Excess Reactant

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Additional Worked Out Examples/ Practice

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Topic Summary & Highlights
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Core Concept

the excess reactant (or excess reagent) is the reactant that remains after the limiting reactant is completely used up. The excess reactant is not entirely consumed, meaning some amount of it will be left over once the reaction stops. Understanding how to identify and calculate the leftover amount of the excess reactant is important for accurately predicting the reaction’s outcome and for practical applications like reducing waste.

Practice Tips

  • Balance the Equation First: Accurate mole ratios are necessary to identify the limiting and excess reactants.

  • Use the Limiting Reactant to Calculate Excess Used: Once the limiting reactant is identified, use it to find how much of the excess reactant is actually consumed.

  • Double-Check Units: Keep track of units (grams, moles) to avoid errors in conversions.

  • Practice with Different Reactions: Excess reactant calculations are useful in various contexts, so practicing with diverse problems helps reinforce the process.

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Key Concepts

  1. Excess Reactant:

    • The reactant that is not fully consumed in a chemical reaction.

    • It is present in a larger quantity than needed to react completely with the limiting reactant.

  2. Limiting Reactant:

    • The reactant that determines the maximum amount of product formed.

    • It is completely consumed in the reaction, causing the reaction to stop.

  3. Importance of Excess Reactant Calculation:

    • Helps understand reaction efficiency and minimize waste.

    • Useful for calculating the actual yield of a reaction in practical applications.

Steps to Determine the Amount of Excess Reactant Left Over

  1. Write and Balance the Chemical Equation:

    • Ensure the reaction equation is balanced, which allows for accurate mole ratios between reactants.

  2. Identify the Limiting and Excess Reactants:

    • Convert the given amounts of reactants (usually in grams) to moles.

    • Use mole ratios from the balanced equation to determine which reactant will be completely used up first (the limiting reactant) and which will remain (the excess reactant).

  3. Calculate Moles of Excess Reactant Used:

    • Use the limiting reactant to determine how much of the excess reactant is required to react with it.

    • Apply mole ratios to calculate the moles of the excess reactant that actually participate in the reaction.

  4. Find the Remaining Amount of Excess Reactant:

    • Subtract the moles of excess reactant used from the total moles of excess reactant initially present.

    • Convert the leftover moles of excess reactant back to grams (if needed) by multiplying by its molar mass.

Example Problem: Finding the Excess Reactant Left Over

Problem: For the reaction below, if 5.0 g of H₂ reacts with 20.0 g of O₂, identify the excess reactant and calculate how much of it is left over after the reaction.

2H₂ + O₂ → 2H₂O

Solution:

  1. Write and Balance the Equation:

    • The equation is balanced as written.

  2. Convert Given Amounts to Moles:

    • Molar mass of H₂ = 2.02 g/mol

    • Moles of H₂ = (5.0 g) / (2.02 g/mol) = 2.48 mol

    • Molar mass of O₂ = 32.00 g/mol

    • Moles of O₂ = (20.0 g) / (32.00 g/mol) = 0.625 mol

  3. Identify the Limiting and Excess Reactants:

    • According to the balanced equation, 2 mol H₂ reacts with 1 mol O₂.

    • To react with 0.625 mol O₂, we need: (0.625 mol O₂) * (2 mol H₂ / 1 mol O₂) = 1.25 mol H₂

    • We have 2.48 mol H₂, which is more than enough to react with 0.625 mol O₂. Therefore, O₂ is the limiting reactant, and H₂ is the excess reactant.

  4. Calculate Moles of Excess Reactant Used:

    • Using 0.625 mol O₂ (the limiting reactant), we find the moles of H₂ used: (0.625 mol O₂) * (2 mol H₂ / 1 mol O₂) = 1.25 mol H₂

  5. Find the Remaining Amount of Excess Reactant:

    • Subtract the moles of H₂ used from the initial moles of H₂: (2.48 mol H₂) - (1.25 mol H₂) = 1.23 mol H₂

  6. Convert Moles of Remaining H₂ to Grams:

    • Mass of remaining H₂ = (1.23 mol H₂) * (2.02 g/mol) = 2.49 g H₂

Answer: After the reaction, 2.49 g of H₂ remains as the excess reactant.

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