Understanding and Calculating the Limiting Reagent

In the realm of chemical reactions, the concept of the limiting reagent is pivotal. It refers to the reactant that determines the maximum amount of product that can be obtained in a reaction. This article will guide you through the process of calculating the limiting reagent with a series of examples and step-by-step explanations.

Steps to Calculate the Limiting Reagent

The process of determining the limiting reagent involves several key steps, ensuring precise stoichiometric calculations. Let's break down the procedure:

Write the Balanced Equation: Begin by ensuring you have a balanced chemical equation for the reaction. Determine Moles of Reactants: Calculate the number of moles of each reactant you have. Use Stoichiometry: Utilize the coefficients from the balanced equation to find out how many moles of each reactant are required for the reaction. Calculate the Amount Required: For each reactant, calculate how many moles are needed based on the available amount of the other reactant. Compare Available Moles with Required Moles: Identify which reactant runs out first. Conclusion: The reactant that is present in the least amount relative to the stoichiometric requirements is the limiting reagent.

Example: Calculating the Limiting Reagent

Let's consider a practical example to illustrate the steps involved in determining the limiting reagent.

Example: Reaction of (S_8), (O_2), and (H_2O)

Given the reaction:

(S_8 12 O_2 8 H_2O rightarrow 8 H_2SO_4)

With the following quantities of reactants:

14.5 g (S_8) 20.8 L (O_2) at STP (Standard Temperature and Pressure) 20.8 mL (H_2O) with a density of 1.00 g/mL

Step-by-step Calculation:

1. Write the Balanced Equation

The equation is already balanced: (S_8 12 O_2 8 H_2O rightarrow 8 H_2SO_4)

2. Determine Moles of Reactants

Substance Given Quantity Moles (mol) (S_8) 14.5 g (frac{14.5 , g}{256.48 , g/mol} 0.05653 , text{mol}) (O_2) 20.8 L (STP) (frac{20.8 , L}{22.4 , L/mol} 0.9285 , text{mol}) (H_2O) 20.8 mL with 1.00 g/mL density (frac{20.8 , mL times 1.00 , g/mL}{18.01 , g/mol} 1.155 , text{mol})

3. Use Stoichiometry

Divide each mole quantity by the coefficient of that reactant in the balanced equation:

Substance Moles / Coefficients Quotient (S_8) 0.05653 mol / 1 mol/reaction 0.05653 "reactions" (O_2) 0.9285 mol / 12 mol/reaction 0.0774 "reactions" (H_2O) 1.155 mol / 8 mol/reaction 0.1444 "reactions"

4. Identify the Smallest Quotient

The smallest quotient, 0.05653 "reactions," is for (S_8), making (S_8) the limiting reactant.

5. Calculating Theoretical Yield

Using the limiting reagent, calculate the theoretical yield of (H_2SO_4):

Theoretical yield 0.05653 "reactions" (times) 8 mol (H_2SO_4) / "reaction" 0.4523 mol (H_2SO_4)

Example: Reaction of (H_2O_2 rightarrow H_2O)

Consider the reaction:

(2 H_2O_2 rightarrow 2 H_2O)

With the following quantities of reactants:

3 moles (H_2O_2) 1 mole (O_2)

Step-by-step Calculation:

Write the Balanced Equation:

(2 H_2O_2 rightarrow 2 H_2O)

Determine Moles of Reactants: 3 moles (H_2O_2) 1 mole (O_2) Use Stoichiometry:

According to the balanced equation, 2 moles of (H_2O_2) react with 1 mole of (O_2).

Calculate the Amount Required: For 3 moles of (H_2O_2), you need (frac{3}{2} 1.5) moles of (O_2). Comparison:

You have 1 mole of (O_2) but you need 1.5 moles.

Conclusion:

(O_2) is the limiting reagent.

By following these steps, you can effectively determine the limiting reagent in any chemical reaction, ensuring optimal usage of reactants and maximizing product yield.