balancing chemical equations worksheet intermediate level answers

Balancing Chemical Equations with Ease (100 Free Worksheets)

• October 2, 2019

In this article, you will learn about how to balance chemical equations easily with simple steps. Feel free to download our free worksheets with answers for your practice.

Parts of a Balanced Chemical Equation

Before you start balancing chemical equations, it is important that you become firmly acquainted with the various part of one. Every balanced chemical equation consists of two parts: the reactant side and the product side. Both of these sides are separated by the means of an arrow.

On the left side of the arrow, you will find the reactant side. This side represents the elements which are used for initiating the experiment. On the right side of the arrow, we have the product side. This side is used to display the elements or compounds which are generated from the chemical process.

The Need for Balancing Chemical Equations

Ever since you started learning about the field of Chemistry, your teachers might often have stressed upon the importance of balancing chemical equations. But have you ever pondered on the reason behind this? Why do you need to respect the law of the conservation of mass? Quite simply, you need to balance your equations to follow the law of conservation of mass.

Law of Conservation of Mass: According to the law of conservation of mass, the mass of products that are derived from a chemical equation should mandatorily equal the mass of the reactants.

Minding the law of conservation of mass while balancing equations is quite important. Not only does it help you to prevent errors, but it also assists scientists in knowing the quantities of reactants to create a particular product that they want to make. Moreover, the law also helps chemical manufacturers to increase the efficiencies of their processes.

When Should You Start To Balance A Chemical Equation?

As soon as you derive a chemical equation stating the reactants and the products, check out if the number of atoms on either side are equal to each other or not. In the event that you detect the numbers to be uncommon, rest assured, you should start balancing the elements and compounds on either side of the chemical equation.

How Should You Balance A Chemical Equation?

The primary aspect that you need to keep in mind while balancing a chemical equation is this; the entire process is completely based on trial and error. When you start to balance a particular chemical equation, you will need to go through several processes before you stumble upon the right coefficients to balance the number of atoms.

Another aspect that you need to remember is that balancing chemical equations requires a lot of practice. Once you perfect the practice of balancing, you can become completely reliant on your intuition to lead you through the complete process.

While balancing your equations, you need to follow certain simple stems. Here’s what you need to do:

• Start by counting the number of atoms, present for each element on the side of the reactants as well as the products.
• When you find out that certain elements are not balanced, place the required coefficient that is needed to balance the elements.
• Once you are done with this, check out if the number of atoms for the other elements is also equal on both the sides.
• Repeat the process until you find out that all the elements on both the sides of the chemical equations are balanced.

As aforementioned, the process in itself is quite simple. However, it takes significant practice before can start to balance these equations completely with your intuition.

An Easy Example To Get You Started With

Now that you know the steps, you are wholly capable of balancing chemical equations. Let’s solve some of them, shall we? With the help of above-mentioned steps and a practical example, you will be better able to understand how the entire process works.

Do not feel anxious if you feel that you are still not ready to solve these problems. With our method, even your toddler sibling will be able to understand how chemical equations are balanced. And if still feel a tad bit confused after solving all these equations, try to solve a few more of such problems. Remember what we advised in a previous section: You will need significant practice before you can confidently start to balance these equations with your intuition.

Let’s start with this example. This equation represents a reaction between two Iron Oxide (Fe 2 O 3 ) and Carbon (C). The products formed are Iron (Fe) and Carbon Dioxide (CO 2 ).

Fe 2 O 3 + C        →       Fe + CO 2

Alright, so we have our equation. Let’s begin to balance the equation with the help of the steps mentioned above.

Step 1: Start by counting the number of atoms present, for each element on the side of the reactants as well as the products.

On the reactants side, we have:

• 2 atoms of Fe
• 3 atoms of O
• 1 atom of C

On the product side, we have:

• 1 atom of Fe
• 2 atoms of O

By comparing the number of atoms present for each element on each side, you might have determined that the reaction is obviously not balanced. Therefore, let’s move on to Step 2.

Step 2: When you find out that certain elements are not balanced, place the required coefficient that is needed to balance the elements.

Let’s start by balancing the oxygen atoms. To do this, make the oxygen atoms as six on either side of the chemical equation.

2Fe 2 O 3 + C      →         Fe + 3CO 2

On towards the next step now.

Step 3: Once you are done with this, check out if the number of atoms for the other elements is also equal on both the sides.

Now that we have an equal number of oxygen atoms on either side of the equation, let’s check out if the other elements of the equation are equal or not.

2Fe 2 O 3 + C      →        Fe + 3CO 2

On the reactant side, we have:

• 4 atoms of Fe.
• 6 atoms of O.
• 1 atom of C.
• 1 atom of Fe.
• 3 atoms of C.

As you can see, the elements of iron and carbon are still not balanced. Therefore, it is time that we move on to the 4 th step.

Step 4: Repeat the process until you find out that all the elements on both the sides of the chemical equations are balanced.

Alright, let’s start balancing the equation again and this time, let’s balance the number of iron atoms first. On the reactant side, we have 4 atoms of Fe while the product side has 1 atom of Fe. To balance them, we need to place 4 atoms of Fe on the product side.

2Fe 2 O 3 + C      →        4Fe + 3CO 2

Now, on the reactant side we have:

And, on the product side, we have:

The only element that remains to be balanced now is carbon. This can be easily done considering the fact that carbon exists only in a singular form on the reactant side. In order to correct this, we need to place 3 atoms of carbon on the reactant side. The chemical reaction, hence, will turn out to be:

2Fe 2 O 3 + 3C    →         4Fe + 3CO 2

And there you go. Perfectly balanced as all things should be (Yes. We are Marvel fanboys as well).

Essential Tips for Beginners

As you become further acquainted with balancing chemical equations, it becomes quite easy for you to solve them. However, it still maintains a certain level of difficulty at the beginner level. As a result of this, you might find yourself shying away from the equations and procrastinating to the level where you get totally and utterly repulsed by them.

However, there are certain tips that help you during such a stage. When you are beginner, you will be solving quite easy problems compared to those you might see in your Chemistry books. During such times, you will need to keep two essential tips in your mind. These tips will help you to easily balance the equations with ease. These tips are:

• Start Balancing With Single Elements – Attempt at balancing those elements first which occur in the form of a single molecule first. Owing to their single nature, they are easily flexible and their coefficient can be easily changed as and when needed in further steps.
• Balance the Hydrogen and Oxygen Molecules at the End – At the beginning, you will come across a lot of equations involving hydrogen and oxygen molecules. Whenever you encounter these, you should interact with these at the end. This is because hydrogen and oxygen molecules often occur together in both the reactant and product side. Once you are done with balancing other elements, focus on these.

Format for Writing a Balanced Equation

Now that you have balanced the assigned chemical reaction, you might be wondering if there is a format for writing these balanced chemical equations. In actuality, there is not said format that you need to mind for arranging the balanced equation. However, it has also been noticed that people in the field of chemistry often prefer to write solid elements and other compounds first, followed by the gaseous elements and single elements. This often acts as an unwritten rule which is followed by a lot of people around the world.

The Coefficients in a Balanced Chemical Equation

Up until this point of balancing your chemical equations, you might have known about the various facets surrounding the chemical equation. But there is still one significant aspect of balancing which we haven’t discussed: The role of coefficients while balancing equations.

At some point or another, you might have certainly wondered how are these coefficients be used while balancing the equation. After all, we cannot magically create or destroy elements during a chemical reaction. The Law of Conservation of Mass prevents this. In actuality, these coefficients define the ratios. For the reactant side, the coefficients define the ratio in which the substances are being used. And for the product side, the coefficients define the ratio in which the substances are being produced.

What a Balanced Chemical Equation Does Not Tell Us

Balanced chemical equations are highly informative in nature. They divulge a lot of information which is implemented for deriving the desired results from the reactions. However, there are certain aspects which balanced chemical equations don’t make you aware of just by solving the equations. The most prominent aspects amongst these are the subscripts used.

Take, for example, the last chemical equation which we balanced.

Now, if you notice, the element Fe has the subscript 2 beside itself, signifying the number of atoms. But if you notice on the product side, element lacks any subscript. This is quite similar to the oxygen element as well. On one hand, it has the subscript 3 while it has the subscript 2 on the other hand.

In spite of all this, the total mass of the individual atoms present on both sides of the equation is equal to each other. This is due to the Law of Conservation of Mass which ensures that matter isn’t created nor destroyed during a chemical reaction. This is also the reason why the total number of individual atoms are equal on both the reactant and product side.

Rules for Balancing Chemical Equations

By this point, you might have become nicely acquainted with balancing chemical equations on your own. Resulting, the rules associated with balancing chemical equations must also have become imprinted within your mind. It is with the help of such rules that you can easily balance the assigned chemical equations. However, it is equally important that you put these rules on paper and revise them once thoroughly.

Here are a few of the most prominent rules, include:

• Keep The Placement of Reactants and Products in Mind – In every chemical equation, there are two parts to an equation. These parts are separated by an arrow. While writing down the chemical equation, take care to list all the reactants on the left-hand side of the arrow. Similarly, you should take care to list all the products on the right side of the arrow.
• Ensure That The Right Arrow Is Placed – In most cases, the reactants and the products are separated by a single-sided arrow. This signifies a reaction which is irreversible or is unchangeable after a certain stage. However, in certain situations, the reactions occur at equilibrium. This means that reaction at any forward rate results in a reverse reaction. In such situations, the arrow used is two-sided, i.e. facing towards the reactants and the products.
• Emphasize on The Law of Conservation of Mass – While balancing the equations, it is of a predominant nature that you keep applying the Law of Conservation of Mass. This is because matter can neither be produced nor destroyed. Keeping this law in mind greatly helps you while balancing equations. Whenever you find an element which has more or less number of molecules, you can easily place a coefficient to balance it.
• Start With Independent Elements – When you start to balance the equation, start by balancing the independent elements. These are the elements which appear in individually in the equation. If there is no such element or if these elements are already balanced, proceed with the elements that exist in conjunction with other elements. Once this particular element is balanced, you should proceed on to balance other elements until all the elements are balanced.
• Balanced Only With Coefficients – While balancing the chemical equations, balance them only by placing coefficients in front of them. By no means should you add subscripts because this will completely change the formula of the particular reactant or compound, causing a change in the entire meaning that the equation wants to render.

Balancing Chemical Equations with Matrices

Up until this point, you have been balancing chemical equations by the means of trial and error. The process was simple, you had to place a coefficient, check if the other elements were balanced or not, and repeat all the steps until you had all the elements balanced.

However, it won’t be long before you face even tougher balancing problems. And you will face innumerous problems while using the trial and error method for such tough equations. Therefore, on such occasions, you will need a more versatile method for solving the problems.

Fortunately, there is one such methodology for solving chemical equations. This method involves a matrix which you can use to easily solve even the toughest of equations. Here are the steps that you should follow while solving chemical equations:

• Start by placing an alphabet which acts as a variable coefficient for your elements.
• Arrange all the elements in a column matrix format, as per the subscript values.
• Solve each of these matrices and generate the various equations.
• Individually equate all these equations and place the values generated into the other equations that you generated in Step 4.
• Assume a particular number for each of the values, such that, neither of the values that you derive appear in the form of a fraction and use this number to find out the values of the other coefficients.
• Finally, place these values into the initial chemical reaction to derive your balance equation.

Let’s use a simple example to understand this process. Take this chemical equation for instance:

NO + O 2          →         NO 2

Now, this is quite a simple equation. In fact, you might have even figured out how to balance this equation. In spite of this, we will use a simple methodology to help you understand how the entire process works.

Step 1: Start by placing an alphabet which acts as a variable coefficient for your elements

You can use any alphabet as a variable coefficient. For our purpose, we will be using alphabets X, Y, and Z. We will be placing them in this order:

X NO + Y O 2   →         Z NO 2

Step 2: Arrange all the elements in a column matrix format, as per the subscript values.

You should always follow a format for arranging the elements in a column matrix format. First, start by counting the number of atoms present for every individual occurrence of each element. From our first equation, we can derive that:

No. of N atoms = 1 + No. of N atoms = 0 → No. of N atoms = 1

No. of O atoms = 1 + No. of O atoms = 2 → No. of O atoms = 2

According to this format, we will separate the values of each of the elements according to the number of atoms present. Each of these positions will display a value depending on the number of those elements at that particular location. Hence, this is how we will display the values of the elements that are separated into the form of matrices:

X  + Y     →         Z

Notice that value signifying the elements show that each of the elements acquires a particular row. In essence, the nitrogen element acquires the first row while oxygen acquires the second row.

Step 3: Solve each of these matrices and generate the various equations

Once you have the matrices, you need to solve them and generate the required equations for them. The equations that you generate, generally, depending on the number of elements present within the equation. In this case, we have two elements. Therefore, the equations that formed are:

• X + Y0 = Z or X = Z (Equation i)
• X + 2Y = 2Z (Equation ii)

Step 4: Individually equate all these equations and place the values generated into the other equations that you generated in Step 4.

We have already generated the value of the coefficient X in Equation i. The value of X that we have generated is Z. Therefore, it is time that we focused on Equation ii.

X + 2Y = 2Z

According to Equation i, X = Z. Therefore,

• Z + 2Y = 2Z
• 2Y = 2Z – Z
• Y = ½Z             (Equation iii)

Step 5: Assume a particular number for each of the values, such that, neither of the values that you derive appears in the form of a fraction and use this number to find out the values of the other coefficients.

Once we have generated the final equations, it is time that we used them to generate the final values for our coefficients. In order to do this, we need to assume a particular value for each of the variable coefficients, such that the result does not turn out to be a fractional value.

Let’s start by assuming that the value of Z = 1. If Z = 1, then Y = 1/2 (according to the Equation iii). However, we don’t want a fractional value as our result. Therefore, let’s assume that Z = 2. Now that Z = 2, therefore Y = 1. Resultingly, the value of X = 2, since X = Z (as per equation i).

Step 6: Finally, place these values into the initial chemical reaction to derive your balance equation.

The equation which we had at the beginning, was:

According to the results generated by, the value of the variable coefficients stand as per the following:

Let’s place these values into the equation. Upon doing so, we get:

2NO + O 2        →         2NO 2

Therefore, on the reactant side, we have:

• 2 atoms of N
• 4 atoms of O

And on the product side, we have:

• 2 atoms of N.
• 4 atoms of O.

As there you have it again. A perfectly balanced chemical equation solved with the help of matrices.

Balancing Chemical Equations with Odd Number of Atoms on Elements

Another area wherein balancing becomes a tricky affair is during the presence of odd subscripts or atoms of an element. Let us take into consideration, this particular equation:

NH­­­ 3 + O 2         →         NO + H 2 O

The first thing that you will want to do in such cases is to balance those elements which are present in odd numbers on one side but are present in even numbers on the other side of the chemical equation. In this case, we have hydrogen following such a suit. Let us balance this out first.

2NH­­­ 3 + O 2       →         NO + 3H 2 O

Now, we need to balance nitrogen to equate the reaction.

2NH­­­ 3 + O 2       →         2NO + 3H 2 O

At this point, all the elements present in our chemical equations are balanced… except for oxygen. Hence, you will need to find out a coefficient which can effectively help you to balance the oxygen molecule present on the left-hand side of the reaction. On the reactant side, we have 2 oxygen molecules while on the product side, we have 5 oxygen molecules. Therefore, we will need to find a number which, when multiplied by 2, gives us 5 as the answer. Let this number be x.

Therefore, let us place this value into the equation.

2NH­­­ 3 + 5/2O 2 →         2NO + 3H 2 O

Finally, we need to eliminate the fractional part of the equation. Let’s do so by multiplying the entire chemical equation with 2.

4NH­­­ 3 + 5O 2     →         4NO + 6H 2 O

And there you have it. The equation becomes perfectly balanced.

A Few Examples worth Mentioning

Now that you have covered everything that is to be learned about the basics of balancing chemical equations, you should get yourself acquainted with certain worthwhile chemical equations.

• Chemical Reaction for Photosynthesis

6CO 2 + 6H 2 O              →        C 6 H 12 O 6 + 6O 2

• Chemical Reaction for Cellular Respiration

C 6 H 12 O 6 + 6O 2                        →         CO 2 + H 2 O + ATP

• Chemical Reaction for Ammonium Nitrate and Water

NH 4 NO 3 + Water       →         NH 4 + NO 3

• Chemical Reaction for Magnesium and Hydrochloric Acid

Mg + HCl       →         MgCl 2 + H 2

• Chemical Reaction for Lithium and Water

2 Li + 2H 2 O    →         2 LiOH + H 2

• Chemical Reaction for Calcium Carbonate and Hydrochloric Acid

CaCO 3 + HCl →         CaCl 2 + H 2 O

Using Games and Apps to Learn About Balancing Chemical Equations

It has not escaped our sights that a technologically-savvy world such as ours often uses technological means to better understand any newer concepts that they come across. Keeping this factor in mind, we have brought for you two of the greatest means by which you can enhance your skills at balancing chemical equations while simultaneously enjoying it via your smartphones of computers. Here are the means:

• Balancing Chemical Equations – We often encounter situations in which, whatever we do, we fail to solve the equation that has presented itself to us. And let’s be honest, all of us have been there at some point in our lives. It is in such situations that you might find the need for additional help. And this is exactly what Balancing Chemical Equations aims to do. With the help of this application, you will be easily able to balance the toughest of chemical equations. All that you need to do is to enter the unbalanced reactants and products and by clicking a button, the app displays the balanced chemical equation. You can find the Balancing Chemical Equations application on Google Play for free. Here’s a link for the same.

2. The Balancing Equations Game from PHET – Now, an application can only go so far as to keep you engaged. But this is completely contrary to what games can do at keeping you engaged. One of the most engaging games comes from PHET. At their website, you will find the Balancing Equations Game. Upon choosing the option, you are redirected towards a screen for choosing the difficulty of the game. This game is quite interesting. Having tried it out ourselves, we can assure you that not only is it engaging and entertaining, but it is quite informative as well. Therefore, it is one game that you should play if you want to get better at balancing chemical equations and get entertained for a while. Here’s a link to their website.

Leave a Reply Cancel Reply

Your email address will not be published. Required fields are marked *

Name  *

Email  *

Add Comment  *

Save my name, email, and website in this browser for the next time I comment.

Post Comment

How to Balance Equations - Printable Worksheets

Balancing Equations Worksheets

• Chemical Laws
• Periodic Table
• Projects & Experiments
• Scientific Method
• Biochemistry
• Physical Chemistry
• Medical Chemistry
• Chemistry In Everyday Life
• Famous Chemists
• Activities for Kids
• Abbreviations & Acronyms
• Weather & Climate
• Ph.D., Biomedical Sciences, University of Tennessee at Knoxville
• B.A., Physics and Mathematics, Hastings College

A balanced chemical equation gives the number and type of atoms participating in a reaction, the reactants, products, and direction of the reaction. Balancing an unbalanced equation is mostly a matter of making certain mass and charge are balanced on the reactants and products side of the reaction arrow. This is a collection of printable worksheets to practice balancing equations. The printable worksheets are provided in pdf format with separate answer keys.

Balancing Chemical Equations - Worksheet #1 Balancing Chemical Equations - Answers #1 Balancing Chemical Equations - Worksheet #2 Balancing Chemical Equations - Answers #2 Balancing Chemical Equations - Worksheet #3 Balancing Chemical Equations - Answers #3 Balancing Equations - Worksheet #4 Balancing Equations - Answer Key #4

I also offer printable worksheets for balancing equations on my personal site. The printables are also available as PDF files:

Balancing Equation Practice Sheet  [ answer sheet ]​ Another Equation Worksheet [ answer sheet ] Yet Another Printable Worksheet [ answer key ]

You may also wish to review the step-by-step tutorial on how to balance a chemical equation .

Online Practice Quizzes

Another way to practice balancing equations is by taking a quiz.

Coefficients in Balanced Equations Quiz Balance Chemical Equations Quiz

• Printable Chemistry Worksheets
• Balancing Chemical Equations
• How to Balance Chemical Equations
• Balanced Equation Definition and Examples
• Mole Relations in Balanced Equations
• Examples of 10 Balanced Chemical Equations
• 5 Steps for Balancing Chemical Equations
• What Is a Chemical Reaction?
• What Is a Chemical Equation?
• Stoichiometry Definition in Chemistry
• Introduction To Stoichiometry
• 20 Practice Chemistry Tests
• Mole Ratio: Definition and Examples
• The Balanced Chemical Equation for Photosynthesis
• Reactant Definition and Examples
• A List of Common General Chemistry Problems

Balancing Chemical Equations Worksheet (Intermediate)

Description

This pdf worksheet contains intermediate level chemical equations to balance along with an answer key.

Use this worksheet to test your students abilities before you move on to balancing more difficult chemical equations.

Questions & Answers

J26 education.

• We're hiring
• Help & FAQ
• Privacy policy
• Student privacy
• Terms of service
• Tell us what you think

404 Not found

• Science Notes Posts
• Contact Science Notes
• Todd Helmenstine Biography
• Anne Helmenstine Biography
• Free Printable Periodic Tables (PDF and PNG)
• Periodic Table Wallpapers
• Interactive Periodic Table
• Periodic Table Posters
• How to Grow Crystals
• Chemistry Projects
• Fire and Flames Projects
• Holiday Science
• Chemistry Problems With Answers
• Physics Problems
• Unit Conversion Example Problems
• Chemistry Worksheets
• Biology Worksheets
• Periodic Table Worksheets
• Physical Science Worksheets
• Science Lab Worksheets
• My Amazon Books

Balancing Chemical Equations Quiz – Questions and Answers

Here is a ten-question balancing chemical equations quiz. Each question presents an unbalanced equation. Select the balanced equation. Find the answer key below the questions. Remember, the number of each type of atoms is the same on both sides of the reaction arrow when an equation is balanced.

Balancing Chemical Equations Quiz Questions

(1) __ Na₃PO₄ + __ HCl → __ NaCl + __ H₃PO₄

• Na₃PO₄ + HCl → NaCl + H₃PO₄
• Na₃PO₄ +3 HCl → 3 NaCl + H₃PO₄
• 3 Na₃PO₄ + HCl → 3 NaCl + H₃PO₄
• Na₃PO₄ + 3 HCl → NaCl + H₃PO₄

(2) __ TiCl₄ + __ H₂O → __ TiO₂ + __ HCl

• TiCl₄ + 2 H₂O → TiO₂ + 2 HCl
• TiCl₄ + 2 H₂O → TiO₂ + 4 HCl
• 2 TiCl₄ + H₂O → 2 TiO₂ + HCl
• TiCl₄ + 4 H₂O → TiO₂ + 4 HCl

(3) __ C₂H₆O + __ O₂ → __ CO₂ + __ H₂O

• 2 C₂H₆O + O₂ → CO₂ + 3 H₂O
• C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O
• C₂H₆O + 2 O₂ → 3 CO₂ + 3 H₂O
• 2 C₂H₆O + O₂ → 2 CO₂ + H₂O

(4) __ FeS + __ O₂ → __ Fe₂O₃ + __ SO₂

• 2 FeS + O₂ → 2 Fe₂O₃ + 2 SO₂
• FeS + O₂ → Fe₂O₃ + SO₂
• 3 FeS + 2 O₂ → 3 Fe₂O₃ + 2 SO₂
• 4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂

(5) __ CaCl₂ + __ Na₃PO₄ → __ Ca₃(PO₄)₂ + __ NaCl

• 2 CaCl₂ + 2 Na₃PO₄ → 2 Ca₃(PO₄)₂ + NaCl
• CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl
• 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 6 NaCl
• 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 3 NaCl

(6) __ KOH + __ H₃PO₄ → __ K₃PO₄ + __ H₂O

• 3 KOH + 2 H₃PO₄ → 3 K₃PO₄ + 3 H₂O
• KOH + H₃PO₄ → K₃PO₄ + 6 H₂O
• 3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
• 2 KOH + H₃PO₄ → K₃PO₄ + H₂O

(7) __ AgI + __ Na₂S → __ Ag₂S + __ NaI

• AgI + Na₂S → Ag₂S + 2 NaI
• 2 AgI + 2 Na₂S → 2 Ag₂S + 2 NaI
• 2 AgI + 2 Na₂S → Ag₂S + 2 NaI
• 2 AgI + Na₂S → Ag₂S + 2 NaI

(8) __ Ba₃N₂ + __ H₂O → __ Ba(OH)₂ + __ NH₃

• 2 Ba₃N₂ + 3 H₂O → 2 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + 6 H₂O → 3 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + 3 H₂O → 3 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + H₂O → Ba(OH)₂ + NH₃

(9) ___ SnO₂ + ___ H₂ → ___ Sn + ___ H₂O

• 2 SnO₂ + 2 H₂ → 2 Sn + H₂O
• SnO₂ + 2 H₂ → Sn + 2 H₂O
• SnO₂ + H₂ → Sn + 2 H₂O
• 2 SnO₂ + H₂ → 2Sn + H₂O

(10) __ KNO₃ + __ H₂CO₃ → __ K₂CO₃ + __ HNO₃

• 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃
• 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 2 HNO₃
• KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃
• 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 3 HNO₃

Balancing Chemical Equations Quiz Answers

Learn more about balancing equations.

If you had trouble with this quiz, it’s a good idea to review the steps of balancing chemical equations.

• Guide to Balancing Chemical Equations – Learn the steps for balancing chemical equations and see how to check your work.
• Examples of Balanced Equations – Not sure what a balanced equation looks like? Here are some examples.
• Practice Worksheet – If you want more practice, try worksheets! Print worksheets out and improve your skills.

Related Posts

• Sign in / Register
• Administration
• Edit profile

The PhET website does not support your browser. We recommend using the latest version of Chrome, Firefox, Safari, or Edge.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

7.1: Balancing Chemical Equations

• Last updated
• Save as PDF
• Page ID 435660

Harper College Chemistry Department

By the end of this section, you will be able to:

• Derive chemical equations from narrative descriptions of chemical reactions.
• Write and balance chemical equations in molecular, total ionic, and net ionic formats.

A preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions or combine with other atoms to form molecules, their symbols are modified or combine to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH 4 ) and two diatomic oxygen molecules (O 2 ) to produce one carbon dioxide molecule (CO 2 ) and two water molecules (H 2 O). The chemical equation representing this process is provided in the upper half of Figure 1 , with space-filling molecular models shown in the lower half of the figure.

This example illustrates the fundamental aspects of any chemical equation:

• The substances undergoing reaction are called reactants and their formulas are placed on the left side of the equation.
• The substances generated by the reaction are called products and their formulas are placed on the right side of the equation.
• Plus signs (+) separate individual reactant and product formulas and an arrow (⟶) separates the reactant and product (left and right) sides of the equation.
• The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4 or 3-6-3-6 and so on ( Figure 2 ). Likewise, these coefficients may be interpreted with regard to any amount (number) unit and so this equation may be correctly read in many ways, including:

• One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.
• One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.
• One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules. The mole concept will be discussed in a different chapter.

Balancing Equations

The chemical equation described in the previous section is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO 2 and H 2 O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H 2 O to H 2 O 2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H 2 O to 2.

$\ce{2 H2O -> H2 + O2} \hspace{4em} \textrm{(unbalanced)}$

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H 2 product to 2.

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides and the balanced equation is, therefore:

Balancing Chemical Equations

Write a balanced equation for the reaction of molecular nitrogen (N 2 ) and oxygen (O 2 ) to form dinitrogen pentoxide.

First, write the unbalanced equation.

Next, count the number of each type of atom present in the unbalanced equation.

Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O 2 and N 2 O 5 to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).

The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N 2 to 2.

$\ce{2 N2 + 5 O2 -> 2 N2O5} \hspace{4em} \textrm{(unbalanced)}$

The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.

Check Your Learning

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)

It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C 2 H 6 ) with oxygen to yield H 2 O and CO 2 , represented by the unbalanced equation:

$\ce{C2H6 + O2 -> H2O + CO2}\hspace{4em}\textrm{unbalanced}$

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:

$\ce{C2H6 + O2 -> 3H2O + 2CO2}\hspace{4em}\textrm{unbalanced}$

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O 2 reactant to yield an odd number, so a fractional coefficient, $\frac{7}{2}$ , is used instead to yield a provisional balanced equation:

$\ce{C2H6 + \frac{7}{2}O2 -> 3H2O + 2CO2}\hspace{4em}$

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

$\ce{2C2H6 + 7O2 -> 6H2O + 4CO2}\hspace{4em}$

Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients . Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,

$\ce{3N2 + 9H2 -> 6NH3}$

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:

$\ce{N2 + 3H2 -> 2NH3}$

Use this interactive tutorial for additional practice balancing equations.

Additional Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases and aq for substances dissolved in water ( aqueous solutions ). These notations are illustrated in the example equation here:

$\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}$

This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).

Key Concepts and Summary

Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products.

End of Chapter Exercises

1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?

2. Balance the following equations:

(a) $\ce{PCl5(s) + H2O(l) -> POCl3(l) + HCl(aq)}$

(b) $\ce{Cu(s) + HNO3(aq) -> Cu(NO3)2(aq) + H2O(l) + NO(g)}$

(c) $\ce{H2(g) + I2(s) -> HI(s)}$

(d) $\ce{Fe(s) + O2(g) -> Fe2O3(s)}$

(e) $\ce{Na(s) + H2O(l) -> NaOH(aq) + H2(g)}$

(f) $\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + H2O(g)}$

(g) $\ce{P4(s) + Cl2(g) -> PCl3(l)}$

(h) $\ce{PtCl4(s) -> Pt(s) + Cl2(g)}$

3. Balance the following equations:

(a) Ag(s) +H 2 S(g) + O 2 (g) → Ag 2 S(s) + H 2 O(l)

(b) P 4 (s) + O 2 (g) → P 4 O 10 (s)

(c) Pb(s) + H 2 O(l) + O 2 (g) → Pb(OH) 2 (s)

(d) Fe(s) + H 2 O(l) → Fe 3 O 4 + H 2 (g)

(e) Sc 2 O 3 + SO 3 → Sc 2 (SO 4 ) 3

(f) Ca 3 (PO 4 ) 2 (aq) + H 3 PO 4 (aq) → Ca(H 2 PO 4 ) 2 (aq)

(g) Al(s) + H 2 SO 4 (aq) → Al 2 SO 4 (aq) + H 2 (g)

(h) TiCl 4 (s) + H 2 O(l) → TiO 2 (s) + HCl

4. Write a balanced molecular equation describing each of the following chemical reactions.

(a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.

(b) Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.

(c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.

(d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.

5. Write a balanced equation describing each of the following chemical reactions.

(a) Solid potassium chlorate, KClO 3 , decomposes to form solid potassium chloride and diatomic oxygen gas.

(b) Solid aluminum metal reacts with solid diatomic iodine to form solid Al 2 I 6 .

(c) When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.

(d) Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.

6. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.

(a) Write the formulas of barium nitrate and potassium chlorate.

(b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.

(c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.

7. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.

(a) The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.

(b) The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.

(c) Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.

(d) The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.

(e) Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.

Beyond GCSE Revision

Gcse-grade revision from beyond, powered by twinkl, balancing equations – a level chemistry revision.

Balanced chemical equations provide a lot of useful information about chemical reactions – to describe a reaction properly you need to be capable of correctly balancing equations. Work through the examples below to revise this important topic, then try the practice questions at the bottom!

As well as content on balancing equations, y ou can also subscribe to Beyond Secondary Resources for access to thousands of worksheets and revision tools. Our site was created with teachers in mind , however, it also contains content for students that will be particularly useful when revising! For more revision blogs, check out our A Level Chemistry section. You can sign up for a free account here and take a look around at our free resources before you subscribe too.

The reaction between solid calcium carbonate, CaCO 3 , and a solution of hydrochloric acid, HCl, produces a solution of calcium chloride, CaCl 2 , water, H 2 O, and carbon dioxide gas, CO 2 . This reaction can be represented by the following equation:

This is known as a full equation because it shows the full chemical formula of all of the reactants (on the left–hand side) and all of the products (on the right–hand side).

The arrow in the equation represents the transformation that occurs during the chemical reaction.

The numbers in front of the formulae of the reactants and products in a balanced equation represent the stoichiometry of the reaction. This is the ratio in which reactants react with each other and products are produced. In the example above, one mole of calcium carbonate would react completely with two moles of hydrochloric acid to produce one mole of calcium chloride, one mole of water and one mole of carbon dioxide. It is the stoichiometry of the reaction that you adjust when balancing equations.

Finally, state symbols indicate the physical state of reactants and products during the chemical reaction: (s) represents a solid, (l) represents a liquid, (g) represents a gas and (aq) represents an aqueous solution (a compound dissolved in water).

Balancing Equations

Atoms are not lost or gained in chemical reactions; they are simply rearranged to form different substances. Therefore, the total number of each type of atom represented on the left–hand side of an equation must equal the total number of each type of atom represented on the right–hand side of an equation. In the full equation above, the number of atoms of each element is the same on both sides of the equation, so the equation is balanced.

If the equation is not balanced, you must change the stoichiometric ratios in the equation until it is balanced.

Example Question 1

Write a balanced full equation for the reaction between sodium metal and oxygen to produce sodium oxide.

Correctly balancing equations requires you to use your knowledge of the periodic table and chemical formulae, structure and bonding to deduce the chemical formulae of the reactants and products. You can assume that all substances are in their standard states unless otherwise stated.

Formulae of Reactants sodium Na(s) oxygen O 2 (g)

Formula of Product sodium oxide Na 2 O(s) (Sodium oxide is an ionic compound containing Na + and O 2– ions.)

Na(s) + O 2 (g) → Na 2 O(s)

Next, check the number of each type of atom on each side of the equation.

The equation is not balanced. Both the number of oxygen atoms on the right-hand side and the number of sodium atoms on the left hand side are wrong. Start by correcting the number of oxygen atoms on the right – increasing the number of molecules of Na 2 O will also change the amount of sodium required on the left-hand side, which can be adjusted later.

Na(s) + O 2 (g) → 2Na 2 O(s)

Four atoms of sodium are now needed on the left–hand side of the equation.

4Na(s) + O 2 (g) → 2Na 2 O(s)

The equation is now balanced: 4Na(s) + O 2 (g) → 2Na 2 O(s)

Ionic Equations

In some reactions, there are ions present that do not take part in the overall reaction. These are known as spectator ions (ions that have the same formula, charge and physical state on both sides of the equation).

An ionic equation can be used to show only the ions that take part in the reaction.

The steps to writing an ionic equation are outlined below.

• Start with the full balanced equation for the reaction. CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g)
• Identify any ions that are present in each reactant and product.
• Write the equation out showing all ions that are present. Ca 2+ (s) + CO 3 2– (s) + 2H + (aq) + 2Cl – (aq) → Ca 2+ (aq) + 2Cl – (aq)+ H 2 O(l) + CO 2 (g)
• Identify the spectator ions. These are ions that have not taken part in the overall reaction and remain unchanged. The spectator ions can be cancelled out of the equation. Ca 2+ (s) + CO 3 2– (s) + 2H + (aq) + 2Cl – (aq) → Ca 2+ (aq) + 2Cl – (aq) + H 2 O(l) + CO 2 (g)
• Rewrite the equation without the spectator ions. At this stage you can write the full chemical formula for compounds where all ions are still present in the equation. CaCO 3 (s) + 2H + (aq) → Ca 2+ (aq) + H 2 O(l) + CO 2 (g)

Example Question 2

Write a balanced ionic equation for the reaction between potassium iodide solution and silver nitrate solution to form a precipitate of silver iodide and soluble potassium nitrate.

Start by writing a balanced, full equation:

Formulae of Reactants

potassium iodide KI(aq) silver nitrate AgNO3(aq)

Formulae of Products

silver iodide AgI(s) potassium nitrate KNO 3 (aq)

Balanced full equation: KI(aq) + AgNO 3 (aq) → AgI(s) + KNO 3 (aq)

In this case, it is easy to see by inspection that the equation is balanced. Next, identify the ions present:

Next, write an equation with ions:

K + (aq) + I – (aq) + Ag + (aq) + NO 3 – (aq) → Ag + (s) + I – (s) + K + (aq) + NO 3 – (aq)

Then, identify the spectator ions and cancel them out:

This leaves a balanced ionic equation of:

I – (aq) + Ag + (aq) → AgI(s)

Practice Questions

1. Write a balanced full equation for the reaction between magnesium metal and oxygen to produce magnesium oxide.

2. Write a balanced full equation for the reaction between lithium hydroxide solution and sulfuric acid to produce soluble lithium sulfate and water.

3. Write a balanced full equation for the reaction between solid aluminium oxide and phosphoric acid to produce insoluble aluminium phosphate and water.

4. Write a balanced ionic equation for the reaction between hydrochloric acid and sodium hydroxide solution to produce sodium chloride and water.

5. Write a balanced ionic equation for the reaction between magnesium metal and iron(II) nitrate solution to produce magnesium nitrate solution and iron metal.

6. Write a balanced ionic equation for the reaction between barium chloride solution and sodium sulfate solution to produce a precipitate of barium sulfate and aqueous sodium chloride.

2Mg(s) + O 2 (g) → 2MgO(s)

2LiOH(aq) + H 2 SO 4 (aq) → Li 2 SO 4 (aq) + 2H 2 O(l)

Al 2 O 3 (s) + 2H 3 PO 4 (aq) → 2AlPO 4 (s) + 3H 2 O(l)

H + (aq) + OH – (aq) → H 2 O(l)

Mg(s) + Fe 2+ (aq) → Mg 2+ (aq) + Fe(s)

Ba 2+ (aq) + SO 4 2– (aq) → BaSO 4 (s)

Share this:

• Click to share on Twitter (Opens in new window)
• Click to share on Facebook (Opens in new window)
• Click to share on LinkedIn (Opens in new window)
• Click to share on Pinterest (Opens in new window)

Leave a Reply Cancel reply

Discover more from beyond gcse revision.

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading