University High School Chemistry: Chemical Reactions

Chemical Reactions

Chemical Reactions and Equations

Lesson Objectives

The student will:

explain what happens during a chemical reaction.
identify the reactants and products in any chemical reaction.
convert verbal descriptions of chemical reactions into chemical equations, and vice versa.
use the common symbols $(s)$ , $(l)$ , $(g)$ , $(aq)$ , and $\rightarrow$ appropriately.

Vocabulary

chemical reaction
products
reactants

Introduction

In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that make up the substances separate and re-arrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances.

In order to describe a chemical reaction, we need to indicate what substances are present at the beginning and what substances are present at the end. The substances that are present at the beginning are called reactants, and the substances present at the end are called products.

Writing Chemical Equations

When sulfur dioxide is added to oxygen, sulfur trioxide is produced. In the chemical equation shown below, sulfur dioxide and oxygen $(\text{SO}_2$ and $\text{O}_2)$ are reactants, and sulfur trioxide $(\text{SO}_3)$ is the product.

The general equation for a reaction is:

: Reactants $\rightarrow$ Products

There are a few special symbols that we need to know in order to communicate in chemical shorthand. In Table below is a summary of the major symbols used in chemical equations. There are other symbols, but these are the main ones that we need to know.

Common Symbols in Chemical Reactions
Symbol	Meaning	Example
$\rightarrow$	separates reactants from products; can be read as “to produce” or “to yield”	$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$
+	separate reactants from each other or products from each other; can be read as “is added to”	$\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$
$(s)$	in the solid state	sodium in the solid state = $\text{Na}_{(s)}$
$(l)$ or $(L)$	in the liquid state	water in the liquid state = $\text{H}_2\text{O}_{(l)}$
$(g)$	in the gaseous state	carbon dioxide in the gaseous state = $\text{CO}_{2(g)}$
$(aq)$	in the aqueous state, dissolved in water	sodium chloride solution = $\text{NaCl}_{(aq)}$

Chemists have a choice of methods for describing a chemical reaction. They could draw a picture of the chemical reaction, like in the image shown below.

Alternatively, they could describe the reaction in words. The image above can be described as two molecules of hydrogen gas reacting with one molecule of oxygen gas to produce two molecules of water vapor.

Chemists could also write the equation in chemical shorthand.

: $2 \text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2 \text{H}_2\text{O}_{(g)}$

In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products, and symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations. For example, we could write out that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. In shorthand, however, we could simply write:

: $\text{Ca(NO}_3)_{2(aq)} + 2 \text{NaOH}_{(aq)} \rightarrow \text{Ca(OH)}_{2(s)} + 2 \text{NaNO}_{3(aq)}$

How much easier is that to read? Let's try it in reverse. Look at the following reaction in shorthand notation and describe the reaction in words.

: $\text{Cu}_{(s)} + \text{AgNO}_{3(aq)} \rightarrow \text{Cu(NO}_3)_{2(aq)} + \text{Ag}_{(s)}$

The description of this reaction might read something like “solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper(II) nitrate and solid silver.”

Example:

Transfer the following symbolic equations into verbal descriptions or vice versa.

$\text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)}$
Gaseous propane, $\text{C}_3\text{H}_8$ , burns in oxygen gas to produce gaseous carbon dioxide and liquid water.
Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride, liquid water, and gaseous carbon dioxide.

Solution:

An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water.
$\text{C}_3\text{H}_{8(g)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$
$\text{HF}_{(g)} + \text{K}_2\text{CO}_{3(aq)} \rightarrow \text{KF}_{(aq)} + \text{H}_2\text{O}_{(l)} + \text{CO}_{2(g)}$

Lesson Summary

A chemical reaction is the process in which one or more substances are changed into one or more new substances.
Chemical reactions are represented by chemical equations.
Chemical equations have reactants on the left, an arrow that symbolizes “yields,” and the products on the right.

Review Questions

Mothballs are commonly used to preserve clothing during off-season. We recognize mothballs by its smell because of a chemical compound known as naphthalene, $\text{C}_{10}\text{H}_8$ . What are the different elements found in naphthalene, and how many atoms of each are found in the formula?
Give the verbal description of the following chemical equations.
1. $\text{H}_2\text{SO}_{4(aq)} + \text{NaCN}_{(aq)} \rightarrow \text{HCN}_{(aq)} + \text{Na}_2\text{SO}_{4(aq)}$
2. $\text{Cu}_{(s)} + \text{AgNO}_{3(aq)} \rightarrow \text{Ag}_{(s)} + \text{Cu(NO}_3)_{2(aq)}$
3. $\text{Fe}_{(s)} + \text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}$
Write the chemical equations for the following reactions.
1. Solid calcium metal is placed in liquid water to produce aqueous calcium hydroxide and hydrogen gas.
2. Aqueous sodium hydroxide is mixed with gaseous chlorine to produce aqueous solutions of sodium chloride and sodium hypochlorite plus liquid water.
3. Solid xenon hexafluoride is mixed with liquid water to produce solid xenon trioxide and gaseous hydrogen fluoride.

Balancing Chemical Equations

Lesson Objectives

The student will:

explain the roles of subscripts and coefficients in chemical equations.
write a balanced chemical equation when given the unbalanced equation for any chemical reaction.
explain the role of the law of conservation of mass in a chemical reaction.

Vocabulary

balanced chemical equation
coefficient
subscript

Introduction

Even though chemical compounds are broken up to form new compounds during a chemical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products. The atoms are merely re-organized into different arrangements. In a complete chemical equation, the two sides of the equation must be balanced. That is, in a balanced chemical equation, the same number of each atom must be present on the reactant and product sides of the equation.

Balancing Equations

The process of writing a balanced chemical equation involves three steps. As a beginning chemistry student, you will not know whether or not two given compounds will react or not. Even if you saw them react, you would not know what the products are without running any tests to identify them. Therefore, for the time being, you will be told both the reactants and products in any equation you are asked to balance.

Step 1: Know what the reactants and products are, and write a word equation for the reaction.

Step 2: Write the formulas for all the reactants and products.

Step 3: Adjust the coefficients to balance the equation.

There are a number of elements shown in Table below that exist as diatomic molecules under normal conditions. When any of these elements appear in word equations, you must remember that the name refers to the diatomic molecule and insert the diatomic formula into the symbolic equation. If, under unusual circumstances, it was desired to refer to the individual atoms of these elements, the text would refer specifically to atomic hydrogen, atomic oxygen, and so on.

Homonuclear Diatomic Molecules
Element	Formula for Diatomic Molecule	Phase at Room Temperature
Hydrogen	$\text{H}_2$	Gaseous
Oxygen	$\text{O}_2$	Gaseous
Nitrogen	$\text{N}_2$	Gaseous
Chlorine	$\text{Cl}_2$	Gaseous
Fluorine	$\text{F}_2$	Gaseous
Bromine	$\text{Br}_2$	Liquid
Iodine	$\text{I}_2$	Solid

There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the chemical formulas of the reactants and products, and there are coefficientsthat are placed in front of the formulas to indicate how many molecules of that substance are used or produced. In the chemical formula shown below, the coefficients and subscripts are labeled.

The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper(II) nitrate and two moles of solid silver. Recall that a subscript of 1 is not written - when no subscript appears for an atom in a formula, it is understood that only one atom is present. The same is true in writing balanced chemical equations. If only one atom or molecule is present, the coefficient of 1 is omitted.

The subscripts are part of the formulas, and once the formulas for the reactants and products are determined, the subscripts may not be changed. The coefficients indicate the mole ratios of each substance involved in the reaction and may be changed in order to balance the equation. Coefficients are inserted into the chemical equation in order to make the total number of each atom on both sides of the equation equal. Note that equation balancing is accomplished by changing coefficients, never by changing subscripts.

Example:

Write a balanced equation for the reaction that occurs between chlorine gas and aqueous sodium bromide to produce liquid bromine and aqueous sodium chloride.

Step 1: Write the word equation (keeping in mind that chlorine and bromine refer to the diatomic molecules).

: chlorine + sodium bromide yields bromine + sodium chloride

Step 2: Substitute the correct formulas into the equation.

: $\text{Cl}_2 + \text{NaBr} \rightarrow \text{Br}_2 + \text{NaCl}$

Step 3: Insert coefficients where necessary to balance the equation.

By placing a coefficient of 2 in front of $\text{NaBr}$ , we can balance the bromine atoms. By placing a coefficient of 2 in front of the $\text{NaCl}$ , we can balance the chlorine atoms.

: $\text{Cl}_2 + 2 \ \text{NaBr} \rightarrow \text{Br}_2 + 2 \ \text{NaCl}$

A final check (always do this) shows that we have the same number of each atom on the two sides of the equation. We have also used the smallest whole numbers possible as the coefficients, so this equation is properly balanced.

Example:

Write a balanced equation for the reaction between aluminum sulfate and calcium bromide to produce aluminum bromide and calcium sulfate. Recall that polyatomic ions usually remain together as a unit throughout a chemical reaction.

Step 1: Write the word equation.

: aluminum sulfate + calcium bromide yields aluminum bromide + calcium sulfate

Step 2: Replace the names of the substances in the word equation with formulas.

: $\text{Al}_2(\text{SO}_4)_3 + \text{CaBr}_2 \rightarrow \text{AlBr}_3 + \text{CaSO}_4$

Step 3: Insert coefficients to balance the equation.

In order to balance the aluminum atoms, we must insert a coefficient of 2 in front of the aluminum compound in the products.

: $\text{Al}_2(\text{SO}_4)_3 + \text{CaBr}_2 \rightarrow 2 \ \text{AlBr}_3 + \text{CaSO}_4$

In order to balance the sulfate ions, we must insert a coefficient of 3 in front of the product $\text{CaSO}_4$ .

: $\text{Al}_2(\text{SO}_4)_3 + \text{CaBr}_2 \rightarrow 2 \ \text{AlBr}_3 + 3 \ \text{CaSO}_4$

In order to balance the bromine atoms, we must insert a coefficient of 3 in front of the reactant $\text{CaBr}_2$ .

: $\text{Al}_2(\text{SO}_4)_3 + 3 \ \text{CaBr}_2 \rightarrow 2 \ \text{AlBr}_3 + 3 \ \text{CaSO}_4$

The insertion of the 3 in front of the reactant $\text{CaBr}_2$ also balances the calcium atoms in the product $\text{CaSO}_4$ . A final check shows that there are two aluminum atoms, three sulfur atoms, twelve oxygen atoms, three calcium atoms, and six bromine atoms on each side. This equation is balanced.

Note that this equation would still have the same number of atoms of each type on each side with the following set of coefficients:

: $2 \ \text{Al}_2(\text{SO}_4)_3 + 6 \ \text{CaBr}_2 \rightarrow 4 \ \text{AlBr}_3 + 6 \ \text{CaSO}_4$

Count the number of each type of atom on either side of the equation to confirm that this equation is “balanced.” While this set of coefficients does “balanced” the equation, they are not the lowest set of coefficients possible. Chemical equations should be balanced with the simplest whole number coefficients. We could divide each of the coefficients in this equation by 2 to get another set of coefficients that still balance the equation and are whole numbers. Since it is required that an equation be balanced with the lowest whole number coefficients, the equation above is not properly balanced. When you have finished balancing an equation, you should not only check to make sure it is balanced, you should also check to make sure that it is balanced with the simplest set of whole number coefficients possible.

Example:

Balance the following skeletal equation. (The term “skeletal equation” refers to an equation that has the correct chemical formulas but does not include the proper coefficients.)

: $\text{Fe(NO}_3)_3 + \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \ \text{NaNO}_3 \ \ \ \ \ \ \text{(skeletal equation)}$

Solution:

We can balance the hydroxide ion by inserting a coefficient of 3 in front of the $\text{NaOH}$ on the reactant side.

: $\text{Fe(NO}_3)_3 + 3 \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaNO}_3$

We can then balance the nitrate ions by inserting a coefficient of 3 in front of the sodium nitrate on the product side.

: $\text{Fe(NO}_3)_3 + 3 \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \ \text{NaNO}_3$

Counting the number of each type of atom on the two sides of the equation will now show that this equation is balanced.

Example:

Given the following skeletal (un-balanced) equations, balance them.

$\text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}$
$\text{H}_2\text{SO}_{4(aq)} + \text{Al(OH)}_{3(aq)} \rightarrow \text{Al}_2(\text{SO}_4)_{3(aq)} + \text{H}_2\text{O}_{(l)}$
$\text{Ba(NO}_3)_{2(aq)} + \text{Na}_2\text{CO}_{3(aq)} \rightarrow \text{BaCO}_{3(aq)} + \text{NaNO}_{3(aq)}$
$\text{C}_2\text{H}_{6(g)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$

Solution:

$\text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}$ (sometimes the skeletal equation is already balanced)
$3 \ \text{H}_2\text{SO}_{4(aq)} + 2 \ \text{Al(OH)}_{3(aq)} \rightarrow \text{Al}_2(\text{SO}_4)_{3(aq)} + 6 \ \text{H}_2\text{O}_{(l)}$
$\text{Ba(NO}_3)_{2(aq)} + \text{Na}_2\text{CO}_{3(aq)} \rightarrow \text{BaCO}_{3(aq)} + 2 \ \text{NaNO}_{3(aq)}$
$2 \ \text{C}_2\text{H}_{6(g)} + 7 \ \text{O}_{2(g)} \rightarrow 4 \ \text{CO}_{2(g)} + 6 \ \text{H}_2\text{O}_{(l)}$

Conservation of Mass in Chemical Reactions

We already know from the law of conservation of mass that mass is conserved in chemical reactions. But what does this really mean? Consider the following reaction.

: $\text{Fe(NO}_3)_3 + 3 \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \ \text{NaNO}_3$

Verify to yourself that this equation is balanced by counting the number of each type of atom on each side of the equation. We can demonstrate that mass is conserved by determining the total mass on both sides of the equation.

Mass of the Reactant Side:

: Total mass of reactants = 241.9 daltons + 120. daltons = 361.9 daltons

: 3 molecules of $\text{NaOH} \times \text{molecular weight} = (3) \cdot (40.0 \ \text{daltons})$ = 120. daltons

: 1 molecule of $\text{Fe(NO}_3)_3 \times \text{molecular weight} = (1) \cdot (241.9 \ \text{daltons})$ = 241.9 daltons

Product Side Mass:

: Total mass of products = 106.9 daltons + 255 daltons = 361.9 daltons

: 3 molecules of $\text{NaNO}_3 \times \text{molecular weight} = (3) \cdot (85.0 \ \text{daltons})$ = 255 daltons

: 1 molecule of $\text{Fe(OH)}_3 \times \text{molecular weight} = (1) \cdot (106.9 \ \text{daltons})$ = 106.9 daltons

As you can see, both the number of atoms and mass are conserved during chemical reactions. This is logically similar to saying that a group of 20 objects stacked in different ways will still have the same total mass no matter how you stack them.

Lesson Summary

Chemical equations must always be balanced.
Balanced chemical equations have the same number and type of each atom on both sides of the equation.
The coefficients in a balanced equation must be the simplest whole number ratio.
Mass is always conserved in chemical reactions.

Review Questions

Explain in your own words why coefficients can change but subscripts must remain constant.
Which set of coefficients will properly balance the following equation: ?
1. 1, 1, 1, 1
2. 1, 3, 2, 2
3. 1, 3.5, 2, 3
4. 2, 7, 4, 6
When properly balanced, what is the sum of all the coefficients in the following chemical equation: ?
1. $4$
2. $7$
3. $9$
4. None of the above
When the following equation is balanced, what is the coefficient found in front of the : ?
1. $1$
2. $3$
3. $5$
4. $7$
Balance the following equations.
1. $\text{XeF}_{6(s)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{XeO}_{3(s)} + \text{HF}_{(g)}$
2. $\text{Cu}_{(s)} + \text{AgNO}_{3(aq)} \rightarrow \text{Ag}_{(s)} + \text{Cu(NO}_3)_{2(aq)}$
3. $\text{Fe}_{(s)} + \text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}$
4. $\text{Al(OH)}_3 + \text{Mg}_3(\text{PO}_4)_2 \rightarrow \text{AlPO}_4 + \text{Mg(OH)}_2$

Types of Reactions

Lesson Objectives

The student will:

describe what is occurring in synthesis, decomposition, single replacement, double replacement, and combustion reactions.
classify a chemical reaction as a synthesis, decomposition, single replacement, double replacement, or a combustion reaction.
predict the products of simple reactions.

Vocabulary

combustion reaction
decomposition reaction
double replacement reaction
hydrocarbon
single replacement reaction
synthesis reaction

Introduction

Chemical reactions are classified into types to help us analyze them and to help us predict what the products of the reaction will be. The five major types of chemical reactions are synthesis, decomposition, single replacement, double replacement, and combustion.

Synthesis Reactions

A synthesis reaction is one in which two or more reactants combine to make one product. The general equation for a synthesis reaction is:

: $A + B \rightarrow AB$

Synthesis reactions occur as a result of two or more simpler elements or molecules combining to form a more complex molecule. We can always identify a synthesis reaction because there is only one product. If you are given elemental reactants and told that the reaction is a synthesis reaction, you should be able to predict the products. For example, consider the equation below. Two elements (hydrogen and oxygen) combine to form one product (water).

: $2 \ \text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(l)}$

You should also be able to write the chemical equation for a synthesis reaction if you are given a product by picking out its elements and writing the equation. As a result, we can write the synthesis reaction for sodium chloride just by knowing the elements that are present in the product.

: $2 \ \text{Na}_{(s)} + \text{Cl}_{2(g)} \rightarrow 2 \ \text{NaCl}_{(s)}$

Example:

Write the chemical equation for the synthesis reaction of silver bromide, $\text{AgBr}$ .
Predict the products for the following reaction: $\text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$ .
Predict the products for the following reaction: $\text{Li}_2\text{O}_{(s)} + \text{CO}_{2(g)}$ .

Solution:

$2 \ \text{Ag}_{(s)} + \text{Br}_{2(l)} \rightarrow 2 \ \text{AgBr}_{(s)}$
$\text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{H}_2\text{CO}_{3(aq)}$
$\text{Li}_2\text{O}_{(s)} + \text{CO}_{2(g)} \rightarrow \text{Li}_2\text{CO}_{3(s)}$

Decomposition Reactions

When one type of reactant breaks down to form two or more products, we have a decomposition reaction. The best way to remember a decomposition reaction is that for all reactions of this type, there is only one reactant. The general equation for a decomposition reaction is:

: $AB \rightarrow A + B$

Look at the equation below for an example of a decomposition reaction. In this reaction, ammonium nitrate breaks down to form dinitrogen oxide and water.

: $\text{NH}_4\text{NO}_{3(s)} \rightarrow \text{N}_2\text{O}_{(g)} + 2 \text{H}_2\text{O}_{(g)}$

Notice that there is only one reactant, $\text{NH}_4\text{NO}_3$ , on the left of the arrow and that there is more than one on the right side of the arrow. This is the exact opposite of a synthesis reaction.

When studying decomposition reactions, we can predict the reactants in a similar manner as we did for synthesis reactions. Look at the formula for magnesium nitride, $\text{Mg}_3\text{N}_2$ . What elements do you see in this formula? You see magnesium and nitrogen. Now we can write a decomposition reaction for magnesium nitride.

: $\text{Mg}_3\text{N}_{2(s)} \rightarrow 3 \ \text{Mg}_{(s)} + \text{N}_{2(g)}$

Notice there is only one reactant.

Example:

Write the chemical equation for the decomposition of the following compounds into their individual elements:

$\text{Al}_2\text{O}_3$
$\text{Ag}_2\text{S}$
$\text{MgO}$

Solution:

$2 \ \text{Al}_2\text{O}_3 \rightarrow 4 \ \text{Al} + 3 \ \text{O}_2$
$\text{Ag}_2\text{S} \rightarrow 2 \ \text{Ag} + \text{S}$
$2 \ \text{MgO} \rightarrow 2 \ \text{Mg} + \text{O}_2$

Single Replacement Reactions

A third type of reaction is the single replacement reaction. In single replacement reactions, one element reacts with one compound to form products. The single element is said to replace an element in the compound when the products form, hence the name single replacement.

There are actually three different types of single replacement reactions: 1) the single element is a metal and replaces the metal in the second reactant, 2) the single element is a metal and replaces the hydrogen in the second reactant, which is always an acid, and 3) the single element is a nonmetal and replaces the nonmetal in the second reactant.

Replacement of a Metal with a Metal

In this section, we will focus on single replacement reactions where an elemental metal reactant replaces the metal (or the cation) of a second compound. The general equation for this reaction is:

: $A + BC \rightarrow B + AC$

Consider the following example. Notice there is only one reactant that is an element and one reactant that is a compound.

: $\text{Zn}_{(s)} + \text{Cu(NO}_3)_{2(aq)} \rightarrow \text{Zn(NO}_3)_{2(aq)} + \text{Cu}_{(s)}$

When studying single replacement reactions, we can predict reactants in a similar manner as we did for synthesis and decomposition reactions. Suppose that we know a single replacement reaction will occur between solid aluminum and solid iron(III) oxide.

: $\text{Al}_{(s)} + \text{Fe}_2\text{O}_{3(s)}$

In order to predict the products, we need to know that aluminum will replace iron and form aluminum oxide. Aluminum has a charge of $+3$ (it is in Group 3A), and oxygen has a charge of $-2$ (it is in Group 6A). The compound formed between aluminum and oxygen, therefore, will be $\text{Al}_2\text{O}_{3(s)}$ . Since iron is replaced in the compound by aluminum, the iron product will now be a single element. The unbalanced equation will be:

: $\text{Al}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow \text{Al}_2\text{O}_{3(s)} + \text{Fe}_{(s)}$ .

The balanced equation will be:

: $2 \ \text{Al}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow \text{Al}_2\text{O}_{3(s)} + 2 \ \text{Fe}_{(s)}$

Example:

Write the chemical equation for the single replacement reaction between zinc solid and lead(II) nitrate solution to produce zinc nitrate solution and solid lead.
Predict the products for the following reaction: $\text{Fe} + \text{CuSO}_4$ .
Predict the products for the following reaction: $\text{Al} + \text{CuCl}_2$ .

Solution:

$\text{Zn}_{(s)} + \text{Pb(NO}_3)_{2(aq)} \rightarrow \text{Pb}_{(s)} + \text{Zn(NO}_3)_{2(aq)}$
$\text{Fe}_{(s)} + \text{CuSO}_{4(aq)} \rightarrow \text{Cu}_{(s)} + \text{FeSO}_{4(aq)}$
$2 \ \text{Al} + 3 \ \text{CuCl}_2 \rightarrow 3 \ \text{Cu} + 2 \ \text{AlCl}_3$

Replacement of Hydrogen with a Metal

These reactions are the same as those studied in the last section, except the compound in the reactant side of the equation is always an acid. Since you may not have studied acids yet, you should consider an acid to be a compound in which hydrogen is combined with an anion. Therefore, in this section, we will consider single replacement reactions where the element reactant replaces the hydrogen in the acid compound. The general reaction is:

: $\mathrm{A} + 2 \ \mathrm{HX} \rightarrow \mathrm{AX}_2 + \mathrm{H}_2$

The chemical equation below is an example of this type of reaction:

: $\text{Zn}_{(s)} + 2 \ \text{HBr}_{(aq)} \rightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)}$

When studying these single replacement reactions, we can predict reactants in a similar manner as we did for the other types of single replacement reactions. Look at the reaction below. Since $\text{HCl}$ is a compound that has hydrogen combined with an anion, it is an acid.

: $\text{Mg}_{(s)} + 2 \text{HCl}_{(aq)}$

In order to predict the products, we need to know that magnesium will replace hydrogen and form magnesium chloride. Magnesium has a charge of $+2$ (it is in Group 2A), and chlorine has a charge of $-1$ (it is in group 7A). Therefore, the compound formed will be $\text{MgCl}_2$ . The replaced hydrogen forms the product elemental hydrogen, $\text{H}_2$ . After the products are determined, all that remains is to balance the equation.

: $\text{Mg}_{(s)} + 2 \ \text{HCl}_{(aq)} \rightarrow \text{MgCl}_{2(s)} + \text{H}_{2(g)}$

Notice that one reactant is the element $\text{Mg}$ and one reactant is an acid compound. The $\text{Mg}$ has replaced the hydrogen in $\text{HCl}$ in the same manner as $\text{Zn}$ replaced the hydrogen in the $\text{HBr}$ in the example above.

Example:

Write the chemical equation for the single replacement reaction between iron solid and hydrochloric acid solution to produce iron(II) chloride solution and hydrogen gas.
Predict the products for the following reaction: $\text{Zn}_{(s)} + \text{H}_2\text{SO}_{4(aq)}$ .
Predict the products for the following reaction: $\text{Al}_{(s)} + \text{HNO}_{3(aq)}$ .

Solution:

$\text{Fe}_{(s)} + 2 \ \text{HCl}_{(aq)} \rightarrow \text{FeCl}_{2(aq)} + \text{H}_{2(g)}$
$\text{Zn}_{(s)} + \text{H}_2\text{SO}_{4(aq)} \rightarrow \text{ZnSO}_{4(aq)} + \text{H}_{2(g)}$
$2 \ \text{Al}_{(s)} + 6 \ \text{HNO}_{3(aq)} \rightarrow 2 \ \text{Al(NO}_3)_{3(aq)} + 3 \ \text{H}_{2(g)}$

Replacement of a Nonmetal with a Nonmetal

In this section, we will focus on the final type of single replacement reactions where the element reactant replaces the nonmetal (or the anion) in a compound. The general equation for this type of reaction is:

: $A + BC \rightarrow C + BA$

Notice in the equation below that the chlorine replaced the iodine to produce solid iodine as a product. We can predict the products for these single replacement reactions in a similar manner as for all other single replacement reactions. The only difference here is that we have to remember that we are replacing the anion of the compound rather than the cation.

: $\text{Cl}_{2(g)} + 2 \ \text{KI}_{(aq)} \rightarrow 2 \ \text{KCl}_{(aq)} + \text{I}_{2(s)}$

Look at the reaction between chlorine gas and sodium bromide shown below. This is an actual method for extracting bromine from ocean water found to contain sodium bromide. Can you complete the reaction?

: $\text{Cl}_{2(g)} + \text{NaBr}_{(aq)} \rightarrow ?$

In order to predict the products of this reaction, we need to know that chlorine will replace bromine and form sodium chloride. Sodium has a charge of $+1$ (it is in Group 1A) and chlorine has a charge of $-1$ (it is in group 7A). The compound formed will be $\text{NaCl}$ .

: $\text{Cl}_{2(g)} + 2 \ \text{NaBr}_{(aq)} \rightarrow 2 \ \text{NaCl}_{(aq)} + \text{Br}_{2(l)}$

Notice, as with all of the other single replacement reactions, that the reactants include one element and one compound, and the products contain one element and one compound. This is the determining factor for identifying whether you have a single replacement reaction.

Example:

Write the chemical equation for the single replacement reaction between sodium iodide solution and liquid bromine to produce sodium bromide solution and solid iodine.
Predict the products for the following reaction: $\text{Br}_{2(aq)} + \text{KI}_{(aq)}$ .
Predict the products for the following reaction: $\text{MgI}_{2(aq)} + \text{Cl}_{2(aq)}$ .

Solution:

$2 \ \text{NaI}_{(aq)} + \text{Br}_{2(l)} \rightarrow 2 \ \text{NaBr}_{(aq)} + \text{I}_{2(s)}$
$\text{Br}_{2(aq)} + 2 \ \text{KI}_{(aq)} \rightarrow 2 \ \text{KBr}_{(aq)} + \text{I}_{2(s)}$
$\text{MgI}_{2(aq)} + \text{Cl}_{2(aq)} \rightarrow \text{MgCl}_{2(aq)} + \text{I}_{2(s)}$

Double Replacement

For double replacement reactions, two reactants will react by having the cations exchange places. The key to identifying this type of reaction is to recognize that it has two compounds as reactants. This type of reaction is more common than any of the others, and there are many different types of double replacement reactions. Some double replacement reactions are more common than others. For example, precipitation and neutralization reactions are two of the most common double replacement reactions. Precipitation reactions are ones where two aqueous reactants combine to form products where one of the products is an insoluble solid. A neutralization reaction is one where the two reactant compounds are an acid and a base and the two products are a salt and water.

Example:

The following is a precipitation reaction because $\text{AgCl}_{(s)}$ is formed.

: $\text{AgNO}_{3(aq)} + \text{NaCl}_{(aq)} \rightarrow \text{AgCl}_{(s)} + \text{NaNO}_{3(aq)}$

The following is a neutralization reaction because the acid, $\text{H}_2\text{SO}_4$ , is neutralized by the base, $\text{NaOH}$ .

: $2 \ \text{NaOH}_{(aq)} + \text{H}_2\text{SO}_{4(aq)} \rightarrow \text{Na}_2\text{SO}_{4(aq)} + 2 \ \text{H}_2\text{O}_{(l)}$

In order to write the products for a double displacement reaction, you must be able to determine the correct formulas for the new compounds. Consider this common laboratory experiment that involves the reaction between lead(II) nitrate and sodium iodide, both clear solutions. Here is the start of the reaction:

: $\text{Pb(NO}_3)_{2(aq)} + \text{NaI}_{(aq)}$

Now, predict the products based on what you know about charges. We know that the cations exchange anions. We now have to look at the charges of each of the cations and anions to see what the products will be.

We should presume the charge of the lead will remain $+2$ , and since iodine forms ions with a charge of $-1$ , one product will be $\text{PbI}_2$ . The other product will form between the sodium ion, whose charge is $+1$ , and the nitrate ion, whose charge is $-1$ . Therefore, the second product will be $\text{NaNO}_3$ . Once the products are written in, the equation can be balanced.

: $\text{Pb(NO}_3)_{2(aq)} + 2 \ \text{NaI}_{(aq)} \rightarrow \text{PbI}_{2(s)} + 2 \ \text{NaNO}_{3(aq)}$

The experiment produces a brilliant yellow precipitate. If you have use of a solubility table, it is easy to determine that the precipitate will be the lead(II) iodide. Even without a solubility table, knowing that lead compounds tend to precipitate and sodium compounds are always soluble, we would still be able to determine that the $\text{PbI}_2$ is the brilliant yellow precipitate.

Look at the reaction between acetic acid and barium hydroxide below.

: $\text{HC}_2\text{H}_3\text{O}_{2(aq)} + \text{Ba(OH)}_{2(aq)} \rightarrow \text{?}$

Try to predict the products by having the cations exchange places and writing the correct formulas for the products formed.

: $\text{HC}_2\text{H}_3\text{O}_{2(aq)} + \text{Ba(OH)}_{2(aq)} \rightarrow \text{Ba(C}_2\text{H}_3\text{O}_2)_{2(aq)} + \text{H}_2\text{O}_{(l)} \ \text{(not balanced)}$

Therefore, the final balanced equation will be:

: $2 \ \text{HC}_2\text{H}_3\text{O}_{2(aq)} + \text{Ba(OH)}_{2(aq)} \rightarrow \text{Ba(C}_2\text{H}_3\text{O}_2)_{2(aq)} + 2 \ \text{H}_2\text{O}_{(l)} \ \text{(balanced)}$

This is an acid-base reaction yielding salt, barium acetate, and water. Notice that $\text{HOH}$ and $\text{H}_2\text{O}$ are the same.

Example:

Write a chemical equation for the double replacement reaction between calcium chloride solution and potassium hydroxide solution that produces potassium chloride solution and a precipitate of calcium hydroxide.
Predict the products for the following reaction: $\text{AgNO}_{3(aq)} + \text{NaCl}_{(aq)}$ .
Predict the products for the following reaction: $\text{FeCl}_{3(aq)} + \text{KOH}_{(aq)}$ .

Solution:

$\text{CaCl}_{2(aq)} + 2 \ \text{KOH}_{(aq)} \rightarrow \text{Ca(OH)}_{2(s)} + 2 \ \text{KCl}_{(aq)}$
$\text{AgNO}_{3(aq)} + \text{NaCl}_{(aq)} \rightarrow \text{AgCl}_{(s)} + \text{NaNO}_{3(aq)}$
$\text{FeCl}_{3(aq)} + 3 \ \text{KOH}_{(aq)} \rightarrow \text{Fe(OH)}_{3(s)} + 3 \ \text{KCl}_{(aq)}$

Combustion

A special type of single replacement reaction deserves some attention. These reactions are combustion reactions. In a combustion reaction, oxygen reacts with another substance to produce carbon dioxide and water.

In a particular branch of chemistry, known as organic chemistry, we study compounds known as hydrocarbons. A hydrocarbon is an organic substance consisting of only hydrogen and carbon. Combustion reactions usually have a hydrocarbon reacting with oxygen to produce $\text{CO}_2$ and $\text{H}_2\text{O}$ . In other words, the only part that changes from one combustion reaction to the next is the actual hydrocarbon involved in the reaction. The general equation for this reaction is given below. Notice oxygen, carbon dioxide, and water are listed in the general equation to show that these reactants and products remain the same from combustion reaction to combustion reaction.

: $\text{hydrocarbon} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$

Look at the the combustion reaction of octane, $\text{C}_8\text{H}_{18}$ , shown below. Octane has 8 carbon atoms, hence the prefix “oct-.”

: $2 \ \text{C}_8\text{H}_{18(l)} + 25 \ \text{O}_{2(g)} \rightarrow 16 \ \text{CO}_{2(g)} + 18 \ \text{H}_2\text{O}_{(l)}$

This reaction is referred to as complete combustion. Complete combustion reactions occur when there is enough oxygen to burn the entire hydrocarbon. This is why the only products are carbon dioxide and water.

Have you ever been in a lab and seen black soot appear on the bottom of a heated beaker? Or, have you ever seen the black puffs of smoke come out from the exhaust pipe of a car? If there is not enough oxygen, the result is an incomplete combustion reaction with $\text{CO}_{(g)}$ and $\text{C}_{(s)}$ (in the form of soot) also formed as products. Incomplete combustion reactions are actually quite dangerous because one of the products in the reaction is carbon monoxide, not carbon dioxide. Carbon monoxide is a gas that prevents oxygen from binding to the oxygen transport proteins in our blood cells. When the concentration of carbon monoxide in the blood becomes too high, not enough oxygen can be transported and the person can die.

Example:

Identify whether each of the following reactions are complete or incomplete combustions, and then balance the equation.

$\text{C}_7\text{H}_{16(l)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$
$\text{C}_3\text{H}_{8(g)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$
$\text{CH}_{4(g)} + \text{O}_{2(g)} \rightarrow \text{CO}_{(g)} + \text{H}_2\text{O}_{(l)}$
$\text{C}_5\text{H}_{12(l)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$
$\text{C}_2\text{H}_{6(g)} + \text{O}_{2(g)} \rightarrow \text{CO}_{(g)} + \text{H}_2\text{O}_{(l)}$

Solution:

Complete; $\text{C}_7\text{H}_{16(l)} + 11 \ \text{O}_{2(g)} \rightarrow 7 \ \text{CO}_{2(g)} + 8 \ \text{H}_2\text{O}_{(l)}$
Complete; $\text{C}_3\text{H}_{8(g)} + 5 \ \text{O}_{2(g)} \rightarrow 3 \ \text{CO}_{2(g)} + 4 \ \text{H}_2\text{O}_{(l)}$
Incomplete; $2 \ \text{CH}_{4(g)} + 3 \ \text{O}_{2(g)} \rightarrow 2 \ \text{CO}_{(g)} + 4 \ \text{H}_2\text{O}_{(l)}$
Complete; $\text{C}_5\text{H}_{12(l)} + 8 \ \text{O}_{2(g)} \rightarrow 5 \ \text{CO}_{2(g)} + 6 \ \text{H}_2\text{O}_{(l)}$
Incomplete; $2 \ \text{C}_2\text{H}_{6(g)} + 5 \ \text{O}_{2(g)} \rightarrow 4 \ \text{CO}_{(g)} + 6 \ \text{H}_2\text{O}_{(l)}$

This video contains classroom demonstrations of several reaction types and then shows the balancing process for the reaction equations (3a) http://www.youtube.com/watch?v=4B8PFqbMNIw (8:40).

Lesson Summary

The Five Types of Chemical Reactions
Reaction Name	Reaction Description
synthesis	two or more reactants form one product.
decomposition	one type of reactant forms two or more products.
single replacement	one element reacts with one compound to form products.
double replacement	two compounds act as reactants.
combustion	a hydrocarbon reacts with oxygen gas.

Review Questions

When balancing combustion reactions, did you notice a consistency relating to whether the number of carbons in the hydrocarbon was odd or even?
Distinguish between synthesis and decomposition reactions.
When dodecane, , burns in excess oxygen, the products would be:
1. $\text{CO}_2 + 2 \ \text{H}_2$
2. $\text{CO} + \text{H}_2\text{O}$
3. $\text{CO}_2 + \text{H}_2\text{O}$
4. $\text{CH}_4\text{O}_2$
In the decomposition of antimony trichloride, which of the following products and quantities will be found?
1. $\text{An} + \text{Cl}_2$
2. $2 \ \text{An} + 3 \ \text{Cl}_2$
3. $\text{Sb} + \text{Cl}_2$
4. $2 \ \text{Sb} + 3 \ \text{Cl}_2$
Acetylsalicylic acid (aspirin), $\text{C}_9\text{H}_8\text{O}_{4(s)}$ , is produced by reacting acetic anhydride, $\text{C}_4\text{H}_6\text{O}_{3(l)}$ , with salicylic acid, $\text{C}_7\text{H}_6\text{O}_{3(s)}$ . The other product in the reaction is acetic acid, $\text{C}_2\text{H}_4\text{O}_{2(l)}$ . Write the balanced chemical equation.
When iron rods are placed in liquid water, a reaction occurs. Hydrogen gas evolves from the container, and iron(III) oxide forms onto the iron rod.
1. Write a balanced chemical equation for the reaction.
2. What type of reaction is this?
A specific fertilizer is being made at an industrial plant nearby. The fertilizer is called a triple superphosphate and has a formula $\text{Ca}(\text{H}_2\text{PO}_4)_2$ . It is made by treating sand and clay that contains phosphate with a calcium phosphate solution and phosphoric acid. The simplified reaction is calcium phosphate reacting with phosphoric acid to yield the superphosphate. Write the balanced chemical equation and name the type of reaction.