Excess Reactants: A Comprehensive Guide

Excess Reactants

Introduction

• Excess reactants are the leftover reactants after a reaction stops due to the limiting reactant being used up.

Understanding Limiting and Excess Reactants

• Limiting Reactant: Determines the amount of product formed because the reaction stops when it is completely used.
• Excess Reactants: The reactants that are left over after the limiting reactant is used up.
• The focus is on determining how much of the excess reactant is used or remains.

Calculating the Amount of Excess Reactant Used

Example 2: Finding the Amount of $O_2$ Used

• Problem: How much $O<em>2$ is used when 65.0g of $Mg$ reacts with an excess of $O</em>2$?
• Balanced Equation: $2Mg + O_2 \rightarrow 2MgO$

Steps:

1. Balance the Chemical Equation:

   • The equation is already balanced: $2Mg + O_2 \rightarrow 2MgO$

2. Find the Moles of the Reactant Not in Excess (Mg):

   • $65.0g \, Mg \times \frac{1 \, mol \, Mg}{24.31 \, g \, Mg} = 2.67 \, mol \, Mg$

3. Multiply by the Mole Ratio from the Balanced Equation to Convert to Moles of $O_2$:

   • $2.67 \, mol \, Mg \times \frac{1 \, mol \, O<em>2}{2 \, mol \, Mg} = 1.34 \, mol \, O</em>2$

4. Convert from Moles of $O<em>2$ to Grams of $O</em>2$:

   • $1.34 \, mol \, O<em>2 \times \frac{32 \, g \, O</em>2}{1 \, mol \, O<em>2} = 42.9 \, g \, O</em>2$
   • Therefore, 42.9 g of $O_2$ is used.

Determining Leftover Excess Reactant

• If finite amounts of both reactants are given, one is the limiting reactant, and the other is the excess reactant.
• The objective is to find out how much of the excess reactant remains.

Steps:

1. Calculate Moles of Product from Each Reactant:

   • Determine the number of moles of product each reactant can produce to identify the limiting reactant (the one that produces the least product).

2. Convert Moles of Limiting Reactant to Moles of Excess Reactant Used:

   • Use mole ratios to convert moles of the limiting reactant to moles of the excess reactant that were actually used in the reaction.

3. Subtract Moles Used from Total Moles Given:

   • Subtract the moles of excess reactant used from the total moles of excess reactant initially given. Then convert the result to grams.
   • Alternatively, convert to grams first and then subtract.

Example 3: Determining Limiting Reactant and Excess Reactant Leftover

• Reaction: $2Li + Br_2 \rightarrow 2LiBr$
• Problem: If you have 25.0 g of each reactant, identify the limiting reactant and calculate the amount of the excess reactant left over.

Steps:

1. Calculate Moles of $LiBr$ Produced by Each Reactant:

   • For Li:

     • $25.0g \, Li \times \frac{1 \, mol \, Li}{6.94 \, g \, Li} = 3.60 \, mol \, Li$
     • $3.60 \, mol \, Li \times \frac{2 \, mol \, LiBr}{2 \, mol \, Li} = 3.60 \, mol \, LiBr$

   • For $Br_2$:

     • $25.0g \, Br<em>2 \times \frac{1 \, mol \, Br</em>2}{159.80 \, g \, Br<em>2} = 0.156 \, mol \, Br</em>2$
     • $0.156 \, mol \, Br<em>2 \times \frac{2 \, mol \, LiBr}{1 \, mol \, Br</em>2} = 0.312 \, mol \, LiBr$

   • Since $Br_2$ produces less $LiBr$, it is the limiting reactant.

2. Calculate Amount of Li Used:

   • $0.156 \, mol \, Br<em>2 \times \frac{2 \, mol \, Li}{1 \, mol \, Br</em>2} = 0.312 \, mol \, Li$
   • $0.312 \, mol \, Li \times \frac{6.94 \, g \, Li}{1 \, mol \, Li} = 2.17 \, g \, Li$

3. Calculate Li Left Over:

   • $25.0 \, g \, Li \, available - 2.17 \, g \, Li \, actually \, used = 22.8 \, g \, Li \, left \, over$
   • Therefore, 22.8 g of Li is left over, unused as the excess reactant.