Limiting Reagent

Limiting reagents are substances that are ultimately consumed during a chemical reaction, determining when the reaction stops. They are also known as limiting reactants or limiting agents. By the stoichiometry of a chemical reaction, a specific amount of reactants is required for the reaction to complete.

1.0Introduction

In a chemical reaction, the limiting reagent determines the maximum product formed. Other reactants may be in excess, meaning they are only partially consumed when the reaction finishes. The theoretical yield is the maximum quantity of product that can be formed based on the limiting reagent. Identifying the limiting reagent is essential for calculating the percentage yield of a reaction. Several methods are available, such as a balanced chemical equation, to determine the limiting reagent and calculate the excess amounts of the other reactants.

2.0Limiting Reagent

In a chemical reaction, the limiting reagent is the reactant that is entirely consumed first, thus limiting the amount of product formed. It is present in the smallest stoichiometric amount relative to the reaction requirements.

• The limiting reagent controls the maximum amount of product that can be produced.
• Any reactant left over after the reaction is called the excess reagent.
• To apply the concept of limiting reagents, the chemical reaction must be balanced.

If the initial moles of reactants are not in the same ratio as their stoichiometric coefficients in the balanced equation, at least one reactant will act as the limiting reagent.

The limiting reagent is the reactant that dictates when a reaction will stop, as it is wholly consumed first. The quantity required to react with another substance is determined by stoichiometry, and the limiting reagent depends on the mole ratio, not on the masses of the reactants present.

3.0Methods to identify the limiting reactant

Consider the reaction:

3H₂+N₂→2NH₃​

According to the balanced equation, 3 moles of hydrogen gas (H₂) are needed to react with 1 mole of nitrogen gas (N₂) to produce 2 moles of ammonia (NH₃).

However, if only 2 moles of hydrogen are available along with 1 mole of nitrogen, the reaction is limited by the hydrogen:

• 3 moles of H₂ are required for every 1 mole of N₂.
• With only 2 moles of H₂, it’s not enough to completely react with the 1 mole of N₂.
• Therefore, hydrogen is the limiting reagent, as it determines the maximum amount of ammonia produced, while nitrogen will remain partially unused.

Step 1: Determine the Ratio
Divide the initial moles of each reactant by its respective stoichiometric coefficient in the balanced equation. Use the formula:

$Ratio=\frac{GivenInitialMolesofReactant}{Stoichiometriccoefficient}$

Step 2: Identify the Limiting Reagent
Compare the ratios obtained for all reactants. The reactant with the smallest ratio is the limiting reagent, as it is the first to be wholly consumed in the reaction.

Step 3: Base Calculations on the Limiting Reagent
Once the limiting reagent is identified, the amounts of products formed and any remaining excess reactants are calculated. All stoichiometric calculations will revolve around this limiting reagent.

Two methods exist to identify the limiting reactant in a chemical reaction. These are outlined below:

Approach 1: Using Mole Ratio 

To identify the limiting reagent by comparing the number of moles of each reactant:

• Write the Balanced Chemical Equation: Ensure the reaction is correctly balanced with the correct stoichiometric coefficients.
• Convert Quantities to Moles: Molar masses convert all given reactants into moles.
• Calculate the Mole Ratio: Based on the quantities provided, determine the mole ratio of the reactants. Compare this calculated ratio to the ratio in the balanced equation.
• Identify the Limiting Reagent: The reactant with a smaller mole ratio than what is required by the balanced equation is the limiting reagent.
• Calculate Product Formation: Use the limiting reagent to determine the amount of product that can be formed.
• Find Excess Reactant: If needed, calculate the leftover quantity of the excess reagent.

Solved Example

Q. Identify the limiting reagent in the reaction between 46 grams of iron and 32 grams of sulfur.

Solution

• Balanced Equation:
  The reaction is represented by:
  Fe(s)+S(s)→FeS(s))
  (One mole of iron reacts with one sulfur to form iron(II) sulfide.)
• Mole Ratio of Reactants:
  According to the equation, the ideal mole ratio of Fe to S is 1:1.
• Convert Mass to Moles:
• • Iron:
    46 g Fe÷55.85 g/mol=0.82 mol Fe
  • Sulfur:
    23 g S÷32.06 g/mol=0.72 mol S
• Compare the Mole Ratio:
  The actual mole ratio is 0.82 mol Fe: 0.72 mol S.
• The Limiting Reactant:
  Since Fe and S react in a 1:1 ratio, and there are fewer moles of S than Fe, sulfur (S) is the limiting reactant.

Approach 2: Using Product Amount  

To determine the limiting reagent by calculating and comparing the amount of product each reactant can form:

• Balance the Chemical Equation: Make sure the chemical reaction is balanced.
• Convert to Moles: Convert the given quantities of reactants into moles.
• Stoichiometric Calculation: Use stoichiometry for each reactant to calculate the amount of product that can be produced.
• Identify the Limiting Reagent: The reactant that yields the smaller quantity of product is the limiting reagent.
• Determine the Excess Reagent: The excess reagent is the reactant that produces more product.
• Calculate Remaining Excess: To find the leftover amount of the excess reactant, subtract the mass of the consumed excess reagent from the initial mass provided.

Solved Example

Let’s revisit the reaction between 46 grams of iron and 32 grams of sulfur by applying approach 2:

1. Balanced Equation:
   The reaction is represented by:
   Fe(s)+S(s)→FeS(s))
   (One mole of iron reacts with one sulfur to form iron(II) sulfide.)
2. Convert Grams to Moles:

• Moles of Fe: 46 g÷55.85 g/mol=0.82 mol
• Moles of S: 23 g÷32.06 g/mol=0.72 mol

3. Determine Theoretical Moles of Product:
   Based on the 1:1 reactant-to-product ratio: 0.82 mol Fe can produce 0.82 mol FeS

0.72 mol S can produce 0.72 mol FeS.

4. Identifying  the Limiting Reactant: Since sulfur yields a smaller amount of iron sulfide, sulfur (S) is the limiting reactant.

4.0Solved Example

Identify the limiting reactant. AgNO_3​+HCl→AgCl+HNO_3​

Step 1: Mole Ratio from the Balanced Equation

The balanced equation shows:

• 1 mol AgNO₃ reacts with 1 mol HCl
• This means the mole ratio of AgNO₃ to HCl is  1:1.

Step 2: Given Quantities

• 2 mol AgNO₃​
• 1 mol HCl

Step 3: Determine the Limiting Reactant

The reaction requires 1 mol HCl for every 1 mol AgNO₃​:

• To completely react 2 mol AgNO₃​, you would need 2 mol HCl
• However, only 1 mol HCl is available.

Since HCl is insufficient to react with all the AgNO₃ HCl is the limiting reactant.

Step 4: Excess Reactant

• AgNO₃ is in excess because only 1 mol AgNO₃ will react with 1 mol HCl.
• The remaining AgNO₃​ is 2−1=1mol.

Conclusion: The limiting reactant is HCl.