Calculating the Molar Mass of Ag and S

Oct 13, 2025 by talynt.org 46 views

Introduction

In this article, we will explore the theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S in the given reaction. The reaction is as follows:

${ 2Ag + S \rightarrow Ag_2S }$

To solve this problem, we need to calculate the limiting reagent and then determine the theoretical yield of Ag2S.

Calculating the Molar Mass of Ag and S

First, let's calculate the molar mass of Ag and S.

The atomic mass of Ag is 107.868 g/mol, and the atomic mass of S is 32.065 g/mol.

Calculating the Number of Moles of Ag and S

Now, let's calculate the number of moles of Ag and S.

Number of moles of Ag = mass of Ag / molar mass of Ag = 10.5 g / 107.868 g/mol = 0.097 mol

Number of moles of S = mass of S / molar mass of S = 7.5 g / 32.065 g/mol = 0.234 mol

Determining the Limiting Reagent

To determine the limiting reagent, we need to compare the mole ratio of Ag and S in the reaction.

The balanced equation is:

${ 2Ag + S \rightarrow Ag_2S }$

The mole ratio of Ag to S is 2:1.

We can see that the number of moles of Ag (0.097 mol) is less than half the number of moles of S (0.234 mol / 2 = 0.117 mol).

Therefore, Ag is the limiting reagent.

Calculating the Theoretical Yield of Ag2S

Now, let's calculate the theoretical yield of Ag2S.

The molar mass of Ag2S is 2(107.868 g/mol) + 32.065 g/mol = 247.701 g/mol.

The number of moles of Ag2S that can be formed from 0.097 mol of Ag is:

Number of moles of Ag2S = number of moles of Ag / 2 = 0.097 mol / 2 = 0.0485 mol

The mass of Ag2S that can be formed is:

Mass of Ag2S = number of moles of Ag2S x molar mass of Ag2S = 0.0485 mol x 247.701 g/mol = 12.06 g

Conclusion

In conclusion, the theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S is 12.06 g.

This is the maximum amount of Ag2S that can be formed from the given reactants, assuming that the reaction goes to completion.

Discussion

The reaction is a simple combination reaction between Ag and S to form Ag2S.

The limiting reagent is Ag, which means that Ag is the reactant that is consumed first in the reaction.

The theoretical yield of Ag2S is calculated based on the number of moles of Ag that is available.

The mass of Ag2S that can be formed is calculated by multiplying the number of moles of Ag2S by its molar mass.

Final Answer

The final answer is 12.06 g.

This is the theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S.

Note: The answer is rounded to two decimal places.

Introduction

In our previous article, we explored the theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S in the given reaction. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the balanced equation for the reaction between Ag and S to form Ag2S?

A: The balanced equation for the reaction between Ag and S to form Ag2S is:

${ 2Ag + S \rightarrow Ag_2S }$

Q: How do you determine the limiting reagent in a chemical reaction?

A: To determine the limiting reagent in a chemical reaction, you need to compare the mole ratio of the reactants in the reaction. The reactant that is consumed first in the reaction is the limiting reagent.

Q: What is the molar mass of Ag2S?

A: The molar mass of Ag2S is 2(107.868 g/mol) + 32.065 g/mol = 247.701 g/mol.

Q: How do you calculate the theoretical yield of Ag2S?

A: To calculate the theoretical yield of Ag2S, you need to multiply the number of moles of Ag2S by its molar mass.

Q: What is the theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S?

A: The theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S is 12.06 g.

Q: Why is Ag the limiting reagent in this reaction?

A: Ag is the limiting reagent in this reaction because the number of moles of Ag (0.097 mol) is less than half the number of moles of S (0.234 mol / 2 = 0.117 mol).

Q: What is the significance of the limiting reagent in a chemical reaction?

A: The limiting reagent in a chemical reaction is the reactant that determines the maximum amount of product that can be formed. In this reaction, Ag is the limiting reagent, which means that Ag determines the maximum amount of Ag2S that can be formed.

Q: How do you calculate the mass of Ag2S that can be formed from a given number of moles of Ag2S?

A: To calculate the mass of Ag2S that can be formed from a given number of moles of Ag2S, you need to multiply the number of moles of Ag2S by its molar mass.

Q: What is the relationship between the number of moles of Ag and the number of moles of Ag2S?

A: The number of moles of Ag2S is half the number of moles of Ag, as shown in the balanced equation:

${ 2Ag + S \rightarrow Ag_2S }$

Q: How do you determine the maximum amount of product that can be formed in a chemical reaction?

A: To determine the maximum amount of product that can be formed in a chemical reaction, you need to identify the limiting reagent and calculate the theoretical yield of the product.

Q: What is the significance of the theoretical yield in a chemical reaction?

A: The theoretical yield in a chemical reaction is the maximum amount of product that can be formed, assuming that the reaction goes to completion.

Conclusion

In conclusion, we have answered some frequently asked questions related to the theoretical quantity of Ag2S that can be obtained from the reaction of 10.5 g of Ag with 7.5 g of S. We hope that this article has provided you with a better understanding of the concept of limiting reagent and theoretical yield in a chemical reaction.