How Many Grams Of $Ag_2S_2O_3$ Form When 125.0 G Of $AgBr$ Reacts Completely According To The Reaction Below?$\[ \begin{array}{c} 2 AgBr + Na_2S_2O_3 \rightarrow Ag_2S_2O_3 + 2 NaBr \\ Ag_2S_2O_3: 327.74 \, \text{g/mol} \\ AgBr:

Feb 28, 2025 by ADMIN 229 views

How Many Grams of $Ag_2S_2O_3$ Form When 125.0 g of $AgBr$ Reacts Completely?

Understanding the Chemical Reaction

The given chemical reaction is: $2 AgBr + Na_2S_2O_3 \rightarrow Ag_2S_2O_3 + 2 NaBr$ . This reaction involves the conversion of silver bromide ( $AgBr$ ) into silver thiosulfate ( $Ag_2S_2O_3$ ) in the presence of sodium thiosulfate ( $Na_2S_2O_3$ ). The reaction is as follows:

$2 AgBr + Na_2S_2O_3 \rightarrow Ag_2S_2O_3 + 2 NaBr$

Calculating the Molar Mass of $AgBr$

To calculate the number of grams of $Ag_2S_2O_3$ formed, we first need to calculate the molar mass of $AgBr$ . The atomic masses of silver (Ag) and bromine (Br) are 107.868 g/mol and 79.904 g/mol, respectively.

Molar mass of $AgBr$ = atomic mass of Ag + atomic mass of Br = 107.868 g/mol + 79.904 g/mol = 187.772 g/mol

Calculating the Number of Moles of $AgBr$

Given that 125.0 g of $AgBr$ reacts completely, we can calculate the number of moles of $AgBr$ using the formula:

Number of moles = mass of substance / molar mass = 125.0 g / 187.772 g/mol = 0.6667 mol

Calculating the Number of Moles of $Ag_2S_2O_3$

From the balanced chemical equation, we can see that 2 moles of $AgBr$ produce 1 mole of $Ag_2S_2O_3$ . Therefore, the number of moles of $Ag_2S_2O_3$ formed is half the number of moles of $AgBr$ .

Number of moles of $Ag_2S_2O_3$ = number of moles of $AgBr$ / 2 = 0.6667 mol / 2 = 0.3333 mol

Calculating the Mass of $Ag_2S_2O_3$ Formed

The molar mass of $Ag_2S_2O_3$ is given as 327.74 g/mol. We can calculate the mass of $Ag_2S_2O_3$ formed using the formula:

Mass of $Ag_2S_2O_3$ = number of moles of $Ag_2S_2O_3$ x molar mass of $Ag_2S_2O_3$ = 0.3333 mol x 327.74 g/mol = 109.3 g

Therefore, when 125.0 g of $AgBr$ reacts completely, 109.3 g of $Ag_2S_2O_3$ is formed.

Conclusion

In this article, we have calculated the number of grams of $Ag_2S_2O_3$ formed when 125.0 g of $AgBr$ reacts completely according to the given chemical reaction. We have used the molar masses of $AgBr$ and $Ag_2S_2O_3$ to calculate the number of moles of $AgBr$ and $Ag_2S_2O_3$ formed, and then used these values to calculate the mass of $Ag_2S_2O_3$ formed. The result is 109.3 g of $Ag_2S_2O_3$ formed when 125.0 g of $AgBr$ reacts completely.

References

CRC Handbook of Chemistry and Physics, 97th Edition
NIST Chemistry WebBook, National Institute of Standards and Technology

Discussion Category: Chemistry

This article is categorized under chemistry, as it involves the calculation of the mass of a chemical compound formed through a chemical reaction. The article uses the principles of stoichiometry and molar masses to calculate the number of moles of $AgBr$ and $Ag_2S_2O_3$ formed, and then uses these values to calculate the mass of $Ag_2S_2O_3$ formed.
Q&A: Calculating the Mass of $Ag_2S_2O_3$ Formed

Q: What is the chemical reaction for the conversion of $AgBr$ to $Ag_2S_2O_3$ ?

A: The chemical reaction for the conversion of $AgBr$ to $Ag_2S_2O_3$ is: $2 AgBr + Na_2S_2O_3 \rightarrow Ag_2S_2O_3 + 2 NaBr$ .

Q: What is the molar mass of $AgBr$ ?

A: The molar mass of $AgBr$ is 187.772 g/mol.

Q: How do I calculate the number of moles of $AgBr$ given a mass of 125.0 g?

A: To calculate the number of moles of $AgBr$ , you can use the formula: Number of moles = mass of substance / molar mass. In this case, the number of moles of $AgBr$ is 125.0 g / 187.772 g/mol = 0.6667 mol.

Q: How do I calculate the number of moles of $Ag_2S_2O_3$ formed?

A: From the balanced chemical equation, we can see that 2 moles of $AgBr$ produce 1 mole of $Ag_2S_2O_3$ . Therefore, the number of moles of $Ag_2S_2O_3$ formed is half the number of moles of $AgBr$ . In this case, the number of moles of $Ag_2S_2O_3$ formed is 0.6667 mol / 2 = 0.3333 mol.

Q: What is the molar mass of $Ag_2S_2O_3$ ?

A: The molar mass of $Ag_2S_2O_3$ is 327.74 g/mol.

Q: How do I calculate the mass of $Ag_2S_2O_3$ formed?

A: To calculate the mass of $Ag_2S_2O_3$ formed, you can use the formula: Mass of $Ag_2S_2O_3$ = number of moles of $Ag_2S_2O_3$ x molar mass of $Ag_2S_2O_3$ . In this case, the mass of $Ag_2S_2O_3$ formed is 0.3333 mol x 327.74 g/mol = 109.3 g.

Q: What is the relationship between the number of moles of $AgBr$ and the number of moles of $Ag_2S_2O_3$ formed?

Q: What is the significance of the molar masses of $AgBr$ and $Ag_2S_2O_3$ in calculating the mass of $Ag_2S_2O_3$ formed?

A: The molar masses of $AgBr$ and $Ag_2S_2O_3$ are used to calculate the number of moles of $AgBr$ and $Ag_2S_2O_3$ formed, respectively. The number of moles of $Ag_2S_2O_3$ formed is then used to calculate the mass of $Ag_2S_2O_3$ formed.

Q: What is the final answer to the problem of calculating the mass of $Ag_2S_2O_3$ formed when 125.0 g of $AgBr$ reacts completely?

A: The final answer is 109.3 g of $Ag_2S_2O_3$ formed when 125.0 g of $AgBr$ reacts completely.