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# 13.2: Balancing Chemical Equations

Difficulty Level: At Grade Created by: CK-12

## 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 \begin{align*}\text{H}_2\end{align*} Gaseous
Oxygen \begin{align*}\text{O}_2\end{align*} Gaseous
Nitrogen \begin{align*}\text{N}_2\end{align*} Gaseous
Chlorine \begin{align*}\text{Cl}_2\end{align*} Gaseous
Fluorine \begin{align*}\text{F}_2\end{align*} Gaseous
Bromine \begin{align*}\text{Br}_2\end{align*} Liquid
Iodine \begin{align*}\text{I}_2\end{align*} 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 coefficients that 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.

\begin{align*}\text{Cl}_2 + \text{NaBr} \rightarrow \text{Br}_2 + \text{NaCl}\end{align*}

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

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

\begin{align*}\text{Cl}_2 + 2 \ \text{NaBr} \rightarrow \text{Br}_2 + 2 \ \text{NaCl}\end{align*}

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.

\begin{align*}\text{Al}_2(\text{SO}_4)_3 + \text{CaBr}_2 \rightarrow \text{AlBr}_3 + \text{CaSO}_4\end{align*}

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.

\begin{align*}\text{Al}_2(\text{SO}_4)_3 + \text{CaBr}_2 \rightarrow 2 \ \text{AlBr}_3 + \text{CaSO}_4\end{align*}

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

\begin{align*}\text{Al}_2(\text{SO}_4)_3 + \text{CaBr}_2 \rightarrow 2 \ \text{AlBr}_3 + 3 \ \text{CaSO}_4\end{align*}

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

\begin{align*}\text{Al}_2(\text{SO}_4)_3 + 3 \ \text{CaBr}_2 \rightarrow 2 \ \text{AlBr}_3 + 3 \ \text{CaSO}_4\end{align*}

The insertion of the 3 in front of the reactant \begin{align*}\text{CaBr}_2\end{align*} also balances the calcium atoms in the product \begin{align*}\text{CaSO}_4\end{align*}. 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:

\begin{align*}2 \ \text{Al}_2(\text{SO}_4)_3 + 6 \ \text{CaBr}_2 \rightarrow 4 \ \text{AlBr}_3 + 6 \ \text{CaSO}_4\end{align*}

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.)

\begin{align*}\text{Fe(NO}_3)_3 + \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \ \text{NaNO}_3 \ \ \ \ \ \ \text{(skeletal equation)}\end{align*}

Solution:

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

\begin{align*}\text{Fe(NO}_3)_3 + 3 \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaNO}_3\end{align*}

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

\begin{align*}\text{Fe(NO}_3)_3 + 3 \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \ \text{NaNO}_3\end{align*}

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.

1. \begin{align*}\text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}\end{align*}
2. \begin{align*}\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)}\end{align*}
3. \begin{align*}\text{Ba(NO}_3)_{2(aq)} + \text{Na}_2\text{CO}_{3(aq)} \rightarrow \text{BaCO}_{3(aq)} + \text{NaNO}_{3(aq)}\end{align*}
4. \begin{align*}\text{C}_2\text{H}_{6(g)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}\end{align*}

Solution:

1. \begin{align*}\text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}\end{align*} (sometimes the skeletal equation is already balanced)
2. \begin{align*}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)}\end{align*}
3. \begin{align*}\text{Ba(NO}_3)_{2(aq)} + \text{Na}_2\text{CO}_{3(aq)} \rightarrow \text{BaCO}_{3(aq)} + 2 \ \text{NaNO}_{3(aq)}\end{align*}
4. \begin{align*}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)}\end{align*}

## 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.

\begin{align*}\text{Fe(NO}_3)_3 + 3 \ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \ \text{NaNO}_3\end{align*}

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:

1 molecule of \begin{align*}\text{Fe(NO}_3)_3 \times \text{molecular weight} = (1) \cdot (241.9 \ \text{daltons})\end{align*} = 241.9 daltons
3 molecules of \begin{align*}\text{NaOH} \times \text{molecular weight} = (3) \cdot (40.0 \ \text{daltons})\end{align*} = 120. daltons
Total mass of reactants = 241.9 daltons + 120. daltons = 361.9 daltons

Product Side Mass:

1 molecule of \begin{align*}\text{Fe(OH)}_3 \times \text{molecular weight} = (1) \cdot (106.9 \ \text{daltons})\end{align*} = 106.9 daltons
3 molecules of \begin{align*}\text{NaNO}_3 \times \text{molecular weight} = (3) \cdot (85.0 \ \text{daltons})\end{align*} = 255 daltons
Total mass of products = 106.9 daltons + 255 daltons = 361.9 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

1. Explain in your own words why coefficients can change but subscripts must remain constant.
2. Which set of coefficients will properly balance the following equation: \begin{align*}\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\end{align*}?
1. 1, 1, 1, 1
2. 1, 3, 2, 2
3. 1, 3.5, 2, 3
4. 2, 7, 4, 6
3. When properly balanced, what is the sum of all the coefficients in the following chemical equation: \begin{align*}\text{SF}_4 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3 + \text{HF}\end{align*}?
1. \begin{align*}4\end{align*}
2. \begin{align*}7\end{align*}
3. \begin{align*}9\end{align*}
4. None of the above
4. When the following equation is balanced, what is the coefficient found in front of the \begin{align*}\text{O}_2\end{align*}: \begin{align*}\text{P}_4 + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4\end{align*}?
1. \begin{align*}1\end{align*}
2. \begin{align*}3\end{align*}
3. \begin{align*}5\end{align*}
4. \begin{align*}7\end{align*}
5. Balance the following equations.
1. \begin{align*}\text{XeF}_{6(s)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{XeO}_{3(s)} + \text{HF}_{(g)} \end{align*}
2. \begin{align*}\text{Cu}_{(s)} + \text{AgNO}_{3(aq)} \rightarrow \text{Ag}_{(s)} + \text{Cu(NO}_3)_{2(aq)}\end{align*}
3. \begin{align*}\text{Fe}_{(s)} + \text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}\end{align*}
4. \begin{align*}\text{Al(OH)}_3 + \text{Mg}_3(\text{PO}_4)_2 \rightarrow \text{AlPO}_4 + \text{Mg(OH)}_2\end{align*}

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