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

This quiz focuses on the process of balancing chemical equations, specifically for reactions involving hydrocarbons and oxygen. You'll encounter examples such as the combustion of pentane, where you will learn how to balance the number of carbon, hydrogen, and oxygen atoms on both sides of the equation.