Limiting Reagents and Stoichiometry

Introduction to Limiting Reagents

In chemical reactions, the limiting reagent (or limiting reactant) is the substance that is completely consumed first, thus determining the maximum amount of product that can be formed. The other reactants are considered excess reagents because they are not fully used up in the reaction.

• Balanced chemical equations are essential for identifying the stoichiometric relationships between reactants and products.

• To determine the limiting reagent, compare the mole ratios of the reactants used to those required by the balanced equation.

Example: Sulfur Dioxide and Oxygen Reaction

Consider the reaction:

• If you start with 3.5 moles of ethanol () and 10.0 moles of oxygen (), you must determine which reactant will be consumed first based on the balanced equation.

• Identify the limiting reagent by calculating how much product can be formed from each reactant and selecting the smaller value.

Stepwise Approach to Limiting Reagent Problems

1. Write the balanced chemical equation.

2. Convert all given quantities to moles.

3. Use stoichiometry to determine the amount of product each reactant can produce.

4. The reactant that produces the least amount of product is the limiting reagent.

Theoretical Yield and Percent Yield

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the limiting reagent, calculated using stoichiometry and the balanced equation.

• It assumes perfect reaction conditions and complete conversion of the limiting reagent.

• Formula:

Percent Yield

The percent yield compares the actual amount of product obtained from a reaction to the theoretical yield.

• Formula:

• Percent yield is always less than or equal to 100% due to losses or incomplete reactions.

Worked Example: Reaction Between Phosphorus and Chlorine

Step 1: Write the Balanced Equation

Elemental phosphorus reacts with chlorine gas to produce phosphorus trichloride:

Step 2: Determine the Limiting Reagent

• Given: 5.2 g of (molar mass = 123.88 g/mol) and 5.7 g of (molar mass = 70.90 g/mol).

• Convert to moles:

  • :

  • :

• Stoichiometry: reacts with , so requires , but only is available. Thus, is the limiting reagent.

Step 3: Calculate Theoretical Yield of

• From the balanced equation: produces , so produces .

• Molar mass of = 137.2 g/mol.

• Theoretical yield:

Step 4: Calculate Percent Yield

• If only 6.9 g of is collected, then:

Step 5: Mass of Limiting and Excess Reagents Remaining

• Limiting reagent () is completely consumed; mass remaining = 0 g.

• Excess reagent ():

  • Used:

  • Initial: , so remaining:

  • Mass remaining:

Example: Combustion of Methane

Given:

• 16 g ( mol), 16 g ( mol).

• Stoichiometry: reacts with , so mol can react with mol .

• Limiting reagent: .

• Water produced: mol produces $0.5$ mol ( per ), so $0.5$ mol 18 g/mol = 9 g $\mathrm{H_2O}$.

• If 8.2 g is collected:

Summary Table: Key Concepts in Limiting Reagent Calculations