Calculation of the Limiting Reagent in a Chemical Reaction

Understanding the Calculation of the Limiting Reagent in a Chemical Reaction

Calculating the limiting reagent determines which reactant controls product formation. This calculation is essential for efficient chemical processes.

This article explores detailed formulas, common values, and real-world examples for precise limiting reagent determination.

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  • Calculate the limiting reagent when 5 moles of H2 react with 3 moles of O2.
  • Determine the limiting reagent in the reaction of 10 g of Na with 15 g of Cl2.
  • Find the limiting reagent for 2 moles of N2 and 6 moles of H2 in ammonia synthesis.
  • Calculate the limiting reagent when 4 moles of Fe react with 3 moles of S to form FeS.

Comprehensive Table of Common Reactants and Their Molar Masses

Reactant Chemical Formula Molar Mass (g/mol) Common States Typical Use in Reactions
Hydrogen H2 2.016 Gas Reduction, synthesis of ammonia
Oxygen O2 31.998 Gas Combustion, oxidation reactions
Nitrogen N2 28.014 Gas Ammonia synthesis, inert atmosphere
Chlorine Cl2 70.906 Gas Halogenation, disinfection
Sodium Na 22.990 Solid Metallic reactions, salt formation
Iron Fe 55.845 Solid Alloy production, sulfide formation
Sulfur S 32.065 Solid Acid production, sulfide synthesis
Carbon Dioxide CO2 44.009 Gas Photosynthesis, acid-base reactions
Water H2O 18.015 Liquid Solvent, hydrolysis reactions
Ammonia NH3 17.031 Gas Fertilizer production, base

Fundamental Formulas for Calculating the Limiting Reagent

Determining the limiting reagent requires understanding stoichiometric relationships and mole conversions. The key formulas are outlined below with detailed explanations.

1. Mole Calculation from Mass

moles = mass / molar mass

  • mass: The mass of the reactant in grams (g).
  • molar mass: The molar mass of the reactant in grams per mole (g/mol), obtained from the periodic table or compound data.
  • moles: The amount of substance in moles (mol).

2. Mole Ratio from Balanced Chemical Equation

For a general reaction:

aA + bB → products

Where a and b are stoichiometric coefficients.

3. Limiting Reagent Determination

Calculate the mole ratio of available reactants to their stoichiometric coefficients:

ratioA = nA / a

ratioB = nB / b

  • nA: moles of reactant A available.
  • a: stoichiometric coefficient of reactant A.
  • nB: moles of reactant B available.
  • b: stoichiometric coefficient of reactant B.

The limiting reagent is the reactant with the smallest mole ratio.

4. Maximum Product Formation

Once the limiting reagent is identified, calculate the maximum moles of product formed:

nproduct = ratiolimiting reagent × c

  • c: stoichiometric coefficient of the product in the balanced equation.

5. Mass of Product Formed

Convert moles of product to mass:

massproduct = nproduct × molar massproduct

Detailed Explanation of Variables and Their Typical Values

  • Mass (g): Measured using analytical balances, typical values range from milligrams to kilograms depending on the scale of the reaction.
  • Molar Mass (g/mol): Derived from atomic masses; for example, H2 = 2.016 g/mol, O2 = 31.998 g/mol.
  • Moles (mol): Calculated from mass and molar mass; represents the amount of substance.
  • Stoichiometric Coefficients: Integers from the balanced chemical equation, e.g., 2H2 + O2 → 2H2O, coefficients are 2, 1, and 2 respectively.
  • Mole Ratios: Ratios of available moles to stoichiometric coefficients, critical for identifying the limiting reagent.

Real-World Application Examples

Example 1: Combustion of Hydrogen Gas

Consider the reaction:

2H2 (g) + O2 (g) → 2H2O (l)

Given 5 moles of H2 and 3 moles of O2, determine the limiting reagent and the maximum amount of water produced.

  • Calculate mole ratios:

    • ratioH2 = 5 mol / 2 = 2.5

    ratioO2 = 3 mol / 1 = 3

  • The limiting reagent is H2 because 2.5 < 3.
  • Calculate moles of H2O produced:

    • nH2O = 2.5 × 2 = 5 mol

  • Calculate mass of water:

    • mass = 5 mol × 18.015 g/mol = 90.075 g

Thus, 90.075 grams of water can be formed, limited by hydrogen availability.

Example 2: Synthesis of Ammonia via Haber Process

Reaction:

N2 (g) + 3H2 (g) → 2NH3 (g)

Given 2 moles of N2 and 6 moles of H2, identify the limiting reagent and calculate the maximum ammonia produced.

  • Calculate mole ratios:

    • ratioN2 = 2 mol / 1 = 2

    ratioH2 = 6 mol / 3 = 2

  • Both ratios are equal; neither reagent limits the reaction.
  • Calculate moles of NH3 produced:

    • nNH3 = 2 × 2 = 4 mol

  • Calculate mass of ammonia:

    • mass = 4 mol × 17.031 g/mol = 68.124 g

In this case, both reactants are perfectly balanced, producing 68.124 grams of ammonia.

Additional Considerations and Advanced Insights

In industrial and laboratory settings, precise limiting reagent calculations optimize resource use and minimize waste. Factors such as purity, reaction conditions, and side reactions can affect actual yields.

Advanced techniques include:

  • Use of Excess Reagent: Intentionally adding one reactant in excess to drive the reaction to completion.
  • Yield Calculations: Comparing theoretical yield (from limiting reagent) to actual yield to determine efficiency.
  • Stoichiometric Adjustments: Modifying reactant ratios based on reaction kinetics and equilibrium considerations.

For further reading on stoichiometry and limiting reagents, authoritative sources include: