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Language of Chemistry - Balancing Equations

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A chemical equation is a shorthand representation of a chemical reaction using symbols and formulae. For example: Zn+H2SO4→ZnSO4+H2Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2.

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Reactants are the substances that take part in a chemical reaction and are written on the left-hand side (LHS). Products are the new substances formed and are written on the right-hand side (RHS).

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According to the Law of Conservation of Mass, matter can neither be created nor destroyed in a chemical reaction. Thus, the number of atoms of each element must remain the same on both sides.

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A Skeletal Equation is an unbalanced equation where the number of atoms of one or more elements is unequal on the two sides. Example: Mg+O2→MgOMg + O_2 \rightarrow MgO.

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A Balanced Equation has an equal number of atoms of each element on both the reactant and product sides. Example: 2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO.

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When balancing, only coefficients (numbers placed before the formula, e.g., 2H22H_2) can be changed. Subscripts (numbers within the formula, e.g., the 22 in O2O_2) must never be altered as they define the chemical identity of the substance.

📐Formulae

Reactants→Products\text{Reactants} \rightarrow \text{Products}

Total Mass of Reactants=Total Mass of Products\text{Total Mass of Reactants} = \text{Total Mass of Products}

2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O

N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3

2KClO3→2KCl+3O22KClO_3 \rightarrow 2KCl + 3O_2

💡Examples

Problem 1:

Balance the skeletal equation for the formation of water: H2+O2→H2OH_2 + O_2 \rightarrow H_2O

Solution:

2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O

Explanation:

Initially, there are 2 Oxygen atoms on the LHS and only 1 on the RHS. To balance Oxygen, we place a coefficient 2 before H2OH_2O. This gives us 4 Hydrogen atoms on the RHS, so we must place a coefficient 2 before H2H_2 on the LHS.

Problem 2:

Balance the reaction between Iron and Chlorine: Fe+Cl2→FeCl3Fe + Cl_2 \rightarrow FeCl_3

Solution:

2Fe+3Cl2→2FeCl32Fe + 3Cl_2 \rightarrow 2FeCl_3

Explanation:

Chlorine atoms are 2 on the LHS and 3 on the RHS. The lowest common multiple of 2 and 3 is 6. We place 3 before Cl2Cl_2 and 2 before FeCl3FeCl_3. Finally, to balance the 2 Iron atoms now on the RHS, we place a 2 before FeFe on the LHS.

Problem 3:

Balance the reaction of Methane with Oxygen: CH4+O2→CO2+H2OCH_4 + O_2 \rightarrow CO_2 + H_2O

Solution:

CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

Explanation:

First, balance Carbon (1 on both sides). Next, balance Hydrogen (4 on LHS, 2 on RHS) by adding 2 before H2OH_2O. Finally, count Oxygen on the RHS (22 in CO2CO_2 + 22 in 2H2O2H_2O = 4) and place a 2 before O2O_2 on the LHS.