Balancing Equations Part 2 Examples

Balancing Chemical Equations

• Balancing equations involves ensuring the same number of each type of atom on both sides of the equation.

Example 1: Combustion of Pentane

• Reactants: C5H12 (pentane) and O2 (oxygen)
• Products: CO2 (carbon dioxide) and H2O (water)
• Balancing Carbons: 5 carbons in C5H12 requires placing a 5 in front of CO2.
• Balancing Hydrogens: 12 hydrogens in C5H12 require a 6 in front of H2O (6 × 2 = 12).
• Balancing Oxygens: 10 oxygens from 5CO2 and 6 from 6H2O total 16; thus, an 8 is needed before O2 (8 × 2 = 16).
• Final balanced equation: C5H12 + 8O2 → 5CO2 + 6H2O

Example 2: Reaction of Zinc with Hydrochloric Acid

• Reactants: Zn (zinc) and HCl (hydrochloric acid)
• Products: ZnCl2 (zinc chloride) and H2 (hydrogen gas)
• Balancing Chlorides: ZnCl2 has 2 Cl atoms, so a coefficient of 2 is needed before HCl to provide the same amount.
• Final balanced equation: Zn + 2HCl → ZnCl2 + H2

Example 3: Sulfur and Fluorine Reaction

• Reactants: S8 (octasulfur) and F2 (fluorine)
• Products: SF6 (sulfur hexafluoride)
• Balancing Sulfurs: 8 sulfur atoms in S8 require an 8 in front of SF6.
• Balancing Fluorines: 8SF6 produces 48 fluorine atoms. Placing a 24 before F2 supplies the necessary 48 fluorines (24 × 2 = 48).
• Final balanced equation: S8 + 24F2 → 8SF6