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Balancing chemical equations

Module by: Free High School Science Texts Project. E-mail the author

Balancing chemical equations

The law of conservation of mass

In order to balance a chemical equation, it is important to understand the law of conservation of mass.

Definition 1: The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the processes acting inside the system. Matter can change form, but cannot be created or destroyed. For any chemical process in a closed system, the mass of the reactants must equal the mass of the products.

In a chemical equation then, the mass of the reactants must be equal to the mass of the products. In order to make sure that this is the case, the number of atoms of each element in the reactants must be equal to the number of atoms of those same elements in the products. Some examples are shown below:

Example 1:

Fe+SFeSFe+SFeS
(1)

Figure 1
Figure 1 (CG10C5_001.png)

Reactants

Atomic mass of reactants=55,8u+32,1u=87,9uAtomic mass of reactants=55,8u+32,1u=87,9u

Number of atoms of each element in the reactants: (1×Fe)(1×Fe) and (1×S)(1×S)

Products

Atomic mass of products=55,8u+32,1u=87,9uAtomic mass of products=55,8u+32,1u=87,9u

Number of atoms of each element in the products: (1×Fe)(1×Fe) and (1×S)(1×S)

Since the number of atoms of each element is the same in the reactants and in the products, we say that the equation is balanced.

Example 2:

H2 + O2 H2OH2+O2H2O
(2)

Figure 2
Figure 2 (CG10C5_002.png)

Reactants

Atomic mass of reactants=(1+1)+(16+16)=34uAtomic mass of reactants=(1+1)+(16+16)=34u

Number of atoms of each element in the reactants: (2×H)(2×H) and (2×O)(2×O)

Product

Atomic mass of product=(1+1+16)=18uAtomic mass of product=(1+1+16)=18u

Number of atoms of each element in the product: (2×H)(2×H) and (1×O)(1×O)

Since the total atomic mass of the reactants and the products is not the same and since there are more oxygen atoms in the reactants than there are in the product, the equation is not balanced.

Example 3:

NaOH + HCl NaCl + H2O NaOH + HCl NaCl + H2O
(3)

Figure 3
Figure 3 (CG10C5_003.png)

Reactants

Atomic mass of reactants=(23+6+1)+(1+35,4)=76,4uAtomic mass of reactants=(23+6+1)+(1+35,4)=76,4u

Number of atoms of each element in the reactants: (1×Na)+(1×O)+(2×H)+(1×Cl)(1×Na)+(1×O)+(2×H)+(1×Cl)

Products

Atomic mass of products=(23+35,4)+(1+1+16)=76,4uAtomic mass of products=(23+35,4)+(1+1+16)=76,4u

Number of atoms of each element in the products: (1×Na)+(1×O)+(2×H)+(1×Cl)(1×Na)+(1×O)+(2×H)+(1×Cl)

Since the number of atoms of each element is the same in the reactants and in the products, we say that the equation is balanced.

We now need to find a way to balance those equations that are not balanced so that the number of atoms of each element in the reactants is the same as that for the products. This can be done by changing the coefficients of the molecules until the atoms on each side of the arrow are balanced. You will see later that these coefficients tell us something about the mole ratio in which substances react. They also tell us about the volume relationship between gases in the reactants and products.

Tip:

Coefficients

Remember that if you put a number in front of a molecule, that number applies to the whole molecule. For example, if you write 2H2O2H2O, this means that there are 2 molecules of water. In other words, there are 4 hydrogen atoms and 2 oxygen atoms. If we write 3HCl3HCl, this means that there are 3 molecules of HClHCl. In other words there are 3 hydrogen atoms and 3 chlorine atoms in total. In the first example, 2 is the coefficient and in the second example, 3 is the coefficient.

Activity: Balancing chemical equations

You will need: coloured balls (or marbles), prestik, a sheet of paper and coloured pens.

We will try to balance the following equation:

Al + O 2 Al 2 O 3 Al + O 2 Al 2 O 3
(4)
Take 1 ball of one colour. This represents a molecule of AlAl. Take two balls of another colour and stick them together. This represents a molecule of O2O2. Place these molecules on your left. Now take two balls of one colour and three balls of another colour to form a molecule of Al 2 O 3 Al 2 O 3 . Place these molecules on your right. On a piece of paper draw coloured circles to represent the balls. Draw a line down the center of the paper to represent the molecules on the left and on the right.

Count the number of balls on the left and the number on the right. Do you have the same number of each colour on both sides? If not the equation is not balanced. How many balls will you have to add to each side to make the number of balls the same? How would you add these balls?

You should find that you need 4 balls of one colour for AlAl and 3 pairs of balls of another colour (i.e. 6 balls in total) for O2O2 on the left side. On the right side you should find that you need 2 clusters of balls for Al 2 O 3 Al 2 O 3 . We say that the balanced equation is:

4 Al + 3 O 2 2 Al 2 O 3 4 Al + 3 O 2 2 Al 2 O 3
(5)

Repeat this process for the following reactions:

  • CH 4 + 2 O 2 CO 2 + 2 H 2 O CH 4 + 2 O 2 CO 2 + 2 H 2 O
  • 2 H 2 + O 2 2 H 2 O 2 H 2 + O 2 2 H 2 O
  • Zn + 2 HCl ZnCl 2 + H 2 Zn + 2 HCl ZnCl 2 + H 2

Steps to balance a chemical equation

When balancing a chemical equation, there are a number of steps that need to be followed.

  • STEP 1: Identify the reactants and the products in the reaction and write their chemical formulae.
  • STEP 2: Write the equation by putting the reactants on the left of the arrow and the products on the right.
  • STEP 3: Count the number of atoms of each element in the reactants and the number of atoms of each element in the products.
  • STEP 4: If the equation is not balanced, change the coefficients of the molecules until the number of atoms of each element on either side of the equation balance.
  • STEP 5: Check that the atoms are in fact balanced.
  • STEP 6 (we will look at this a little later): Add any extra details to the equation e.g. phase.

Exercise 1: Balancing chemical equations 1

Balance the following equation:

Mg + HCl MgCl2+ H2 Mg + HCl MgCl2+ H2
(6)

Solution
  1. Step 1. Because the equation has been written for you, you can move straight on to counting the number of atoms of each element in the reactants and products :

    Reactants: Mg=1atomMg=1atom; H=1atomH=1atom and Cl=1atomCl=1atom

    Products: Mg=1atomMg=1atom; H=2atomsH=2atoms and Cl=2atomsCl=2atoms

  2. Step 2. Balance the equation :

    The equation is not balanced since there are 2 chlorine atoms in the product and only 1 in the reactants. If we add a coefficient of 2 to the HClHCl to increase the number of HH and ClCl atoms in the reactants, the equation will look like this:

    Mg +2 HCl MgCl2+ H2 Mg +2 HCl MgCl2+ H2
    (7)

  3. Step 3. Check that the atoms are balanced :

    If we count the atoms on each side of the equation, we find the following:

    Reactants: Mg=1atomMg=1atom; H=2atomH=2atom and Cl=2atomCl=2atom

    Products: Mg=1atomMg=1atom; H=2atomH=2atom and Cl=2atomCl=2atom

    The equation is balanced. The final equation is:

    Mg +2 HCl MgCl2+ H2 Mg +2 HCl MgCl2+ H2
    (8)

Exercise 2: Balancing chemical equations 2

Balance the following equation:

CH 4 + O 2 CO 2 +H2O CH 4 + O 2 CO 2 +H2O
(9)

Solution
  1. Step 1. Count the number of atoms of each element in the reactants and products :

    Reactants: C=1C=1; H=4H=4 and O=2O=2

    Products: C=1C=1; H=2H=2 and O=3O=3

  2. Step 2. Balance the equation :

    If we add a coefficient of 2 to H2OH2O, then the number of hydrogen atoms in the reactants will be 4, which is the same as for the reactants. The equation will be:

    CH 4 + O 2 CO 2 +2H2O CH 4 + O 2 CO 2 +2H2O
    (10)

  3. Step 3. Check that the atoms balance :

    Reactants: C=1C=1; H=4H=4 and O=2O=2

    Products: C=1C=1; H=4H=4 and O=4O=4

    You will see that, although the number of hydrogen atoms now balances, there are more oxygen atoms in the products. You now need to repeat the previous step. If we put a coefficient of 2 in front of O2O2, then we will increase the number of oxygen atoms in the reactants by 2. The new equation is:

    CH 4 +2 O 2 CO 2 +2H2O CH 4 +2 O 2 CO 2 +2H2O
    (11)

    When we check the number of atoms again, we find that the number of atoms of each element in the reactants is the same as the number in the products. The equation is now balanced.

Exercise 3: Balancing chemical equations 3

Nitrogen gas reacts with hydrogen gas to form ammonia. Write a balanced chemical equation for this reaction.

Solution
  1. Step 1. Identify the reactants and the products and write their chemical formulae :

    The reactants are nitrogen (N2N2) and hydrogen (H2H2) and the product is ammonia (NH3NH3).

  2. Step 2. Write the equation so that the reactants are on the left and products on the right of the arrow :

    The equation is as follows:

    N 2 + H 2 NH 3 N 2 + H 2 NH 3
    (12)

  3. Step 3. Count the atoms of each element in the reactants and products :

    Reactants: N=2N=2 and H=2H=2

    Products: N=1N=1 and H=3H=3

  4. Step 4. Balance the equation :

    In order to balance the number of nitrogen atoms, we could rewrite the equation as:

    N 2 + H 2 2 NH 3 N 2 + H 2 2 NH 3
    (13)

  5. Step 5. Check that the atoms are balanced :

    In the above equation, the nitrogen atoms now balance, but the hydrogen atoms don't (there are 2 hydrogen atoms in the reactants and 6 in the product). If we put a coefficient of 3 in front of the hydrogen ( H 2 H 2 ), then the hydrogen atoms and the nitrogen atoms balance. The final equation is:

    N 2 +3 H 2 2 NH 3 N 2 +3 H 2 2 NH 3
    (14)

Exercise 4: Balancing chemical equations 4

In our bodies, sugar (C6H12O6C6H12O6) reacts with the oxygen we breathe in to produce carbon dioxide, water and energy. Write the balanced equation for this reaction.

Solution
  1. Step 1. Identify the reactants and products in the reaction and write their chemical formulae. :

    Reactants: sugar (C6H12O6C6H12O6) and oxygen (O2O2)

    Products: carbon dioxide (CO2CO2) and water (H2OH2O)

  2. Step 2. Write the equation by putting the reactants on the left of the arrow and the products on the right :
    C6H12O6+O2CO2+H2OC6H12O6+O2CO2+H2O
    (15)
  3. Step 3. Count the number of atoms of each element in the reactants and the number of atoms of each element in the products :

    Reactants: C=6C=6; H=12H=12 and O=8O=8

    Products: C=1C=1; H=2H=2 and O=3O=3

  4. Step 4. Change the coefficents of the molecules until the number of atoms of each element on either side of the equation balance. :

    It is easier to start with carbon as it only appears once on each side. If we add a 6 in front of CO2CO2, the equation looks like this:

    C6H12O6 + O2 6 CO2 + H2O C6H12O6+O26CO2+H2O
    (16)

    Reactants: C=6C=6; H=12H=12 and O=8O=8

    Products: C=6C=6; H=2H=2 and O=13O=13

  5. Step 5. Change the coefficients again to try to balance the equation. :

    Let's try to get the number of hydrogens the same this time.

    C6H12O6 + O2 6 CO2 +6 H2O C6H12O6+O26CO2+6H2O
    (17)

    Reactants: C=6C=6; H=12H=12 and O=8O=8

    Products: C=6C=6; H=12H=12 and O=18O=18

  6. Step 6. Now we just need to balance the oxygen atoms. :
    C6H12O6 +12 O2 6 CO2 +6 H2O C6H12O6+12O26CO2+6H2O
    (18)

    Reactants: C=6C=6; H=12H=12 and O=18O=18

    Products: C=6C=6; H=12H=12 and O=18O=18

This simulation allows you to practice balancing simple equations.

Figure 4
Figure 4 (balancing-chemical-equations-screenshot.png)
run demo

Balancing simple chemical equations

Balance the following equations:

  1. Hydrogen fuel cells are extremely important in the development of alternative energy sources. Many of these cells work by reacting hydrogen and oxygen gases together to form water, a reaction which also produces electricity. Balance the following equation: H2(g)+O2(g)H2O(l)H2(g)+O2(g)H2O(l)Click here for the solution
  2. The synthesis of ammonia (NH3NH3), made famous by the German chemist Fritz Haber in the early 20th century, is one of the most important reactions in the chemical industry. Balance the following equation used to produce ammonia: N2(g)+H2(g)NH3(g)N2(g)+H2(g)NH3(g) Click here for the solution
  3. Mg+P4Mg3P2Mg+P4Mg3P2 Click here for the solution
  4. Ca+H2OCa(OH)2+H2Ca+H2OCa(OH)2+H2Click here for the solution
  5. CuCO3+H2SO4CuSO4+H2O+ CO2 CuCO3+H2SO4CuSO4+H2O+ CO2 Click here for the solution
  6. CaCl2+Na2CO3CaCO3+NaCl CaCl2+Na2CO3CaCO3+NaClClick here for the solution
  7. C12H22O11+O2H2O+CO2C12H22O11+O2H2O+CO2
     
    Click here for the solution
  8. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid. Click here for the solution
  9. Ethane (C2H6C2H6) reacts with oxygen to form carbon dioxide and steam. Click here for the solution
  10. Ammonium carbonate is often used as a smelling salt. Balance the following reaction for the decomposition of ammonium carbonate: (NH4)2CO3(s)NH3(aq)+CO2(g)+H2O(l)(NH4)2CO3(s)NH3(aq)+CO2(g)+H2O(l)
     
    Click here for the solution

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