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Lesson 12 Balancing Chemical Reactions

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Course: Science 10 [5 cr] - AB Ed copy 1
Book: Lesson 12 Balancing Chemical Reactions
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Date: Sunday, 7 September 2025, 6:44 PM

Table of contents

  Introduction

The ability to balance equations and to predict what products will be formed when substances are mixed together is a fundamental chemistry skill.



B12.1 Writing a chemical reaction
The law of conservation of mass states that for a closed system, all matter must remain constant. This principle will be applied to balancing chemical reaction equations.

In this lesson, you will learn how to balance chemical reaction equations. You will analyze the number of atoms of each element and, by adding in coefficients, ensure there are the same number of atoms of each element on the reactant side and the product side of the chemical equation.

  Targets

By the end of this lesson, you will be able to

  • balance chemical reaction equations
  • translate word equations into balanced chemical equations
  • predict products for formation, decomposition, hydrocarbon combustion, single replacement, and double replacement reactions when given the reactants

  Simulation

Balancing Chemical Equations @ Explore Learning


Please review this balancing chemical equation simulation to help you visualize the process of balancing. Please note that they use the term “synthesis” instead of “formation.” This will help get you in the right mindset for this lesson.

Click on the procedure tab to continue.

Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.

  1. Click on the play icon to open the virtual lab. You can also access this lab from the Online Resources for Print Students section of your course.
  2. Ensure “Choose a reaction:” has “Synthesis” selected.
  3. Select “Show summary.”
  4. Take a screenshot of the equation and reaction vessel.
  5. Determine what atoms are not balanced.
  6. Add molecules to balance the number of atoms from before the reaction to after the reaction.
  7. Repeat with all atoms until the equation is balanced.
  8. Once balanced, take a screenshot of the equation and reaction vessel.
  9. Repeat steps 1 to 6 with “Single replacement,” “Decomposition,” and “Double replacement” reactions.
  10. Click on the analysis tab to complete the analysis questions.
  1. Take a screenshot of the initial setup of the model of the “Synthesis” reaction. Identify how you can tell it is not balanced.

    You can identify that it is not balanced because there are not the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.2 Model of formation chemical reaction


  2. Take a screenshot of the balanced equation of the model of the “Synthesis” reaction. Identify how you can tell it is balanced.

    You can identify that it is balanced because there are the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.3 Model of balanced formation chemical reaction


  3. Take a screenshot of the initial setup of the model of the “Decomposition” reaction. Identify how you can tell it is not balanced.

    You can identify that it is not balanced because there are not the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.4 Model of decomposition chemical reaction


  4. Take a screenshot of the balanced equation of the model of the “Decomposition” reaction. Identify how you can tell it is balanced.

    You can identify that it is balanced because there are the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.5 Model of balanced decomposition chemical reaction



  5. Take a screenshot of the initial setup of the model of the “Single replacement” reaction. Identify how you can tell it is not balanced.

    You can identify that it is not balanced because there are not the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.6 Model of single replacement chemical reaction


  6. Take a screenshot of the balanced equation of the model of the “Single replacement” reaction. Identify how you can tell it is balanced.

    You can identify that it is balanced because there are the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.7 Model of balanced single replacement chemical reaction


  7. Take a screenshot of the initial setup of the model of the “Double replacement” reaction. Identify how you can tell it is not balanced.

    You can identify that it is not balanced because there are not the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.8 Model of double replacement chemical reaction


  8. Take a screenshot of the balanced equation of the model of the “Double replacement” reaction. Identify how you can tell it is balanced.

    You can identify that it is balanced because there are the same number of atoms of each element on both sides of the arrow.

    © Explore Learning
    B12.9 Model of balanced double replacement chemical reaction


  Balancing Chemical Reactions

Balancing a chemical reaction equation ensures that the law of conservation of mass is followed.



B12.10 Writing a chemical reaction
A balanced chemical reaction equation shows that the atoms are conserved during a chemical reaction. Coefficients are added to skeleton equations to ensure that the number of atoms of each element on the reactant side equals the number of atoms of each element on the product side.

Remember, you cannot change the subscripts in a formula to balance an equation. That would alter the identity of the compound. Instead, use coefficients.

For example, given the skeleton equation for the production of water from hydrogen and oxygen,

H2(g) + O2(g) → H2O(l)

you can immediately notice that the oxygen atoms are not balanced. By placing a coefficient of 2 in front of the water, the oxygen atoms will be balanced:

H2(g) + O2(g) → 2H2O(l)

This, however, causes the H atoms to not be balanced:

H2(g) + O2(g) → 2H2O(l)

But by placing a 2 in front of the hydrogen gas, they will be balanced:

2H2(g) + O2(g) → 2H2O(l)

By convention, no 1 is placed in front of the oxygen gas; it is assumed to be there.

You can check your balancing by making a chart that compares the number of atoms on the reactant side to the number of atoms on the product side.

Entity Reactants Products
H 4 4
O 2 2


Watch this video to see a teacher work through this example.  https://adlc.wistia.com/medias/5q00dqu3j7
 


Examples


Work though the following examples to ensure you have strong comprehension of balancing chemical reaction equations. You will be balancing by inspection; so the best place to start is with the element on the far left of the equation and then working your way through all elements in order. Each example has a video to go with it. To play the video, click on the play icon next to the example.

Balance the following skeleton reaction equation:

Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)

Balance by inspection:

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Mg atoms are balanced, as there is 1 Mg on both sides of the equation.

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H atoms are not balanced, as there is 1 H atom on the reactant side and 2 H atoms on the product side. Add a coefficient of 2 in front of the HCl.

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Cl atoms are balanced, as there are 2 Cl atoms on both sides of the arrow. The balanced chemical reaction equation is

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Entity Reactants Products
Mg 1 1
H 2 2
Cl 2 2


Tip: If a polyatomic ion stays together as a unit, it can be balanced as one entity instead of looking at each individual atom within the polyatomic.

Balance the following skeleton reaction equation:

AlI3(aq) + AgNO3(aq) → AgI(s) + Al(NO3)3(aq)

Balance by inspection:

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https://adlc.wistia.com/medias/8wf9ckx0z4
Al atoms are balanced as there is 1 AL atom on both sides of the reaction.

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I atoms are not balanced, as there are 3 I atoms on the reactant side and only 1 I atom on the product side. Add a coefficient of 3 in front of AgI.

«math»«mtable»«mtr»«mtd»«msub»«mi»AlI«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»AgNO«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«mi»AgI«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»Al«/mi»«mo»(«/mo»«msub»«mi»NO«/mi»«mn»3«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
 
Ag atoms are not balanced, as there is 1 Ag atom on the reactant side and 3 Ag atoms on the product side. Add a coefficient of 3 in front of the AgNO3.

«math»«mtable»«mtr»«mtd»«msub»«mi»AlI«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«msub»«mi»AgNO«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«mi»AgI«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»Al«/mi»«mo»(«/mo»«msub»«mi»NO«/mi»«mn»3«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»aq«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»
NO3– ion is balanced, as there are 3 NO3– ions on either side of the equation.

AlI3(aq) + 3AgNO3(aq) → 3AgI(s) + Al(NO3)3(aq)

Entity Reactants Products
Al 1 1
I
 3 3
Ag 3 3
NO3 3 3


Balance the following skeleton reaction equation:

Fe(s) + Br2(l) → FeBr3(s)

Balance by inspection:

«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»

Fe atoms are balanced, as there is 1 Fe atom on both sides of the reaction.

«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»

Br atoms are not balanced, as there are 2 Br atoms on the reactant side and 3 Br atoms on the product side. The lowest common multiple between 2 and 3 is 6; so multiple each substance by the correct factor.

«math»«mtable»«mtr»«mtd»«mi»Fe«/mi»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»3«/mn»«msub»«mi»Br«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»l«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mn»2«/mn»«msub»«mi»FeBr«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mo mathcolor=¨#FF0000¨»§#8593;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«/mtd»«/mtr»«/mtable»«/math»

Fe atoms are no longer balanced, as there is 1 atom on the reactant side and 2 atoms on the product side. Add a coefficient of 2 in front of Fe.

2Fe(s) + 3Br2(l) → 2FeBr3(s)

Entity Reactants Products
Fe
 2 2
Br
 6 6


Balance the following skeleton reaction equation:

C4H10(g) + O2(g) → CO2(g) + H2O(g)

Watch this video to learn how to balance hydrocarbon combustions by inspection.


2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

  Read This

Please read pages 87 to 88 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the technique for balancing chemical reaction equations. Remember, if you have any questions or you do not understand something, ask your teacher!

  Practice Questions

Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.

Balance the following skeleton reaction equations.

  1. __Fe(s) + __S8(s) → __FeS(s)

    8Fe(s) + S8(s) → 8FeS(s)
  2. __KClO3(s) → __KCl(s) + __O2(g)

    2KClO3(s) → 2KCl(s) + 3O2(g)
  3. __H3PO4(aq) + __KOH(aq) → __K3PO4(aq) + __HOH(l)

    H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3HOH(l)
  4. __Fe2O3(s) + __C(s) → __Fe(s) + __CO(g)

    Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g)
  5. __K2O(s) + __H2O(l) → __KOH(aq)

    K2O(s) + H2O(l) → 2KOH(aq)
  6. __C6H6(l) + __O2(g) → __CO2(g) + __H2O(g)

    2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(g)
  7. __K3PO4(aq) + __MgCl2(aq) → __Mg3(PO4)2(s) + __KCl(aq)

    2K3PO4(aq) + 3MgCl2(aq) → Mg3(PO4)2(s) + 6KCl(aq)
  8. __SO2(g) + __H2O(l) → __H2SO3(aq)

    SO2(g) + H2O(l) → H2SO3(aq)
  9. __LiCl(aq) + __F2(g) → __LiF(aq) + __Cl2(g)

    2LiCl(aq) + F2(g) → 2LiF(aq) + Cl2(g)
  10. __NiBr2(aq) + __Al(s) → __AlBr3(aq) + __Ni(s)

    3NiBr2(aq) + 2Al(s) → 2AlBr3(aq) + 3Ni(s)

  Predicting Products

If you know what the reactants are for a chemical reaction, can you predict the products?



B12.11 A student writing a chemical reaction equation on a chalk board
For all reactions so far, both the reactants and products have been identified for you. The next skill that you are going to develop is the ability to predict the products when you are given the identity of the reactants. You will look at each of the five types of chemical reactions. Using the patterns for each reaction will aide you in predicting the products.

Each example has a video to go with it. To play the video, click on the play icon next to the example.

In simple formation reactions, the product formed will be the compound that results from the combination of the two elements reacting. You will need to recall from Section 2 how to balance charges to create neutral compounds.

For example, predict the product and write a balanced chemical reaction equation when copper on the roof of the parliament building oxidizes (reacts with oxygen) to create a green patina.


Write a word equation. Keep in mind that copper is a multivalent metal; so the most common charge, which is written first on the periodic table, is the one that will be used.

copper + oxygen → copper(II) oxide
Write the skeleton equation, remembering to write correct formulas for all compounds.

Cu(s) + O2(g) → CuO(s)
Balance the equation.

2Cu(s) + O2(g) → 2CuO(s)

In simple decomposition reactions, a compound is broken down into its constituent elements. (Tip: You need to recall the seven diatomic elements.)

For example, predict the products and write a balanced chemical reaction equation when molten aluminum chloride is electrolyzed (decomposed).

Word equation:

aluminium chloride → aluminium + chlorine
Skeleton equation:
AlCl3(l) → Al(s) + Cl2(g)
Balanced equation:
2AlCl3(l) → 2Al(s) + 3Cl2(g)

Tip: Remember that states of elements are listed on the periodic table.


The key to predicting products for a single replacement reaction is to carefully identify cations and anions and remember that ionic compounds need to be composed of a cation bonded to an anion. By first identifying entities as being able to form a cation or an anion, you can then determine which entitles switch places.

Predict the products and write a balanced chemical reaction equation when lithium iodide solution is reacted with liquid bromine.

Write the word equation and identify the entities as being able to form a cation or an anion:

«math»«mtable»«mtr»«mtd»«mi»lithium«/mi»«/mtd»«mtd»«mi»iodide«/mi»«/mtd»«mtd»«mo»+«/mo»«/mtd»«mtd»«mi»bromine«/mi»«/mtd»«mtd»«mo»§#8594;«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi mathcolor=¨#FF0000¨»cation«/mi»«/mtd»«mtd»«mi mathcolor=¨#FF0000¨»anion«/mi»«/mtd»«mtd»«/mtd»«mtd»«mi mathcolor=¨#FF0000¨»anion«/mi»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»
Determine which entities switch places:

Iodine and bromine are both anions, so they will switch places.

«math»«mtable»«mtr»«mtd»«mi»lithium«/mi»«/mtd»«mtd»«mi»iodide«/mi»«/mtd»«mtd»«mo»+«/mo»«/mtd»«mtd»«mi»bromine«/mi»«/mtd»«mtd»«mo»§#8594;«/mo»«/mtd»«mtd»«mi»lithium«/mi»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mi»bromide«/mi»«/mtd»«/mtr»«/mtable»«mtable»«mtr»«mtd»«mo»+«/mo»«/mtd»«mtd»«mi»iodine«/mi»«/mtd»«/mtr»«/mtable»«mspace linebreak=¨newline¨»«/mspace»«mtable mathcolor=¨#FF0000¨»«mtr»«mtd»«mo»§#160;«/mo»«mi»cation«/mi»«/mtd»«mtd»«mi»anion«/mi»«/mtd»«mtd»«/mtd»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»anion«/mi»«/mtd»«mtd»«/mtd»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»cation«/mi»«/mtd»«/mtr»«/mtable»«mtable mathcolor=¨#FF0000¨»«mtr»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»anion«/mi»«/mtd»«/mtr»«/mtable»«mtable mathcolor=¨#FF0000¨»«mtr»«mtd»«/mtd»«mtd»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»anion«/mi»«/mtd»«/mtr»«/mtable»«/math»
Skeleton equation:
LiI(aq) + Br2(l) → LiBr(aq) + I2(s)
Balanced equation:
2LiI(aq) + Br2(l) → 2LiBr(aq) + I2(s)



Predict the products and write a balanced chemical reaction equation when magnesium metal is reacted with hydrofluoric acid.

Write the word equation and identify the entities as being able to form a cation or an anion:

«math»«mtable»«mtr»«mtd»«mi»magnesium«/mi»«/mtd»«mtd»«mo»+«/mo»«/mtd»«mtd»«mi»hydrofluoric«/mi»«mo»§#160;«/mo»«mi»acid«/mi»«/mtd»«mtd»«mo»§#8594;«/mo»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mi mathcolor=¨#FF0000¨»cation«/mi»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«/mrow»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«mi»cation«/mi»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mi»anion«/mi»«/mrow»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«/mrow»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«/mrow»«/mtd»«/mtr»«/mtable»«/math»
Determine which entities switch places:

Magnesium and hydrogen are both cations, so they will switch places.

«math»«mtable»«mtr»«mtd»«mi»magnesium«/mi»«/mtd»«mtd»«mo»+«/mo»«/mtd»«mtd»«mi»hydrofluoric«/mi»«mo»§#160;«/mo»«mi»acid«/mi»«/mtd»«mtd»«mo»§#8594;«/mo»«/mtd»«mtd»«mi»magnesium«/mi»«/mtd»«mtd»«mi»flouride«/mi»«/mtd»«mtd»«mo»+«/mo»«/mtd»«mtd»«mi»hydrogen«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi mathcolor=¨#FF0000¨»cation«/mi»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«/mrow»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«mi»cation«/mi»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mi»anion«/mi»«/mrow»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«/mrow»«/mtd»«mtd»«mi mathcolor=¨#FF0000¨»cation«/mi»«/mtd»«mtd»«mi mathcolor=¨#FF0000¨»anion«/mi»«/mtd»«mtd»«mrow mathcolor=¨#FF0000¨»«/mrow»«/mtd»«mtd»«mi mathcolor=¨#FF0000¨»cation«/mi»«/mtd»«/mtr»«/mtable»«/math»
Skeleton equation:
Mg(s) + HF(aq) → MgF2(aq) + H2(g)
Balanced equation:
Mg(s) + 2HF(aq) → MgF2(aq) + H2(g)


The key concept to keep in mind for double replacement reactions is that the cations will switch places. Remember: In ionic compounds, a cation must be bonded to an anion. You will also need to check the solubility table to determine the states of your products. Quite often, a precipitate (solid) will be produced during a double replacement reaction.

Predict the products and write a balanced chemical reaction equation when solutions of sodium chloride and lead(II) chlorate are mixed.

Word equation:



Skeleton equation:
NaCl(aq) + Pb(ClO3)2(aq) → PbCl2(s) + NaClO3(aq)
Balanced equation:
2NaCl(aq) + Pb(ClO3)2(aq) → PbCl2(s) + 2NaClO3(aq)



Predict the products and write a balanced chemical reaction equation when solutions of iron(III) nitrate and potassium carbonate are mixed.

Word equation:



Skeleton equation:
Fe(NO3)3(aq) + K2CO3(aq) → KNO3(aq) + Fe2(CO3)3(s)
Balanced equation:
2Fe(NO3)3(aq) + 3K 2CO3(aq) → 6KNO3(aq) + Fe2(CO3)3(s)



Predicting products for hydrocarbon combustion is quite straightforward. When it is a complete combustion reaction, the hydrocarbon reacts with oxygen to produce carbon dioxide and water vapour.

For example, write the balanced chemical reaction equation for the combustion of butane, C4H10(g).

Notice that butane is a hydrocarbon and that this is a combustion reaction. Therefore, the other reactant will be oxygen gas, and the products will be carbon dioxide and water vapour.
Word equation:
butane + oxygen → carbon dioxide + water vapour
Skeleton equation:
C4H10(g) + O2(g) → CO2(g) + H2O(g)
Balanced equation:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)



  Read This

Please read pages 102 to 105 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on determining products and writing balanced chemical reaction equations. Remember, if you have any questions or you do not understand something, ask your teacher!

  Practice Questions

Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.

Balance the following skeleton reaction equations.

  1. Write a word equation and a balanced chemical reaction equation for the formation of zinc nitride from its elements.

    zinc + nitrogen → zinc nitride
    3Zn(s) + N2(g) → Zn3N2(s)
  2. Write a word equation and a balanced chemical reaction equation for the complete combustion of pentane, C5H12(l).

    pentane + oxygen → carbon dioxide + water vapour
    C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(g)
  3. Write a word equation and a balanced chemical reaction equation for the reaction of aqueous magnesium sulfate with aqueous strontium bromide.



    MgSO4(aq) + SrBr2(aq) → SrSO4(s) + MgBr2(aq)

  4. Write a word equation and a balanced chemical reaction equation for the reaction of zinc metal with a solution of tin(IV) chloride.



    2Zn(s) + SnCl4(aq) → Sn(s) + 2ZnCl2(aq)


    Note: When tin is written by itself as an element, no designation of charge is included (i.e., it is not written as tin(IV)).
  5. Write a word equation and a balanced chemical reaction equation for the decomposition of iron(III) oxide into its elements.

    iron(III) oxide → iron + oxygen
    2Fe2O3(s) → 4Fe(s) + 3O2(g)

  Conclusion

Properly written and balanced chemical reaction equations are essential for future studies in chemistry.



B12.12 Chemical equation
In this lesson, you learned how to balance chemical reaction equations and how to predict products when given the reactant species.

Writing balanced chemical reaction equations requires you to recall from Section 2 how to properly write chemical formulas for compounds.

In the next lesson, you will look at how to calculate amounts of chemical substances.

  Interactive

Balancing Chemical Equations @ PhET


Work through this simulation to help you visualize the number of atoms and molecules in a chemical reaction.

Click on the procedure tab to continue.

  1. Click on the play icon to open the interactive. This interactive can also be accessed at https://phet.colorado.edu/sims/html/balancing-chemical-equations/latest/balancing-chemical-equations_en.html
  2. Select “Game.”
  3. Select “Level 1.”
  4. Adjust the coefficient for each reactant and product until you think the reaction is balanced (note that you need to include a 1 if required for the balancing) and then click “Check.”
  5. You can only check your answer twice before you will need to move on to the next equation.
  6. Once you have completed Level 1, move on to Level 2 and Level 3.


2.6 Assignment

Unit 2 Assignment Lessons 10-13

It is now time to complete the Lesson 12 portion of 2.6 Assignment. This assignment has two parts.

  1. Part 1 Written Portion: Select the preferred document type from the options below. Download and save the assignment on your desktop (or documents folder).

    PDF Document       
  2. Open and print this saved document.
  3. Record your responses in the appropriate textboxes.
  4. When you have completed the assignment, scan it and save it on your desktop (or documents folder).
  5. Once you have completed the written portion of your assignment, click on the button below to go to the submission page.

    Written Portion Submission Page
  6. Part 2 Online Portion: It is now time to complete the Lesson 5 questions of the online portion of this assignment. Click on the button below to go to the online questions of this assignment.

    Online Questions

This assignment is worth ___% of your final grade.