4.2 Limiting Reactant and Excess Reactant

This topic explains the concepts of limiting reactant and excess reactant in chemical reactions. It shows how to identify the limiting reactant and calculate the amount of product formed.

Definition

Limiting Reactant: The reactant that is completely consumed in a chemical reaction and limits the amount of product formed.

Excess Reactant: The reactant that is not completely consumed in a chemical reaction and remains after the reaction stops.

Main Idea

In many chemical reactions, reactants are not present in the exact stoichiometric amounts. One reactant is used up completely before the others. This reactant is called the limiting reactant because it limits the amount of product that can be formed. The other reactants are called excess reactants because they are left over after the reaction stops.

Core Concept

• Calculate the number of moles of each reactant.
• Compare the mole ratio of the reactants with the mole ratio from the balanced equation.
• The reactant that produces the least amount of product is the limiting reactant.

Detailed Explanation

What is a Limiting Reactant?

• The reactant that completely reacts in the reaction and limits the amounts of product formed is called the limiting reactant.
• It is the reactant that is used up first.
• The reaction stops when the limiting reactant is consumed.

What is an Excess Reactant?

• The reactant that is not completely consumed in the reaction is called the excess reactant.
• It is left over after the reaction stops.

Example: Reaction Between H₂ and O₂

Consider the reaction:

2H₂ + O₂ → 2H₂O

• For 1 mole of H₂, 12 mole of O₂ is required.
• For 1 mole of O₂, 2 moles of H₂ are required.

Since only 1 mole of H₂ is available (less than 2 moles needed), H₂ is the limiting reactant. O₂ is the excess reactant because it is not completely consumed.

Example 4.11: Calcium Nitride Formation

Problem: Calculate the mass of calcium nitride (Ca₃N₂) prepared from 54.9 grams of Ca and 43.2 grams of N₂.

Chemical Equation: 3Ca + N₂ → Ca₃N₂

Solution:

Step 1: Calculate moles of each reactant

Moles of Ca = 54.940 = 1.37 moles

Moles of N₂ = 43.228 = 1.54 moles

Step 2: Calculate moles of product from each reactant

From the equation: 3 moles of Ca produce = 1 mole of Ca₃N₂

1 mole of Ca produces = 13 moles of Ca₃N₂

1.37 moles of Ca produce = 13 × 1.37 = 0.457 moles of Ca₃N₂

From the equation: 1 mole of N₂ produces = 1 mole of Ca₃N₂

1.54 moles N₂ produce = 1.54 moles of Ca₃N₂

Step 3: Identify the limiting reactant

Ca produces the least amount of Ca₃N₂ (0.457 moles), so Ca is the limiting reactant.

Step 4: Calculate mass of product

Mass of Ca₃N₂ = moles × molar mass

Molar mass of Ca₃N₂ = 3 × 40 + 2 × 14 = 120 + 28 = 148 g/mol

Mass of Ca₃N₂ = 0.457 × 148 = 67.636 g

Final Answer: Mass of Ca₃N₂ = 67.636 g

Steps to Identify Limiting Reactant

• Write the balanced chemical equation.
• Convert the given masses to moles.
• Use the mole ratio to calculate the moles of product from each reactant.
• The reactant that gives the least moles of product is the limiting reactant.
• Calculate the mass of product using the limiting reactant.

Intermediate Level

Identifying the Limiting Reactant

• Balance the equation.
• Convert masses to moles.
• Calculate moles of product from each reactant.
• The reactant giving the least product is limiting.

Example 4.11

• Mass of Ca = 54.9 g
• Mass of N₂ = 43.2 g

Equation: 3Ca + N₂ → Ca₃N₂

Solution: n_Ca = 54.940 = 1.37 mol n_N2 = 43.228 = 1.54 mol

From Ca: n_Ca3N2 = 1.37 × 13 = 0.457 mol

From N₂: n_Ca3N2 = 1.54 × 11 = 1.54 mol

Ca gives less product → Ca is limiting.

m_Ca3N2 = 0.457 × 148 = 67.636 g

Final Answer: 67.636 g

Key Point

• The limiting reactant determines the maximum product.
• The excess reactant is left over.

Important Points

• 1. Limiting reactant = completely consumed; limits product.
• 2. Excess reactant = left over after reaction.
• 3. To identify limiting reactant: calculate moles of product from each reactant.
• 4. The reactant giving the least product is limiting.
• 5. Example: 3Ca + N₂ → Ca₃N₂.
• 6. Ca is limiting; N₂ is excess.
• 7. Mass of Ca₃N₂ = 67.636 g.

Remember

• - Limiting reactant: Runs out first
• - Excess reactant: Left over
• - Identify: Calculate product from each reactant
• - Least product: Limiting reactant
• - Example: 3Ca + N₂ → Ca₃N₂

Common Mistakes

• 1. Thinking the reactant with less mass is limiting: It depends on moles and mole ratio.
• 2. Forgetting to balance the equation: Always balance first.
• 3. Using the wrong mole ratio: Check coefficients.
• 4. Forgetting to convert mass to moles: Always convert to moles first.

Quick Revision

• Limiting reactant = completely consumed; limits product.
• Excess reactant = left over after reaction.
• To identify limiting reactant: calculate moles of product from each reactant.
• The reactant giving the least product is limiting.
• Example: 3Ca + N₂ → Ca₃N₂.
• Ca is limiting; N₂ is excess.
• Mass of Ca₃N₂ = 67.636 g.

60-Second Revision

• Limiting reactant: Runs out first
• Excess reactant: Left over
• Identify: Calculate product from each reactant
• Least product: Limiting reactant
• Example: 3Ca + N₂ → Ca₃N₂

QUALITY VALIDATION

• Topic: 4.2 Limiting Reactant and Excess Reactant
• Chapter: 4 (Stoichiometry)

• Source re-read: PASS
• Source scope followed: PASS
• Knowledge level matched: PASS
• Important concepts covered: PASS
• Definitions checked: PASS
• Equations checked: PASS
• Derivations checked: N/A
• Examples checked: PASS
• Unsupported knowledge added: NO
• Unnecessary repetition: NO
• Notes format maintained: PASS
• Important source content omitted: NO

• Source limitations:
• NONE

NEXT TOPIC