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Understanding Limiting Reactants in Chemistry

Jun 3, 2025

Lecture Notes: Limiting Reactants and Calculations

Introduction to Limiting Reactants

  • Limiting Reactant: The substance that is completely used up first in a chemical reaction and limits the amount of product formed.
  • Excess Reactant: Reactant that remains after the limiting reactant is completely consumed.

Reaction Example: Hydrochloric Acid and Calcium Carbonate

  • Reaction: Hydrochloric acid reacts with calcium carbonate.
    • Products: Calcium chloride, water, carbon dioxide.
    • Observation: Solid calcium carbonate disappears and bubbles of carbon dioxide gas are formed.
  • Determining Limiting Reactant:
    • When calcium carbonate is completely consumed, the reaction stops.
    • Calcium carbonate is the limiting reactant; hydrochloric acid remains in excess.

Importance of Limiting Reactants in Calculations

  • The amount of product formed is determined by the limiting reactant.
    • Example: Adding more calcium carbonate results in more product until it is used up.
    • Adding more hydrochloric acid has no effect if no calcium carbonate is left.

Example Calculation: Sodium Oxide Production

  • Problem: Calculate the mass of sodium oxide from 115 grams of sodium.
    • Assumption: Oxygen is in excess (abundant in air); sodium is the limiting reactant.

Steps for Calculation

  1. Write and Balance the Equation

    • Balancing equations is essential. (Reference to video on balancing equations for more help.)
  2. Determine Moles of Sodium

    • Use the formula: ( \text{moles} = \frac{\text{mass}}{\text{molar mass (Mr)}} )
    • Sodium: 115 grams divided by ( 23 \text{g/mol} ) = 5 moles.
  3. Mole Ratio and Sodium Oxide Calculation

    • From balanced equation: Sodium to Sodium Oxide ratio is 4:2 or simplified to 2:1.
    • 5 moles of sodium produce 2.5 moles of sodium oxide.
  4. Calculate Mass of Sodium Oxide

    • Use formula: ( \text{mass} = \text{moles} \times \text{molar mass (Mr)} )
    • Molar mass of ( \text{Na}_2\text{O} ): ( 2\times23 + 16 = 62 \text{g/mol} )
    • Mass: 2.5 moles ( \times 62 \text{g/mol} = 155 \text{grams} )._

Conclusion

  • Understanding and identifying the limiting reactant is crucial for accurate product yield calculations in chemical reactions.
  • Practice with different chemical equations and scenarios helps improve proficiency in these calculations.