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Balanced Equations Simplified Revision Notes

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1.2.3 Balanced Equations

Balancing chemical equations is fundamental to ensuring the conservation of mass and the correct stoichiometric relationships between reactants and products. This process applies to both full molecular equations and ionic equations.

Writing Balanced Equations

To balance a chemical equation:

  1. Write the unbalanced equation: Identify the reactants and products, then write their correct chemical formulas.
  2. Balance atoms: Adjust coefficients to ensure the same number of atoms of each element is present on both sides of the equation.
  3. Balance charges (for ionic reactions): Ensure that the total charge is the same on both sides of the equation, particularly in ionic reactions.
  4. Double-check your work: Ensure both mass and charge are conserved.
infoNote

Example: For the combustion of methane CH4CH_4

CH4+O2CO2+H2OCH_4 + O_2 \rightarrow CO_2 + H_2O

To balance:

CH4+2O2CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

Explanation:

This shows that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

Ionic Equations

Some chemical reactions, especially those involving ionic compounds, only involve specific ions. Ionic equations focus on the ions that participate in the reaction, while spectator ions (ions that do not change) are left out.

Steps for Writing Ionic Equations

  1. Write the full molecular equation.
  2. Separate aqueous compounds into their ions.
  3. Identify and cancel out spectator ions (those that appear unchanged on both sides of the equation).
infoNote

Example: Consider the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):


Step 1: Write the full molecular equation.

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)HCl (aq) + NaOH (aq) \rightarrow NaCl (aq) + H_2O (l)

Step 2: Separate aqueous compounds into their ions.

Ionic equation (showing ions in aqueous solution):

H+(aq)+Cl(aq)+Na+(aq)+OH(aq)Na+(aq)+Cl(aq)+H2O(l)H^+ (aq) + Cl^- (aq) + Na^+ (aq) + OH^- (aq) \rightarrow Na^+ (aq) + Cl^- (aq) + H_2O (l)

Step 3: Identify and cancel out spectator ions

Spectator ions Na+Na^+ and ClCl^- do not participate in the reaction, so they are removed, leaving the net ionic equation:

H+(aq)+OH(aq)H2O(l)H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)

Using Balanced Equations to Calculate Masses, Volumes, and Concentrations

Balanced equations not only help you understand how substances react but also allow you to calculate important quantities like masses, volumes of gases, and concentrations of solutions involved in chemical reactions.

Calculating Masses from Balanced Equations

Once you have a balanced equation, you can use it to determine the mass of a reactant or product by following these steps:

  1. Write the balanced equation.
  2. Calculate the moles of the known substance.
  3. Use the molar ratio from the balanced equation to find moles of the unknown substance.
  4. Convert moles to mass using the formula:
Mass=Moles×Mr\text{Mass} = \text{Moles} \times M_r
infoNote

Example: Mass of Calcium Carbonate in an Indigestion Tablet Balanced equation for the reaction of calcium carbonate (CaCO₃) with hydrochloric acid (HCl):

CaCO3+2HClCaCl2+CO2+H2OCaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O

Suppose an indigestion tablet contains 0.50 g of calcium carbonate. How many grams of carbon dioxide (CO₂) are produced?


Step 1: Calculate moles of calcium carbonate:

Moles of CaCO3=0.50g100.09g/mol=0.005mol\text{Moles of } CaCO_3 = \frac{0.50 \, \text{g}}{100.09 \, \text{g/mol}} = 0.005 \, \text{mol}

Step 2: Use the molar ratio (1:1) to find moles of CO₂.

Moles of CO2=0.005mol\text{Moles of } CO_2 = 0.005 \, \text{mol}

Step 3: Calculate mass of CO₂:

Mass of CO2=0.005×44.01g/mol=0.22g\text{Mass of } CO_2 = 0.005 \times 44.01 \, \text{g/mol} = 0.22 \, \text{g}

So, 0.22 g of carbon dioxide is produced.

Calculating Volumes of Gases

When a gas is involved, the balanced equation can be used to calculate its volume. At room temperature and pressure (RTP), 1 mole of any gas occupies 24 dm³.

Formula:

Volume of gas=Moles of gas×24dm3\text{Volume of gas} = \text{Moles of gas} \times 24 \, \text{dm}^3
infoNote

Example: Volume of CO₂ Produced from Calcium Carbonate Using the same reaction as above:

CaCO3+2HClCaCl2+CO2+H2OCaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O

With 0.005 mol of CO₂ produced, the volume at RTP is:

Volume of CO2=0.005mol×24dm3=0.12dm3\text{Volume of } CO_2 = 0.005 \, \text{mol} \times 24 \, \text{dm}^3 = 0.12 \, \text{dm}^3

Therefore, 0.12 dm³ of carbon dioxide is produced.

Calculating Concentrations and Volumes for Reactions in Solutions

For reactions in solution, you can calculate concentrations using the relationship between moles, concentration, and volume:

Moles=Concentration×Volume\text{Moles} = \text{Concentration} \times \text{Volume}
infoNote

Example: Concentration of Ethanoic Acid in Vinegar Consider the reaction between ethanoic acid (CH₃COOH) and sodium hydroxide (NaOH):

CH3COOH+NaOHCH3COONa+H2OCH_3COOH + NaOH \rightarrow CH_3COONa + H_2O

If 25.0 cm³ of vinegar (containing ethanoic acid) is titrated with 0.1 mol/dm³ NaOH, and 30.0 cm³ of NaOH is required to neutralize the vinegar, calculate the concentration of ethanoic acid in the vinegar.


Step 1: Calculate moles of NaOH:

Moles of NaOH=0.1mol/dm3×30.0cm31000=0.003mol\text{Moles of NaOH} = 0.1 \, \text{mol/dm}^3 \times \frac{30.0 \, \text{cm}^3}{1000} = 0.003 \, \text{mol}

Step 2: Use the molar ratio (1:1) to find moles of ethanoic acid:

Moles of CH3COOH=0.003mol\text{Moles of } CH_3COOH = 0.003 \, \text{mol}

Step 3: Calculate concentration of ethanoic acid:

Concentration of CH3COOH=0.003mol25.0cm3/1000=0.12mol/dm3\text{Concentration of } CH_3COOH = \frac{0.003 \, \text{mol}}{25.0 \, \text{cm}^3/1000} = 0.12 \, \text{mol/dm}^3

So, the concentration of ethanoic acid in vinegar is 0.12 mol/dm³.

Molar Mass Calculations Using Balanced Equations

Balanced equations can also help find molar masses (MrM_r) of unknown compounds when you have other relevant data, such as mass and moles.

infoNote

Example: Molar Mass of MHCO₃ If 0.050 moles of MHCO₃ weigh 4.2 g, calculate its molar mass.


Use the formula:

Mr=MassMolesM_r = \frac{\text{Mass}}{\text{Moles}} =4.2g0.050mol=84g/mol= \frac{4.2 \, \text{g}}{0.050 \, \text{mol}} = 84 \, \text{g/mol}

So, the MrM_r of MHCO₃ is 84 g/mol.

infoNote

Example: Molar Mass of Succinic Acid Suppose 1.18 g of succinic acid (C₄H₆O₄) reacts completely with NaOH and the number of moles is 0.01 mol. The molar mass is:

Mr=1.18g0.01mol=118g/molM_r = \frac{1.18 \, \text{g}}{0.01 \, \text{mol}} = 118 \, \text{g/mol}

Thus, the M(_r) of succinic acid is 118 g/mol.

infoNote

Example: Mass of Aspirin in a Tablet To determine the mass of aspirin (C₉H₈O₄) in a tablet, we can use its molar mass (180 g/mol) and the amount in moles. If a tablet contains 0.005 mol of aspirin:

Mass of aspirin=0.005×180g/mol=0.90g\text{Mass of aspirin} = 0.005 \times 180 \, \text{g/mol} = 0.90 \, \text{g}

Thus, the tablet contains 0.90 g of aspirin.

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