Chemistry Class 11: Limiting Reagent

Introduction

• Previous topic: Stoichiometry
  • Discussed stoichiometric problems with four varieties.
• Current topic: Limiting Reagent
  • Important for competitive exams (NEET, JEE)
  • Concept often appears in class 11 and 12 exams.
  • Learn a trick to solve limiting reagent problems easily.

Limiting Reagent

• Definition: Reagent that limits or stops a reaction.
  • Example: A + B gives products; among A and B, the one that limits the reaction is the limiting reagent.

Real-life Example

• Making Chapatis
  • Required: 2 bowls of flour and 1 glass of water for 5 chapatis.
  • If only 1 bowl of flour is available (but sufficient water), flour becomes the limiting reagent.
  • Flour limits the ability to make 5 chapatis, resulting in only 2-3.

Chemical Reactions

• A reactant combines with B to get a product.
  • Both A and B should be in sufficient amounts for the desired product.
  • The reagent less in amount is the limiting reagent.
  • The other is the excess reagent.

Problem Solving

• Example Problem: Identify the limiting reagent and calculate the amount of product.
  • Given: 5g Nitrogen, 10g Hydrogen mixed to form Ammonia.

Steps to Solve

1. Write the Chemical Equation
   • Balance the equation.
2. Calculate Moles
   • Convert mass to moles using the formula: ( \text{moles} = \frac{\text{mass}}{\text{molar mass}} )
3. Apply the Trick
   • Calculate ( \frac{\text{number of moles}}{\text{stoichiometric coefficient}} ) for each reactant.
   • Select the smallest value as the limiting reagent.

Calculating Product Amount

• Use the limiting reagent to calculate product formation.
• Cross-multiply using stoichiometric ratios from the balanced equation.
• Convert moles of product to mass if needed.

Example Problems

Problem 1

• Reactants: 5g Nitrogen and 10g Hydrogen.
• Chemical Equation: ( N_2 + 3H_2 \rightarrow 2NH_3 )
• Solution:
  • Number of moles of ( N_2 ): ( \frac{5}{28} )
  • Number of moles of ( H_2 ): ( \frac{10}{2} = 5 )
  • Limiting Reagent: ( H_2 )
  • Amount of Ammonia formed: Calculate using the moles of ( H_2 ).

Problem 2

• Reactants: 1g Mg and 0.5g O2.
• Chemical Equation: ( 2Mg + O_2 \rightarrow 2MgO )
• Solution:
  • Number of moles of ( Mg ): ( \frac{1}{24} )
  • Number of moles of ( O_2 ): ( \frac{0.5}{32} )
  • Limiting Reagent: ( O_2 )
  • Excess Reagent: ( Mg )
  • Amount of MgO formed: Calculate using the moles of ( O_2 ).

Conclusion

• Limiting reagent questions are crucial for competitive exams.
• Understand and apply the calculation trick efficiently to solve quickly.
• Next topic: Concentration in solutions.