Chemical Equations

What is a Chemical Equation?

A chemical equation represents a chemical reaction using symbols for the elements and compounds involved. It shows the reactants (substances that start the reaction) on the left and the products (substances formed) on the right, separated by an arrow ( $→$ ), which indicates the direction of the reaction.

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Example:

2H_2 + O_2 \rightarrow 2H_2O

This equation shows hydrogen gas reacting with oxygen gas to form water.

Balancing Chemical Equations

A balanced chemical equation has the same number of atoms of each element on both sides, ensuring the law of conservation of mass is obeyed (matter is neither created nor destroyed).

Steps to Balance:

Write the unbalanced equation with correct formulas for all reactants and products.
Count the number of atoms for each element on both sides.
Adjust coefficients (numbers in front of compounds or elements) to make the number of atoms for each element equal on both sides.
Double-check the balance, ensuring coefficients are in the lowest possible ratio.

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Example: Unbalanced:

C_3H_8 + O_2 \rightarrow CO_2 + H_2O

Balanced:

C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O

Balancing Redox Equations (Ionic Equations)

In redox reactions, one species is oxidised (loses electrons) and another is reduced (gains electrons). Ionic equations show only the particles involved in the reaction.

Steps to Balance:

Write separate half-equations for oxidation and reduction.
Balance atoms and charges (add electrons where necessary).
Combine the half-equations, ensuring the total number of electrons lost equals the total gained.
Cancel out electrons and any other species that appear on both sides.

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Example: Oxidation of $Fe^{2+}$ by $MnO_4^-$ Oxidation half-equation:

Fe^{2+} \rightarrow Fe^{3+} + e^-

Reduction half-equation:

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

Balanced equation:

5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O

Stoichiometric Calculations Using Balanced Equations

Once an equation is balanced, we can use the mole concept to perform calculations based on the amounts of reactants or products.

Moles and Masses:

Moles can be calculated from the mass of a substance using:

\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}

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Example: For the reaction

2H_2 + O_2 \rightarrow 2H_2O

How many grams of water are produced from 4 g of hydrogen?

Moles of $H_2$

= \frac{4}{2} = 2 \, \text{mol}

Molar mass of $H_2$ is 2 g/mol

From the balanced equation, 2 mol of $H_2$ produces 2 mol of $H_2O$ .

Mass of $H_2O$ produced

= 2 \times 18 = 36 \, \text{g}

Molar mass of $H_2O$ is 18 g/mol.

Calculations Involving Excess Reactants

In reactions, sometimes one reactant is in excess, meaning it won't completely react. The limiting reactant is the one that runs out first and determines the amount of product formed.

Calculation Steps:

Calculate the moles of each reactant.
Use the balanced equation to identify the limiting reactant.
Perform calculations based on the limiting reactant.

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Example: If 10 g of $H_2$ reacts with 100 g of $O_2$ , which is the limiting reactant in the reaction:

2H_2 + O_2 \rightarrow 2H_2O

Moles of $H_2$

= \frac{10}{2} = 5 \, \text{mol}

Moles of $O_2$

= \frac{100}{32} = 3.125 \, \text{mol}

The balanced equation shows 2 mol $H_2$ reacts with 1 mol $O_2$ .

Therefore $H_2$ is in excess, and $O_2$ is the limiting reactant.

Percentage Yields

In reality, reactions don't always proceed perfectly, so the percentage yield compares the actual amount of product obtained to the theoretical amount predicted by stoichiometry.

Formula:

\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

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Example: If the theoretical yield of water is 36 g, but only 30 g is produced:

\text{Percentage yield} = \frac{30}{36} \times 100 = 83.3\%

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Exam Tip:

Always check that your chemical equation is balanced before starting any calculations.
For redox reactions, make sure to balance charges as well as atoms.
Be careful with units in calculations (convert g to kg where necessary).
Identify the limiting reactant before calculating the mass of products.

Chemical Equations Simplified Revision Notes for Leaving Cert Chemistry

Chemical Equations

What is a Chemical Equation?

Balancing Chemical Equations

Steps to Balance:

Balancing Redox Equations (Ionic Equations)

Steps to Balance:

Stoichiometric Calculations Using Balanced Equations

Moles and Masses:

Calculations Involving Excess Reactants

Calculation Steps:

Percentage Yields

Formula:

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