NaHCO3(aq) + CH3CO2H (aq) NaCH3CO2(aq) + CO2(g) + H2O(l) Molecular equation


ions. Note that HCH3CO2 is a weak acid and does not completely ionize, however when it is ionized,

H+ and CH3CO2

are formed and we use this information to predict what products are formed.

Na+(aq) + HCO3

(aq) + HCH3CO2(aq) Na

+(aq) + CH3CO2

(aq) + CO2(g) + H2O(l)

Finally, we eliminate the spectator ions.

Na+(aq) + HCO3

(aq) + HCH3CO2(aq) Na

+(aq) + CH3CO2

(aq) + CO2(g) + H2O(l)

HCO3(aq) + CH3CO2H (aq) CH3CO2

(aq) + CO2(g) + H2O(l) Net Ionic Equation




(a) Reactants: (NH4)2CO3 and CrCl3

Ions: NH4+, CO3

2, Cr

3+, Cl

Products: NH4Cl and Cr2(CO3)3

Solubility: (NH4)2CO3(aq), CrCl3(aq), NH4Cl(aq), Cr2(CO3)3(s)

3(NH4)2CO3(aq) + 2CrCl3(aq) 6NH4Cl(aq) + Cr2(CO3)3(s) Molecular equation


ions.

6NH4+(aq) + 3CO3

2(aq) + 2Cr

3+(aq) + 6Cl

(aq) 6NH4

+(aq) + 6Cl

(aq) + Cr2(CO3)3(s)


531

6NH4+(aq) + 3CO3

2(aq) + 2Cr

3+(aq) + 6Cl

(aq) 6NH4

+(aq) + 6Cl

(aq) + Cr2(CO3)3(s)

3CO32

(aq) + 2Cr3+

(aq) Cr2(CO3)3(s) Net ionic equation

(b) Reactants: Ba(OH)2 and HCl(aq)

Ions: Ba2+

, OH, H

+, Cl

Products: BaCl2 and H2O

Solubility (Table 5.4, Table 3.10): Ba(OH)2(aq), HCl(aq), BaCl2(aq), H2O(l)

Ba(OH)2(aq) + 2HCl(aq) BaCl2(aq) + 2H2O(l) Molecular equation

Note that this is an acid-base reaction. The driving force for this reaction is the formation of water,

which is a molecular compound. Next we write the ionic equation by separating all the aqueous ionic

compounds and strong acids into their respective ions.

Ba2+

(aq) + 2OH(aq) + 2H

+(aq) + 2Cl

(aq) Ba

2+(aq) + 2Cl

(aq) + 2H2O(l)


Ba2+

(aq) + 2OH(aq) + 2H

+(aq) + 2Cl

(aq) Ba

2+(aq) + 2Cl

(aq) + 2H2O(l)

2OH(aq) + 2H

+(aq) 2H2O(l)

We can reduce the coefficients by dividing each by 2.

OH(aq) + H

+(aq) H2O(l) Net ionic equation

(c) Reactants: FeCl3 and NaOH

Ions: Fe3+

, Cl, Na

+, OH

Products: Fe(OH)3 and NaCl

Solubility (Table 5.4): FeCl3(aq), NaOH(aq), Fe(OH)3(s), NaCl(aq)

FeCl3(aq) + 3NaOH(aq) Fe(OH)3(s) + 3NaCl(aq) Molecular equation


ions.

Fe3+

(aq) + 3Cl

(aq) + 3Na

+(aq) + 3OH

(aq) Fe(OH)3(s) + 3Na

+(aq) + 3Cl

(aq)


Fe3+

(aq) + 3Cl

(aq) + 3Na

+(aq) + 3OH

(aq) Fe(OH)3(s) + 3Na

+(aq) + 3Cl

(aq)

Fe3+

(aq) + 3OH

(aq) Fe(OH)3(s) Net ionic equation

(d) Reactants: Ca(HCO3)2 and HNO3

Ions: Ca2+

, HCO3, H

+, NO3

Products: Ca(NO3)2 and H2CO3 which decomposes to CO2(g) and H2O(l))

Solubility (Table 5.4): Ca(HCO3)2(aq), HNO3(aq), Ca(NO3)2(aq); Acids are generally soluble. Other

solubilities are obtained from Table 5.3.

A reaction takes place because of the formation of liquid water and carbon dioxide gas

Ca(HCO3)2(aq) + 2HNO3(aq) Ca(NO3)2(aq) + 2CO2(g) + 2H2O(l) Molecular equation


ions.

Ca2+

(aq) + 2 HCO3(aq) + 2H

+(aq) + 2NO3

(aq) Ca

2+(aq) + 2NO3

(aq)+ 2CO2(g) + 2H2O(l)


Ca2+

(aq) + 2 HCO3(aq) + 2H

+(aq) + 2NO3

(aq) Ca

2+(aq) + 2NO3

(aq) + 2CO2(g) + 2H2O(l)

2 HCO3(aq) + 2H

+(aq) 2CO2(g) + 2H2O(l)

This coefficients can be reduced by dividing by 2.

532

HCO3(aq) + H

+(aq) CO2(g) + H2O(l) Net ionic equation

5.113 Because we are starting with 2 reactants and forming 1 product, this is a combination reaction (Table 5.1).

5.114 Malachite’s formula suggests that we consider the decomposition of each of its components. The

decomposition of CuCO3 produces the metal oxide and carbon dioxide gas (Table 5.2). The balanced

equation for this reaction is:

CuCO3(s) CuO(s) + CO2(g)

The decomposition of Cu(OH)2 produces copper oxide and water vapor.

Cu(OH)2 CuO(s) + H2O(g)

Combining the two reactions we have:

CuCO3Cu(OH)2(s) 2CuO(s) + CO2(g) + H2O(g)




(a) Reactants: ZnSO4(aq) and Ba(NO3)2(aq)

Ions: Zn2+

, SO42–

, Ba2+

, NO3–

Products: BaSO4 and Zn(NO3)2

Solubility (Table 5.4): BaSO4(s) and Zn(NO3)2(aq)

ZnSO4(aq) + Ba(NO3)2(aq) BaSO4(s) + Zn(NO3)2(aq) Balanced

(b) Reactants: Ca(NO3)2(aq) and K3PO4(aq)

Ions: Ca2+

, NO3–, K

+, PO4

3-

Products: Ca3(PO4)2 and KNO3

Solubility (Table 5.4): Ca3(PO4)2(s) and KNO3(aq)

3Ca(NO3)2(aq) + 2K3PO4(aq) Ca3(PO4)2(s) + 6KNO3(aq) Balanced

(c) Reactants: ZnSO4(aq) and BaCl2(aq)

Ions: Zn2+

, SO42–

, Ba2+

, Cl

Products: ZnCl2 and BaSO4

Solubility (Table 5.4): ZnCl2(aq) and BaSO4(s)

ZnSO4(aq) + BaCl2(aq) BaSO4(s) + ZnCl2(aq) Balanced

(d) Reactants: KOH(aq) and MgCl2(aq)

Ions: K+, OH

, Mg

2+, Cl

Products: KCl and Mg(OH)2

Solubility (Table 5.4): KCl(aq) and Mg(OH)2(s)

2KOH(aq) + MgCl2(aq) 2KCl(aq) + Mg(OH)2(s) Balanced

(e) Reactants: CuSO4(aq) and BaS(aq)

Ions: Cu2+

, SO42–

, Ba2+

, S2

Products: CuS and BaSO4

Solubility (Table 5.4): CuS(s) and BaSO4(s)

CuSO4(aq) + BaS(aq) CuS(s) + BaSO4(s) Balanced

5.116 (a) NaOH is a base, so it will react with HCl to form a salt and water in an acid-base reaction.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

(b) Zn is a metal that is more active than H2. Therefore, it reacts with HCl to form an ionic compound and

hydrogen gas in a single displacement reaction. Water is not formed during the reaction.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

533

(c) CaCO3 reacts with HCl to form carbonic acid, which decomposes to form water and carbon dioxide

gas.

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2CO3(aq) H2CO3 decomposes

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

(d) Cu(OH)2 is a base, so it will react with HCl to form a salt and water in an acid-base reaction.

Cu(OH)2(s) + 2HCl(aq) CuCl2(aq) + H2O(l)

5.117 Potassium and sodium are in the same family (Group IA (1)), so we expect that they will react similarly

with water. Water reacts with potassium to form potassium hydroxide and hydrogen gas.

2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

5.118 Both are solids prior to adding to water. Let’s describe the dissolving of NaCl in detail. The solid is placed

in the water. As the solid dissolves, the Na+ and Cl

ions separate and become aqueous. For every formula

unit of NaCl that dissolves one sodium ion and one chloride ion form. Overall, the solution remains

neutral. As Cu(NO3)2 dissolves, the Cu2+

and NO3 ions separate and become aqueous. For every formula

unit of Cu(NO3)2 that dissolves, one copper ion and two nitrate ions form. Although the numbers of ions

that form are different, the overall charge in the solution is still balanced. Because compounds containing

nitrate and/or chloride ions are generally soluble, we don’t expect to see any precipitation from the

solution.

5.119 A precipitate of silver chloride will form if a compound which forms chloride ions is added to the solution.

(a) Molecular chlorine does not form chloride ions when placed in water, so we don’t expect to see a

precipitate.

(b) When sodium chloride is added to water, it dissolves, forming sodium and chloride ions. A precipitate

of AgCl(s) should form.

(c) Lead chloride is not soluble in water, so we cannot expect there to be enough Cl– ions added to the

solution to cause AgCl to precipitate.

(d) Carbon tetrachloride is a molecular compound; it does not form chloride ions when added to water.

The only molecular compounds that produce ions are acids and bases.

5.120 We could carefully add sodium sulfate, Na2SO4, to the solution. As we add this solution, we would expect

the following precipitation reaction to take place:

BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq) Molecular equation

Since the sulfate ions precipitate out of the solution, the end result is a solution composed of sodium and

chloride ions. This is clearer if we develop the net ionic equation:

Ba2+


+(aq) + SO4

2(aq) BaSO4(s) + 2Na

+(aq) + 2Cl

(aq) Ionic equation

Next we identify and eliminate the spectator ions:

Ba2+


+(aq) + SO4

2(aq) BaSO4(s) + 2Na

+(aq) + 2Cl

(aq)

Finally, we write the net ionic equation:

Ba2+

(aq) + SO42

(aq) BaSO4(s) Net ionic equation

There are two important conclusions to make. First, the sulfate precipitates and does not remain in the

solution. Second, the spectator ions, Na+ and Cl

, remain in the solution. As a result, no new ions remain

in the solution following precipitation.

5.121 Certain combinations of strong acid and strong base produce the net ionic equation:

H+(aq) + OH

(aq) H2O(l)

Some examples are solutions of HCl and NaOH, HNO3 and KOH, HCl and KOH. This reaction occurs

because strong acids and bases completely ionize in aqueous solution. For example:

534

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Molecular equation

H+(aq) + Cl

(aq) + Na

+(aq) + OH

(aq) Na

+(aq) + Cl

(aq) + H2O(l) Ionic equation

H+(aq) + OH

(aq) H2O(l) Net ionic equation

5.122 Any metal that is higher than cobalt in the activity series (Figure 5.21) can be used to plate cobalt from

solution. However, if we use metals that are too active, they will also react with water (which is not very

efficient). Good choices would be iron, zinc, or aluminum.

5.123 No reaction takes place because copper is less active than iron (Figure 5.21).

5.124 We can look at Table 5.4 to find an anion that will make an insoluble compound with lead(II) but be

soluble with copper(II). Most sulfates are soluble but lead(II) sulfate is insoluble. If we add sodium sulfate,

the lead(II) ions will precipitate as lead(II) sulfate leaving the copper(II) ions in solution. Next we look for

an insoluble compound of copper(II). There are several including copper(II) hydroxide. Adding sodium

hydroxide to the solution will precipitate the copper(II) ions as copper(II) hydroxide.

5.125 To show that KI is a strong electrolyte, we must show that it is an ionic solid that separates into its ions in

solution. To show that CO is a nonelectrolyte, we must show that the CO molecules do not form ions in

solution. Some representative images are shown below.

Solution of KI Solution of CO

K+, I

CO

5.126 See Table 5.2.

(a) Decomposition reactions can be used to produce oxygen gas.

2HgO(s) 2Hg(l) + O2(g)

2H2O2 (aq) 2H2O(l) + O2(g)

(b) Single displacement reactions with very active metals with water produce hydrogen gas.

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

(c) Decomposition of metal chlorides can be used to produce chlorine gas.

PtCl4(s) Pt(s) + 2Cl2(g)

5.127 We can select any metal higher in the activity series (Figure 5.21). (a) Al; (b) Zn; (c) Mg; (d) Sn

5.128 We can identify the solutions by testing them with various active metals. Lead has the lowest activity.

When we place a drop of each solution on a piece of lead foil, only the copper sulfate solution will react.

Lead nitrate does not react. Next, we can place a drop of the three remaining solutions on a piece of tin foil

(Sn). Of the three remaining solutions, only the lead nitrate solution will react. Finally, we can place a

535

drop of each of the two remaining solutions on a piece of zinc. Only the iron sulfate solution will react.

The solution that remains is aluminum sulfate.

5.129 We can apply the solubility rules (Table 5.4) to separate ions.

(a) Adding a potassium chromate solution will cause iron(III) chromate to precipitate, leaving Al3+

in

solution.

(b) Adding a sodium sulfate solution will cause barium sulfate to precipitate, leaving Mg2+

in solution.

(c) Adding a silver perchlorate solution will cause silver chloride to precipitate, leaving NO3 in solution.

(d) Adding water will cause MgSO4 to dissolve, separating it from solid barium sulfate.

5.130 Copper probably reacts with oxygen to form copper(II) oxide as described by the following equation:

2Cu(s) + O2(g) 2CuO(s)

In the presence of water, the oxide forms the hydroxide:

CuO(s) + H2O(l) Cu(OH)2(s)

The reaction of CuO with CO2 produces CuCO3:

CuO(s) + CO2(g) CuCO3(s)

5.131 We can look at Table 5.4, the solubility rules, to find an anion that will make an insoluble compound with

Ag+ but be soluble with Ba

2+. Most chlorides are soluble but AgCl is insoluble. If we add sodium chloride

(NaCl), the Ag+ ions will precipitate as AgCl leaving the Ba

2+ ions in solution. Next we look for an

insoluble compound of Ba2+

. There are several, including BaSO4. Adding sodium sulfate (Na2SO4) to the

solution will precipitate the Ba2+

ions as BaSO4.

5.132 The reaction of a base, in this case Mg(OH)2, with an acid, HCl, is a double- displacement, neutralization

reaction. The products are a salt (i.e. ionic compound) and water. Just as with precipitation reactions we:

1) determine the ion formulas, 2) predict the products based on ion charges, and 3) determine the states of

the products. Acid-base reactions are usually driven by the formation of water (a molecular compound).

We must use the solubility rules to determine whether or not the salt will dissolve in water.

Reactants: Mg(OH)2(s) and HCl(aq)

Ions: Mg2+

, OH, H

+, Cl

Products: MgCl2(aq) and H2O(l)

Balanced equation: Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

5.133 In a single-displacement reaction, one free element displaces another element from a compound. For

example, a metal displaces an ion of another metal, or a nonmetal displaces an ion of another nonmetal. To

determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the

metal is more active than the metal whose ion appears in the compound, a reaction occurs. Nickel is more

active than copper, so we expect a reaction to occur. In the lab we would observe the formation of copper

on the surface of the nickel metal. The Cu2+

ions in solution give it a blue color while nickel ions give no

color to the solution. As the Cu2+

ions react to form solid copper, the blue color of the solution would

become lighter. As the nickel atoms form nickel ions they are removed from the surface of the metal.

Over time we might observe that the nickel metal will appear to disintegrate.


example, a metal displaces an ion of another metal, or a nonmetal displaces an ion of another nonmetal. To

determine whether a reaction occurs, we compare the activities of the metals involved (Figure 5.21). If the

metal is more active than the metal whose ion appears in the compound, a reaction occurs. Silver is less

reactive than copper, so we expect no reaction to occur.

5.135 Hydrochloric acid and sodium hydroxide solutions are both clear and colorless. They react in a double-

displacement, neutralization reaction in which aqueous sodium chloride and water are formed. Aqueous

536

sodium chloride is also clear and colorless so there wouldn’t be an observable change unless we inserted a

thermometer. We would notice a change in temperature due to the formation of water. The formation of

more water molecules in an environment that already has a lot of water would not produce a noticeable

change in volume.

5.136 Using the solubility rules (Table 5.4) as a guide, we could identify a soluble compound containing the

barium ion and one containing the sulfate ion such that when combined they would react in a double-

displacement, precipitation reaction in which insoluble barium sulfate is formed along with a soluble

compound. Since all nitrates are soluble, an aqueous solution of barium nitrate could serve as a source of

barium ion. Sodium sulfate could serve as a source of sulfate ion. Mixing the two solutions would produce

solid barium sulfate and aqueous sodium nitrate:

Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)

The solid barium sulfate can be separated from the solution by filtration.

5.137 We start by writing the skeletal equation:



3 C 8 H 2 O 1 C 2 H 3 O

In the combustion reactions of hydrocarbons it is usually easiest to begin by balancing the carbon and leave

the balancing of oxygen for the last step. This follows the general concepts of balancing the atoms that

appear in the smallest number of substances first (you could have also started with hydrogen) and balancing

the substances appearing as pure elements last. Start with a coefficient of 3 in front of CO2.

C3H8(g) + O2(g) 3CO2(g) + H2O(g) Unbalanced


3 C 8 H 2 O 3 C 2 H 5 O

To balance the hydrogen, we use a coefficient of 4 for H2O

C3H8(g) + O2(g) 3CO2(g) + 4H2O(g) Unbalanced


3 C 8 H 2 O 3 C 8 H 10 O

To balance the oxygen, we use a coefficient of 5 for O2

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Balanced


3 C 8 H 10 O 3 C 8 H 10 O


example, a metal displaces an ion of another metal, or a nonmetal displaces an ion of another nonmetal. In

this case a metal displaces hydrogen from water producing an ionic compound and elemental hydrogen.

To write a balanced equation we: 1) determine the ion formulas if ionic compounds are involved, 2) predict

the products based on ion charges if applicable, and 3) determine the states of the products.

Reactants: Mg(s) and H2O(l)

Ions produced: Mg2+

, OH

Products: Mg(OH)2(s) and H2(g)

Balanced equation: Mg(s) + 2H2O(l) Mg(OH)2(s) + H2(l)

537

5.139 During this reaction atoms of elemental copper with a charge of zero are converted to Cu2+

ions. Each

copper atom that reacts loses two electrons. As a copper atom loses electrons, Ag+ ions are converted to

neutral silver atoms. Each silver ion that reacts gains one electron.

5.140 Barium sulfate is insoluble in water according to the solubility rules (Table 5.4). As an insoluble

compound, toxic barium ions are not freely moving through our bodies to be absorbed into cells. Instead,

the compound passes through the body with very little residual barium left behind.

5.141 (a) C4H10(g) + O2(g) CO2(g) + H2O(g)

Start with an atom count: Reactants: 4 C, 10 H, 2 O Products: 1 C, 2 H, 3 O

First balance the carbon atoms by placing a coefficient in front of carbon dioxide:

C4H10(g) + O2(g) 4CO2(g) + H2O(g)

New atom count: Reactants: 4 C, 10 H, 2 O Products: 4 C, 2 H, 9 O

Now balance the hydrogen atoms by changing the coefficient of water:

C4H10(g) + O2(g) 4CO2(g) + 5H2O(g)


Now balance the oxygen atoms by changing the coefficient of oxygen:

C4H10(g) + 13/2O2(g) 4CO2(g) + 5H2O(g)


Finally, we multiply through by 2 to remove the fractions:

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g)


The equation is now balanced.

(b) C2H5OH(l) + O2(g) CO2(g) + H2O(g)



C2H5OH(l) + O2(g) 2CO2(g) + H2O(g)



C2H5OH(l) + O2(g) 2CO2(g) + 3H2O(g)



C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)



(c) HCO2H(l) + O2(g) CO2(g) + H2O(g)


The carbon and hydrogen atoms are balanced, so we balance the oxygen atoms by placing a coefficient in

front of water:

HCO2H(l) + O2(g) CO2(g) + 2H2O(g)


The hydrogen atoms are no longer balanced, so change the coefficient of formic acid:

2HCO2H(l) + O2(g) CO2(g) + 2H2O(g)


Now balance the carbon and oxygen atoms by changing the coefficient of carbon dioxide:

2HCO2H(l) + O2(g) 2CO2(g) + 2H2O(g)



(d) C5H12(l) + O2(g) CO2(g) + H2O(g)



C5H12(l) + O2(g) 5CO2(g) + H2O(g)

538



C5H12(l) + O2(g) 5CO2(g) + 6H2O(g)



C5H12(l) + 8O2(g) 5CO2(g) + 6H2O(g)



(e) CH3OCH3(g) + O2(g) CO2(g) + H2O(g)



CH3OCH3(g) + O2(g) 2CO2(g) + H2O(g)



CH3OCH3(g) + O2(g) 2CO2(g) + 3H2O(g)



CH3OCH3(g) + 3O2(g) 2CO2(g) + 3H2O(g)



CONCEPT REVIEW

5.142 Answer: D; both occur on the right side of the arrow so they are products

A. both are reactants

B. sulfuric acid is a reactant; iron(II) sulfate is a product

C. iron(II) oxide is a reactant; water is a product

E. iron(II) oxide is a reactant; iron(II) sulfate is a product

5.143 Answer: A; Conservation of matter requires that the total number of atoms does not change.

B. 2H2(g) + O2(g) 2H2O(g) produces heat (Figure 5.5), so the temperature changes.

C. Fe(s) + CuCl2(aq) FeCl2(aq) + Cu(s) changes the color of the solution from blue to colorless (Figure

5.2).

D. 2H2(g) + O2(g) 2H2O(g) has three molecules as reactants and two molecules as products.

E. 2HgO(s) 2Hg(l) + O2(g) has a solid changing to a liquid plus a gas.

5.144 Answer: C; the color change results from dilution of the green substance to give a lighter green.

A. Based on the solubility rules, the white solid is AgCl:

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

B. The gray coating is silver metal and the blue solution contains copper(II) nitrate:

2AgNO3(aq) + Cu(s) 2Ag(s) + Cu(NO3)2(aq)

D. The colorless gas is hydrogen (see the activity series in Figure 5.21):

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

E. This is a combination reaction, forming water: 2H2(g) + O2(g) 2H2O(g)

5.145 Answer: A; the image shows an equal number of the two reactants.

B. There should be a 1:2 ratio of the two reactants. Possibilities are 1 N2, 2 Cl2 or 2 N2, 4 Cl2 or 3 N2, 6

Cl2, etc.

539

C. There should be a 1:2 ratio of the two reactants. Possibilities are 1 N2, 2 O2 or 2 N2, 4 O2 or 3 N2, 6 O2,

etc.

D. There should be a 1:3 ratio of the two reactants. Possibilities are 1 N2, 3 H2 or 2 N2, 6 H2 or 3 N2, 9 H2,

etc.

E. There should be a 1:3 ratio of the two reactants. Possibilities are 1 N2, 3 O2 or 2 N2, 6 O2 or 3 N2, 9 O2,

etc.

5.146 Answer: B; this is the correct formula for sodium chloride and the atoms are balanced.

A. The formula for sodium chloride is incorrect since the sodium ion has a charge of +1 and the chloride

ion has a charge of –1. The equation is not balanced.

C. The formula for sodium chloride is incorrect since the sodium ion has a charge of +1 and the chloride


D. The formula for sodium chloride is incorrect since the sodium ion has a charge of +1 and the chloride


E. The equation is balanced, but the formula for sodium chloride is not the simplest formula.

5.147 Answer: E since a compound is converted to another compound and an element.

The reverse of this reaction would be a combination reaction.

5.148 Answer D; a metal reacting with a metal compound can potentially undergo a single-displacement reaction;

the activity series in Figure 5.21 indicates that magnesium is a more active metal than nickel, so the

reaction should occur.

A. The formulas of these two compounds are incorrect; the common charge of these metal ions is +2, not

+1.

B. It would be impossible to balance the reaction with these products, since there would be more ionic

charge on one side of the equation than on the other.

C. If the nickel(II) ion is going to change to the metal, the magnesium must change to its cation.

E. The arguments for the correct answer eliminate this possibility.

5.149 Answer: A, B, and D; all of these choices represent the reaction of a substance with oxygen gas.

C. Either change S to O2 or change H2 to O2.

E. Either change H2 to O2 or change Cl2 to O2.

5.150 Answer: D since CaCO3 is an insoluble compound according to the solubility rules.

A. Lithium carbonate is soluble, while calcium carbonate is insoluble, so this does not represent the

chemical change.

B. Lithium chloride is soluble, while calcium carbonate is insoluble, so this does not represent the

chemical change. In addition, the equation does not use the smallest possible coefficients.

C. Lithium chloride is soluble, while calcium carbonate is insoluble, so this does not represent the

chemical change.

E. The equation is balanced, but it is a “molecular” equation, not a net ionic equation.

5.151 Answer: A since the net ionic equation is Br–(aq) + Ag

+(aq) AgBr(s).

B. Br– is not a spectator ion since it is involved in the formation of the precipitate.

C. Ag+ is not a spectator ion since it is involved in the formation of the precipitate.

D. Br– is not a spectator ion since it is involved in the formation of the precipitate.

E. NO3– is a spectator ion, but this answer ignores that Co

2+ is also a spectator ion.

chapter 5 – chemical reactions and equationschemweb/chem1000... · chapter 5 – chemical...

Documents