Determining Excess Reactants in Zinc and Sulfur Reaction

The reaction between zinc (Zn) and sulfur (S) to form zinc sulfide (ZnS) is a classic example of a stoichiometric chemical reaction. The balanced equation for this reaction is

Equation: Zn S → ZnS

This article will walk you through the process of determining the amount of excess reactant remaining after the reaction between 25.0 grams of zinc and 30.0 grams of sulfur. By understanding the limiting reactant and the concept of excess reactants, we can predict the outcome of the reaction and help improve chemical processes in industries such as material synthesis, pharmaceuticals, and environmental remediation.

Step 1: Calculating Moles of Each Reactant

To determine which reactant is the limiting reactant and the amount of excess reactant remaining, we first need to calculate the moles of each reactant. The molar masses of zinc and sulfur are as follows:

Molar Mass of Zinc (Zn):

65.38 g/mol

Molar Mass of Sulfur (S):

32.07 g/mol

The moles of each reactant can be calculated using the formula:

Moles Mass / Molar Mass

Moles of Zinc (Zn):

[ n_{text{Zn}} frac{25.0 text{ g}}{65.38 text{ g/mol}} approx 0.382 text{ mol} ]

Moles of Sulfur (S):

[ n_{text{S}} frac{30.0 text{ g}}{32.07 text{ g/mol}} approx 0.936 text{ mol} ]

Step 2: Determining Stoichiometry of the Reaction

The balanced chemical equation is:

Zn S → ZnS

From the equation, it is clear that 1 mole of Zn reacts with 1 mole of S.

Step 3: Identifying the Limiting Reactant

By comparing the moles of each reactant:

Moles of Zn 0.382 mol

Moles of S 0.936 mol

Since the reaction requires a 1:1 mole ratio, we compare the availability:

For Zn: 0.382 moles of Zn will require 0.382 moles of S.

For S: We have 0.936 moles of S available, which is more than enough.

Thus, zinc (Zn) is the limiting reactant.

Step 4: Calculating the Remaining Amount of Sulfur (Excess Reactant)

Since zinc (Zn) is the limiting reactant, we can calculate how much sulfur is consumed:

Moles of S consumed Moles of Zn 0.382 mol

Step 5: Calculating the Moles of Sulfur Remaining

The moles of sulfur remaining can be calculated as follows:

[ text{Moles of S remaining} text{Initial moles of S} - text{Moles of S consumed} 0.936 text{ mol} - 0.382 text{ mol} 0.554 text{ mol} ]

Step 6: Converting Moles of Sulfur Remaining to Grams

The mass of sulfur remaining can be calculated using the molar mass of sulfur:

[ text{Mass of S remaining} text{Moles of S remaining} times text{Molar mass of S} 0.554 text{ mol} times 32.07 text{ g/mol} approx 17.74 text{ g} ]

Conclusion: After the reaction is complete, approximately 17.74 grams of sulfur will remain as the excess reactant.

Understanding the limiting reactant and excess reactants is crucial in various industries and scientific applications. By applying the principles discussed here, you can optimize chemical processes to maximize efficiency and minimize waste. Whether you're working in a laboratory setting or in an industrial plant, mastering the fundamentals of stoichiometry and chemical reactions can greatly enhance your understanding and performance.