Chapter 5: Unit 1. Balancing Chemical Equations

Balancing Chemical Equations

The study of how and why a chemical reaction occurs is a very important part of chemistry. In this section we will examine how to describe a chemical change or a reaction through equation and how to balance an equation. Balancing of chemical equation is required because of law of conservation of mass and energy. According to law of conservation of mass, matter of the universe is conserved and it can never be destroyed or created. Therefore, the total atoms present in a reaction before and after must be same. The bonded atoms are rearranged in a chemical reaction but the total mass remains same.

Just like a cooking recipe, A chemical equation contains the starting materials on left side of an arrow and the products on the right. The coefficients are written in front of the reactants and products and they tell how many molecules or moles of a substance react or are formed. States of the reactant and product substances are written with parenthesis. Solid substances are written as (s), liquid substances are written as (l) and gaseous substances are written as (g), Solutions are mainly expressed as (aq) means aqueous.

A chemical equation is balanced by placing coefficients in front of chemical formulas one at a time. Beginning with most complex formula, so that the number of atoms of each element is the same on both sides. Sometimes fractional coefficients are used to balance equation which then converted to an integer by multiplying the entire equation by a whole number. Any special reaction condition is mentioned on the top of the arrow like heat (∆) of light (hv) etc.

Here are some examples of balanced chemical Equation.

Let’s take the first example: Unbalanced equation

Fe (s) + O2(g) → Fe2O3(s)

We will count total number of atoms on each side first

Left hand side Fe-1                   Right hand side Fe-2
O-2                                                               O-3

Let’s balance Fe first with putting coefficient 2 in front of Fe on the reactant side.

2Fe (s) + O2(g) → Fe2O3(s)

But the equation is still unbalanced, because the number of oxygen is not same. To make same number of oxygen on both sides, we must multiply left hand side by 3 and right hand side by 2.

2Fe(s) + 3O2(g) → 2Fe2O3(s)

Left hand side Fe-1                      Right hand side Fe-4
O-6                                                                                O-6

To make total number of Fe atom equals 4 on the left hand side, we put coefficient 4.

So the final balanced equation is:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

C₂H­₆(g) + O₂(g)  → CO₂ + H₂O

Sometimes we have to use fractional coefficients to balance an equation. For example, in the following equation, After balancing C with coefficient 2 and H with coeff. 3 on the product side, the total number of oxygen is 2*2 from CO₂ and 3*1=3 from H₂O, total seven atoms of O.

On the left hand side we have 2 atoms O. To balance O atoms we have to put coefficient 7/2.

Balance equation is:

• But fractional coefficients are not used.
• Complete balanced equation:
• 2C2H6 + 7O2 → 4CO2 + 6H2O

Try this Balancing app(free): “Reactions” with more than 400 equations for practicing balancing. Look for the app in app store on any mobile device.

https://phet.colorado.edu/en/simulation/balancing-chemical-equations

Go to the above link and click on the activity “Introduction”.

We will play with the topic, separation of water. The screen should appear like below.

On the left hand side we have H-2 O-1

On the right hand side H-2, O-2. Apparently Hydrogen is balanced but no oxygen. To balance oxygen we have use coefficient 2 in front of water . Now we have 2*2 =4 atoms H and 2 atoms of O. We adjust the number of H atoms on the right hand side by changing the coefficient equal 2 for H₂ and 1 for O₂. The equation is balanced.

In case of polyatomic ion containing equations, it is always recommended to consider them as unit rather than broken down to individual atoms if they appear on both sides of the equation.

LiOH  + H₃PO4   → Li₂HPO4 +H₂O (unbalanced)

To balance Li, we have to put coefficient 2 in front of LiOH. That will give total number H on the left hand side= 5 and O=2, keeping PO₄ unit separate  from other elements. We can put coefficient 2 in front of H₂O to form 4 H and 2 O, remaining one H is present in Li₂HPO4. PO₄ unit is same on left and right hand side. So the final number of atoms look like :

Left side: Li -2                      Right side: Li-2

O-2                                           O-2

H-5                                          H-5

PO4-1                                       PO4-1

Balanced equation is

2LiOH  + H₃PO4   → Li₂HPO4 +2H₂O (balanced)

Always lowest number of coefficients are used to balance a chemical equation. For example, 2H₂O → 2H₂ + O_2, this equation can also be balanced by 4H₂O → 4H₂ + O₂

But we always use lowest set of whole numbers to balance an equation.

The following video might help you to understand more this topic.

Rules for balancing chemical equations:

1. Mentally, draw a box or circle around chemical formulas – you cannot change any symbol or subscript in the formula to balance equation.
2. Example 1: You cannot change a subscript – H₂O is different than H₂O₂!!
3. Example 2: You cannot insert coefficients in middle of formula – H₂2O is not correct.
4. Count up the number of each type of atom on both sides of the equation. You might want to make a simple table to keep track as you learn how to balance equation.
5. Add coefficients to the front of the boxes to balance the equation and update your element count. Coefficients must be whole numbers.
6. These tips will help you balance equations

Remember to write the seven diatomic elements (H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂) with the subscript₂. Once they react, they will exist as individual atoms in a molecule.

1. If the same polyatomic ion appears both side of the reaction, put a mental box around it and treat it as a single unit
2. In some types of ionic reactions it will help to write water as H–OH instead of H₂
3. Balance the elements in compounds first. Start with metals and then balance nonmetals.
4. Leave the reactants and products that are elements until the end.
5. When the number of atoms of each element is the same before and after the reaction, equation is balanced.

Sample Problems

Example 1

Step 1: Determine the number of each type of atom that are on the reactant and the product side of the equation:

___Mg (s) + ___HCl (aq) → ___H₂ (g) + ___MgCl₂ (aq)

Reactant	Atom	Product	Mg is balanced but H and Cl are not.  and Cl are not.There are twice as many products as reactants for H and Cl.dd a coefficient of 2 in front of HCl.
1	Mg	1
1*	H	2*
1*	Cl	2*

       (*) indicates that the atoms are not balanced in the equation.

                                           Coefficients must be used.

Step 2: Add coefficients to balance the equation. In this case, a coefficient of 2 in front of HCl will balance the equation. The rest of the coefficients are 1.

        1 Mg (s) + 2 HCl (aq) → 1 H₂ (g) + 1 MgCl₂ (aq)

Reactant	Atom	Product	All atoms are balanced.The equations is balanced.
1	Mg	1
2	H	2
2	Cl	2

Example 2

Step 1: Determine the number of each type of atom that are on the reactant and the product side of the equation:

___AgNO₃ (aq) + ___MgCl₂ (aq) → ___AgCl (s) + ___Mg(NO₃)₂ (aq)

Reactant	Atom or Unit	Product	Ag and Mg are balanced but NO3 and Cl are not. Start with the NO3.
1	Ag	1
1*	NO3	2*
1	Mg	1
2*	Cl	1*

  (*) indicates that the atoms are not balanced in the equation.

                                          Coefficients must be used.

Step 2: Add coefficients to balance the equation. In this case, add a coefficient of 2 to balance the NO₃ and add a coefficient of 1 to the others at this point. The number of atoms have changed, updated values appear in the table.

                         2 AgNO₃ (aq) + 1 MgCl₂ (aq) → 1 AgCl (s) + 1 Mg(NO₃)₂ (aq)

Reactant	Atom or Unit	Product	NO3 and Mg are balanced butAg and Cl are not.There are twice as many reactants as product for both Ag and Cl.
2*	Ag	1*
2	NO3	2
1	Mg	1
2*	Cl	1*

(*) indicates that the atoms are not balanced in the equation.

                                           Coefficients must be changed.

Step 3: Change the coefficient for AgCl to 2 to balance the equation. The number of atoms have again changed, the updated values appear in the table.

2 AgNO₃ (aq) + 1 MgCl₂ (aq) → 2 AgCl (s) + 1 Mg(NO₃)₂ (aq)

Reactant	Atom or Unit	Product	All atoms are balanced.The equation is balanced.
2	Ag	2
2	NO3	2
1	Mg	1
2	Cl	2

Pre-Activity Questions:

1. Write the name and then determine the number of atoms of each element in the following chemical formulas:

H3PO4	N5O3	Al2(SO3)3
Name:	Name:	Name:
No. of H  ___________ No. of P  ___________ No. of O  ___________	No. of N  ___________ No. of O  ___________	No. of Al  ___________ No. of S  ___________ No. of O  ___________

2. Use this equation to answer the following questions:

1 Ca (s) + 2 HF (aq) → 1 CaF₂ (s) +1 H₂ (g)

Write a word equation for this reaction.
Is the reaction balanced?
Identify the coefficient(s).
Identify the subscript(s).
What is the state of each of the reactants?
What will serve as evidence that a reaction has occurred?

Simulation

1. Log onto the simulation: http://bit.ly/BalancingEquationsPhET
2. Select “Introduction”.
3. Select “Make Ammonia” and then select the balance “Tool”. Complete the table to count the number of each type of atom in the space provided below. Write the balanced equation next to the table.

Reaction 1Make Ammonia	Total Number of Atoms
Reactants	Products
H
N

4. Reset the simulation.
5. Select “Separate Water” and then select the bar graph “Tool”.
6. Add coefficients until the equation is balanced. Complete the table to count the number of each type of atom in the space provided below. Write the balanced equation next to the table.

Reaction 2Separate Water	Total Number of Atoms
Reactants	Products
H
O

7. Reset the simulation.
8. Select “Combust Methane” and then select “None” for “Tools”.
9. Add coefficients until the equation is balanced. Complete the table to count the number of each type of atom in the space provided below. Write the balanced equation next to the table.

Reaction 3Combust Methane	Total Number of Atoms
Reactants	Products
C
H
O

10. Select “Game” at the bottom of the page and then select “Level 1”.
11. Balance the first equation, and then use the “Check” button to see if your equation is balanced.
    1. If it is, move onto the next equation.
    2. If not, use the “Show Why” button to see the number of each type of atom and then “Try Again”.
    3. Show all of your work and write your final equation in the space provided below.

12. Continue on to Level 2 and follow the directions given for Level 1. Show all of your work and write your final equation in the space provided below.

1

2

3

4

5

13. Continue on to Level 3 and follow the directions given for Level 1. Show all of your work and write your final equation in the space provided below.

1

2

3

4

5

Questions

1. Balance the following equations:

1. Al + O₂ → Al₂O₃
2. C₃H₈ + O₂ → CO₂ + H₂O
3. BaCl₂ + NaNO₃ → Ba(NO₃)₂ + NaCl
4. Fe₂O₃(s) + H₂(g) → Fe(s) + H₂O(g)
5. CaO(s) + C(s) → CaC₂(s) + CO₂(g)
6. Na₂CO₃(s) + H₂CO₃(aq) → Na(HCO₃)₂(aq)
7. KNO₃(aq) + H₂SO₄(aq) → K₂SO₄(aq) + HNO₃(aq)

Ans: a) 4, 3, 2
b) 1, 5, 3, 4
c) 1, 2, 1, 2
d) 1, 3, 2, 3
e) 2, 5, 2, 1
f) 1, 1, 2
g) 2, 1, 1, 2