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Lesson 1: Describing Chemical Reactions

Part c: Writing Balanced Chemical Equations

Part a: What is a Chemical Reaction?
Part b: Chemical Equations
Part c: Writing Balanced Chemical Equations

 

Atoms, and Therefore Mass, are Conserved

Chemical reactions involve a change of reactant chemicals into product chemicals. In Lesson 1b, a variety of means of representing such changes were discussed. The most fundamental of these representations is the balanced chemical equation. A balanced chemical equation is a shorthand, symbolic means of …

  • Identifying the reactants and the products of the reaction.
  • Identifying the physical state of the reactants and products.
  • Identifying the relative number of particles (atoms, molecules, ions, or formula units) of reactants and products involved in the reaction.
 
These three items of information are conveyed by chemical formulae (plural for formula), state symbols, and coefficients.

 
At the heart of a balanced chemical equation is the understanding that reactions involve the rearrangement of atoms. Not only are atoms rearranged; they are also conserved. The number of atoms of each element on the reactant side of the equation is equal to the number of atoms of each element on the product side. This principle is conveyed by the word balanced.


 
The mass of reactants and products is directly related to the number and the type of atoms. Since the number of atoms of each type is conserved, it is reasonable to conclude that the mass of the system of reactants and products is also conserved. That is, the total mass of all reactants is equal to the total mass of all products. This is referred to as the law of conservation of mass.
 

 

 

Balancing Act

The act of balancing an equation relies upon the use of coefficients. Coefficients are the integers placed in front of the formulae that identify the reactants and products. Rules associated with formula writing are used to write the reactant and product formulae. When balancing a chemical equation, it is important that the formulae not be changed.

 
As an example, suppose that you are asked to write the balanced chemical equation for…
 

Solid aluminum reacts with chlorine gas to produce solid aluminum chloride.
 
The process would begin by writing the formulae of the two reactants and the one product. Aluminum is represented as Al. Chlorine is one of the seven diatomic elements and is represented as Cl2. Aluminum chloride is an ionic compound. Its formula is AlCl3, consistent with the rules for formula writing for ionic compounds. Writing these formulae and separating them by a plus sign and the yields symbol is the starting point. The result is a skeleton equation. A skeleton equation is not balanced; it simply identifies reactants and products and their physical state. It is shown below.

 
Once the skeleton equation is written, the balancing act begins. The formulae cannot change at this point. The equation is balanced by adding coefficients in front of the formulae.


 
The equation above on the left is the balanced chemical equation. It is the skeleton equation, except with coefficients placed in front of the formulae. The equation above on the right is an act of heresy. The formula for chlorine was changed. If the heretic is ever caught, there is no telling what type of punishment the gods of Chemistry will inflict on the accused. The moral of the story is to balance chemical equations by adding coefficients and NOT by changing the formulae.
 
 

 

Atom Counting

The importance of counting atoms was discussed in Lesson 1b of this chapter. Atom counting refers to the skill of counting the number of atoms of each element on both the reactant side and the product side of a balanced chemical equation. If needed, review our discussion of atom counting and put it to practice with our Concept Builder.

 
Having professional-quality, atom counting skills is the key to making quick and easy work of balancing chemical equations. Once a coefficient is placed in front of a formula, it is important to do a quick atom count to determine if its use has served to balance the atoms of a target element. And once all coefficients are in place, the careful chemistry student will do a final atom count to ensure the equation is balanced. The table below is our version of an atom count for the balanced chemical equation discussed earlier. Note how the number of atoms of aluminum and of chlorine are the same on each side of the equation. This is a sign that the equation is balanced and earns the green check award.





 
 

 

Method for Writing Balanced Chemical Equations

Students of chemistry will need to acquire the skill of writing balanced chemical equations. A typical question or task will begin by either providing the skeleton equation (formulae, but no coefficients) or providing a word equation. Our suggested step-by-step method for writing balanced chemical equations is shown below. If a word equation is provided, you will need to use formula-writing rules to write the skeleton equation (step 1). If a skeleton equation is provided, you can skip step 1 and start on step 2.
 
  1. Write a skeleton equation for the reaction. This includes accurate formulae for all reactants and products. Use the formula writing rules for molecular compounds, ionic compounds, acids, and our seven diatomic elements. Insert state symbols after each formula.
  2. Identify an element that is present in only one reactant and one product. Place a coefficient in front of these formula to balance the number of atoms of this element.
  3. Repeat step 2 for all elements with the goal of balancing atoms. Always do the easiest elements first; these are elements located in only one reactant and one product. Proceed to more difficult elements after the easy elements are completed.
  4. Ensure that the equation is balanced by conducting a final atom count. If the number of atoms of each element is the same on the reactant and product side, then the equation is balanced.

 
There are four things to note about the above process. First, if the skeleton equation is not given and you must write formulae from a word equation or verbal description, then you must understand formula writing. This was discussed in Chapter 4 of our Chemistry Tutorial. If you need to review formula writing, re-visit Chapter 4.

 
Second, this is a trial-and-error process. Scratch-outs and erasures should be common. “That didn’t work … let me try this!” should be frequent. You’re solving a puzzle. The pieces don’t always fit together on the first attempt. Balancing equations is a process. And it’s a process that takes some time and effort. Be patient with the process.

 
Third, like any process that involves trial-and-error, practice is essential. You will need lots of practice. It’s the process that you are trying to become skilled at. Reflect as your practice. Make mistakes and make a mental note of how you fixed the mistake. Make wrong turns and make a mental note of how you got back on track. We have provided seven examples below. Simply reading how we balanced the equation is not considered practice. Try them yourself and rely on our step-by-step commentary if you get stuck. There is more balancing equation practice in the Check Your Understanding section. And we have a link to a Concept Builder in the Before You Leave section.

 
Fourth and finally, remember that after step 1 is done, the formulae cannot be changed. Don’t disturb the gods of Chemistry!
 

 
 

Example 1

Balance the chemical equation:    N2(g)   +   H2(g)  →   NH3(g)
 
Solution:
Step 1: The skeleton equation is provided. Step 1 can be skipped. We will start on step 2.
 
Step 2: You can pick either element to balance first. Let’s pick N. Placing a 2 in front of NH3 will balance the nitrogen count with two atoms on each side. We now have this:
 
N2(g)   +   H2(g)   →   2 NH3(g)
 
Step 3: H is not balanced yet. With the 2 added in front of NH3, there are 6 H atoms on the product side. To balance H, add a coefficient of 3 in front of H2.
 
N2(g)   +   3 H2(g)   →   2 NH3(g)
 
This puts 6 H atoms on the reactant side. The equation should now be balanced. But do step 4 – check to make sure.
 
Step 4: Here is the atom count for the two elements. As you can see, the equation is now balanced.
 


 
 
 
 

 

Example 2

Balance the chemical equation:    Li2O(s)   +   H2O(l)   →   LiOH(aq)
 
Solution:
Step 1: The skeleton equation is provided. Step 1 can be skipped. We will start on step 2.
 
Step 2: The easiest elements to balance will be Li or H. Let’s pick Li. Placing a 2 in front of LiOH will balance the lithium count with two atoms on each side. We now have this:
 
Li2O(s)   +   H2O(l)   →   2 LiOH(aq)
 
Step 3: With the 2 added in front of LiOH, it appears that both the H and the O were balanced. This often happens; balancing one or two elements often results in the balancing of all elements. We will do an atom count using the following equation:
 
Li2O(s)   +   H2O(l)   →   2 LiOH(aq)
 
 
Step 4: Here is the atom count for the three elements. As you can see, the equation is indeed balanced.
 
 

 
 

Example 3

Balance the chemical equation:    P4O10(s)   +   H2O(l)   →   H3PO4(aq)
 
Solution:
Step 1: The skeleton equation is provided. Step 1 can be skipped. We will start on step 2.
 
Step 2: The easiest element to balance will be P or H. Let’s pick P. Placing a 4 in front of H3PO4 will balance the phosphorus count with four atoms on each side. We now have this:
 
P4O10(s)   +   H2O(l)   →   4 H3PO4(aq)
 
Step 3: Let’s balance the H. There are 12 atoms of H on the product side. Placing a 6 in front of the H2O will balance the hydrogen count with 12 atoms on each side.
 
P4O10(s)   +   6 H2O(l)   →   4 H3PO4(aq)
 
Let’s now proceed to the O. As written immediately above, there are 16 O atoms on both the reactant and the product side. So, O is balanced and no additional coefficients need to be added. It’s time for the atom count.
 
Step 4: Here is the atom count for the three elements. The analysis indicates that the equation is definitely balanced.
 
 
 

 

Example 4

Balance the chemical equation:    C4H10(g)   +   O2(g)   →   CO2(g)  +   H2O(g)
 
Solution:

Step 1: The skeleton equation is provided. Step 1 can be skipped. We will start on step 2.
 
Step 2: The easiest element to balance will be C or H. Let’s pick C. Placing a 4 in front of CO2 will balance the carbon count with four atoms on each side. We now have this:
 

 

C4H10(g)   +   O2(g)   →   4 CO2(g)  +   H2O(g)
 
Step 3: Let’s balance the H. There are 10 atoms of H on the reactant side. Placing a 5 in front of the H2O will balance the hydrogen count with 10 atoms on each side.
 
C4H10(g)   +   O2(g)   →   4 CO2(g)  +   5 H2O(g)
 
Let’s now proceed to the O. As written immediately above, there are 13 O atoms on the product side. To balance the O, a 6.5 must be placed in front of O2. This results in the equation below.
 
C4H10(g)   +   6.5 O2 (g)   →   4 CO2(g)  +   5 H2O(g)
 
While it appears that all elements have been balanced, the use of 6.5 is generally frowned upon. Whole number integers are preferred. So, we will multiply all coefficients by two to turn the 6.5 into a whole number. This results in …
 
2 C4H10(g)   +   13 O2(g)   →   8 CO2(g)  +   10 H2O(g)
 
Step 4: Here is the atom count for the three elements. The analysis indicates that the equation is definitely balanced.
 
 

 
 

Example 5

Write a balanced equation for the following reaction:
 
Aqueous iron(III) chloride reacts with aqueous potassium hydroxide to produce solid iron(III) hydroxide and aqueous potassium chloride.
 
Solution:
Step 1: Here is our first situation involving the need to write formulae. Using the rules for formula writing for ionic compounds and the polyatomic ion list (for hydroxide), we can start with this skeleton equation:
 
FeCl3(aq)   +   KOH(aq)   →   Fe(OH)3(s)   +   KCl(aq)
 
Step 2: We can start the balancing process with any of the elements. The Fe and the K are both balanced as is. Let’s start with Cl. There are three atoms of Cl on the reactant side. Inserting a 3 as a coefficient in front of KCl on the product side will balance the Cl. We now have:
 
FeCl3(aq)   +   KOH(aq)   →   Fe(OH)3(s)   +   3 KCl(aq)
 
Step 3: Let’s balance the O. There are three atoms of O on the product side. Placing a 3 in front of the KOH will balance the oxygen count with three atoms on each side. We now have:
 
FeCl3(aq)   +   3 KOH(aq)   →   Fe(OH)3(s)   +   3 KCl(aq)
 
It appears that the act of using these two coefficients has balanced all the elements. Let’s do an atom count to insure this is the case.
 
Step 4: Here is the atom count for the five elements. The analysis indicates that the equation is definitely balanced.
 
 
It’s worth noting that in cases that include polyatomic ions that remain intact, they are generally balanced as a unit. In this example, Step 3 could begin by balancing OH as opposed to balancing the O.
 
 

 

Example 6

Write a balanced equation for the following reaction:
 
Propane (C3H8) gas reacts with oxygen gas to produce carbon dioxide gas and dihydrogen monoxide gas.
 
Solution:

Step 1: We must begin by writing the skeleton equation. The propane formula is given. Oxygen is diatomic. Carbon dioxide and dihydrogen monoxide are both molecular compounds. We can write their formulae as shown below.
 

 

C3H8(g)   +   O2(g)   →   CO2(g)  +   H2O(g)
 
Step 2: The easiest element to balance will be C or H. Let’s pick C. Placing a 3 in front of CO2 will balance the carbon count with three atoms on each side. We now have this:
 
C3H8(g)   +   O2(g)   →   3 CO2(g)  +   H2O(g)
 
Step 3: Let’s balance the H next. There are eight atoms of H on the reactant side. Placing a 4 in front of the H2O will balance the hydrogen count with eight atoms on each side.
 
C3H8(g)   +   O2(g)   →   3 CO2(g)  +   4 H2O(g)
 
Let’s now proceed to the O. As written immediately above, there are 10 O atoms on the product side. To balance the O, a 5 must be placed in front of O2. This results in the equation below.
 
C3H8(g)   +   5 O2(g)   →   3 CO2(g)  +   4 H2O(g)
 
Let’s do an atom count to ensure that all elements are balanced.
 
Step 4: Here is the atom count for the three elements. The analysis indicates that the equation is balanced.
 
 
 

 

Example 7

Write a balanced equation for the following reaction:
 
Solid silicon dioxide reacts with solid carbon to produce solid silicon carbide (SiC) and carbon monoxide gas.
 
Solution:
Step 1: We must begin by writing the skeleton equation. The formula for silicon carbide is given. Silicon dioxide and carbon monoxide are both molecular compounds. Carbon is a monatomic element. We can write their formulae as shown below.
 
SiO2(s)   +   C(s)   →   SiC(s)  +   CO(g)
 
Step 2: The easiest element to balance will be Si or O. The Si is balanced as is. Let’s start with O. There are two atoms of O on the reactant side. Placing a 2 in front of CO will balance the carbon count with two atoms on each side. We now have this:
 
SiO2(s)   +   C(s)   →   SiC(s)  +   2 CO(g)
 
Step 3: The Si is still balanced. Let’s balance the C. There are three atoms of C on the product side. Placing a 3 in front of the C will balance the carbon count with three atoms on each side.
 
SiO2(s)   +   3 C(s)   →   SiC(s)  +   2 CO(g)
 
It appears that all elements are balanced. Let’s do an atom count to insure this is the case.
 
Step 4: Here is the atom count for the three elements. The equation is balanced.
 
 
 
 
 
 

Before You Leave

  • Download our Study Card on Balancing Chemical Equations. Save it to a safe location and use it as a review tool. 
  • Balancing chemical equations is not a great spectator sport. Get yourself out of the stands and into the game with this Concept Builder on Balancing Chemical Equations. Progress through the three levels of difficulty.
  • The Check Your Understanding section below includes questions with answers and explanations. It provides a great chance to self-assess your understanding.
 
 

 

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.
 
1. The statement:   CH4   +   O2  →   CO2  +    H2O    is not a balanced chemical equation because _____.
  1. there should be more reactant chemicals than product chemicals.
  2. there should be more product chemicals than reactant chemicals.
  3. there are different numbers of atoms of each element on the two sides of the equation.
  4. ... nonsense!  This is a balanced chemical equation.
 
Check Answer



 
2. The smallest integers found in front of formulae of balanced chemical equations are _____.
  1. coefficients
  2. subscripts
  3. balancers
  4. state symbols
 
Check Answer


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3. Lying at the heart of the balancing process is the fact that ... .
  1. chemistry is a quantitative science.
  2. chemistry is an empirical science.
  3. reactions are the heart and soul of chemistry.
  4. atoms are conserved in a chemical reaction.
 
Check Answer


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4. TRUE or FALSE: Equations can be easily and correctly balanced by altering the subscripts present within the molecular formulas of the substances involved.
 
If false, then explain what is wrong or correct the statement.
 
Check Answer


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5. Balancing chemical equations demands that a student be able to atom-count. Count the number of oxygen atoms represented by each of the following sets of coefficients and formulas.
  1. Na3PO4: 
    Check Answer

     
  2. 2 CO2: 
    Check Answer

     
  3. 2 Fe2O3: 
    Check Answer

     
  4. 2 Al2(SO4)3: 
    Check Answer

     
 
 
6. Use coefficients to balance the following skeleton equations. Once done, display an atom count to demonstrate that the equation is balanced.
  1. Sb(s)   +   Cl2(g)   →   SbCl3(s) 
    Check Answer

     
  2. P2O5(s)   +   H2O(l)   →   H3PO4(aq) 
    Check Answer

     
  3. Na(s)   +   O2(g)   →  Na2O(s) 
    Check Answer

     
 
 
7. Write balanced chemical equations for the following reactions. Display an atom count to demonstrate that the equation is balanced.
  1. Solid titanium reacts with nitrogen gas to produce solid titanium(IV) nitride. 
    Check Answer

     
  2. Solid aluminum reacts with aqueous zinc chloride to produce solid zinc and aluminum chloride. (Zinc ions are 2+ ions.) 
    Check Answer

     
  3. Solid aluminum hydroxide decomposes to produce solid aluminum oxide and liquid water. 
    Check Answer