Balancing Equations

 

This is part of Year 11 HSC Chemistry course under the topic of Chemical Reactions and Stoichiometry

HSC Chemistry Syllabus

  • Relate stoichiometry to the law of conservation of mass in chemical reactions by investigating:
    • Balancing chemical equations (ACSCH039)
    • Solving problems regarding mass changes in chemical reactions (ACSCH046)

    How to Balance Chemical Equations?

    What Are Chemical Equations?

    A chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulae. They describe a chemical change which occurs as a result of chemical reaction.

    An example of this is the formation of water from hydrogen and oxygen gases (unbalanced equation):

     

    $$H_2(g) + O_2(g) \rightarrow H_2O(l)$$

    In an expression such as this, the formulas and symbols on the left-hand side are the reacting substances which are called the reactants. These chemical species are used up in the chemical reaction and become transformed into new species which are shown on the right-hand side. These chemical species on the right-hand side of the equation are called the products and this is because they are produced by the chemical reaction. The arrow is read as ‘moves towards’, ‘gives’ or ‘forms’, and the plus sign is read as ‘and’. When the plus sign (+) appears between the formulas for the two reactants, it can also be read as “reacts with”. Thus the equation above can be read as hydrogen gas and/reacts with oxygen gas gives water. 

    Sometimes energy may be required to start a reaction or become formed by the reaction and these would be included just as a word equation.

    State Symbols

    Next to the chemical formula for any reactant or product, you will find the state of the substance in brackets.

    • Solid `(s)`
    • Liquid `(l)`
    • Gaseous `(g)`
    • Aqueous `(aq)` – This means that the substance was dissolved in water or is in solution form

    Balancing Chemical Equations

    As they are written, chemical equations indicate in a qualitative way what substances are consumed in the reaction and what new substances are formed. To have quantitative information about the reaction, the equation must be balanced so that it conforms to the rules of the law of conservation of matter. That is, there must be an equal number of atoms of each element on the right-hand side of the equation as there are on the left-hand side. 

    $$H_2(g) + O_2(g) \rightarrow H_2O(l)$$

    If the number of atoms of each element in the equation above are counted, we observe that there are 2 hydrogen and 2 oxygen atoms on the left side, and 2 hydrogen and 1 oxygen atom on the right hand side. 



    Left Side

    Right Side

    2 atoms of H

    2 atoms of H

    2 atoms of O

    1 atom of O

     

    Since the number of atoms on the left and right hand sides of the equation are unequal, the equation is said to be unbalanced. The balancing of the equation is accomplished by introducing the proper number or coefficient before the chemical formula of each substance. To balance the number of O atoms, there must either be a half added to the number of O on the left, or a 2 must be added in front of the H2 and the H2O

    $$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$ 

    or

    $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$$ 

     

    Left Side

    Right Side

    2 or 4 atoms of H

    2 or 4 atoms of H

    1 or 2 atoms of O

    1 or 2 atoms of O

     

    Since the number of atoms on both the left and right hand sides are balanced after adding coefficients, the equation can now described as being balanced.   

     

    Stoichiometry & Stoichiometric Equivalence

     

    In a balanced chemical equation, the coefficients in front of each formula represent the stoichiometric ratio, or more simply, the molar ratio of the substances involved ('moles' are typically used to describe the quantity of substances). These coefficients tell you how many moles of each reactant are needed to produce a given amount of product, and vice versa. For example, in the balanced equation for the formation of water:



    the coefficients show that 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water. This means the molar ratio between hydrogen and oxygen is 2:1, and between hydrogen and water it is 1:1.

     

    This relationship is called stoichiometric equivalence — it allows us to use mole ratios to calculate unknown quantities in a chemical reaction. For example, if you know the amount of one reactant, you can use the stoichiometric ratio to determine the amounts of other reactants required or products formed.

    Thus using the stoichiometric ratio, we could predict that 4 moles of hydrogen gas would react with 2 moles of oxygen gas to form 4 moles of water

     

    In summary:

     

    • Stoichiometry = the calculation of quantities in chemical reactions using balanced equations

    • Stoichiometric ratio = the whole-number ratio of moles of substances from the balanced equation

    • Stoichiometric equivalence = the idea that these ratios allow conversion between amounts of reactants and products


    Using these principles, chemists can predict yields, identify limiting reactants, and analyse reaction efficiency in practical laboratory and industrial settings.

     

    Calculations Using Balanced Equations

    Example 1 

    What mass of barium sulfate are produced when 75 g of barium nitrate is fully reacted with aluminium sulfate?

    Answer: (detailed explanation in video)

    67 g (2 s.f.) of barium sulfate are produced.

     

    Example 2

    When 12.71 g of copper was added to excess nitric acid, copper nitrate, nitrogen dioxide and water were formed. 

    (a) Calculate how many moles of copper were added.

    Answer: (detailed explanation in video)

    0.2000 moles (4 s.f.) 

     

    (b) Calculate how many moles of nitric acid were reacted.

    Answer: (detailed explanation in video)

    0.8000 moles (4 s.f.) 

     

    Example 3

    3.45 g of calcium carbonate is decomposed to form calcium oxide and carbon dioxide gas.

    (a) Calculate the mass of calcium oxide produced.

    Answer: (detailed explanation in video)

    1.93 g (3 s.f.) 

     

    (a) Calculate the mass of carbon dioxide produced.

    Answer: (detailed explanation in video)

    1.52 g (2 d.p.) 

     

    RETURN TO MODULE 2: INTRODUCTION TO QUANTITATIVE CHEMISTRY