If 73.0 G Of $AgNO_3$ (FW 169.87 G/mol) Is Added To A Solution Containing 45.0 G Of $CaBr_2$ (FW 199.89 G/mol), What Mass, In Grams, Of $AgBr$ (FW 187.77 G/mol) Precipitate Will Be Formed? The Balanced Equation Is:$\[

Feb 26, 2025 by ADMIN 218 views

**Precipitation Reaction: Calculating the Mass of AgBr Formed**

Introduction

In a precipitation reaction, a solid forms from a solution when two solutions containing ions combine to form an insoluble compound. In this reaction, silver nitrate ( $AgNO_3$ ) reacts with calcium bromide ( $CaBr_2$ ) to form silver bromide ( $AgBr$ ) and calcium nitrate ( $Ca(NO_3)_2$ ). The balanced equation for this reaction is:

$AgNO_3 + CaBr_2 \rightarrow AgBr + Ca(NO_3)_2$

Calculating the Number of Moles of AgNO_3 and CaBr_2

To determine the mass of $AgBr$ formed, we need to calculate the number of moles of $AgNO_3$ and $CaBr_2$ first. We can use the formula:

Number of moles = mass of substance / molar mass

For $AgNO_3$ :

Number of moles of $AgNO_3$ = 73.0 g / 169.87 g/mol = 0.430 mol

For $CaBr_2$ :

Number of moles of $CaBr_2$ = 45.0 g / 199.89 g/mol = 0.225 mol

Determining the Limiting Reactant

To determine which reactant is the limiting reactant, we need to compare the mole ratio of the reactants to the coefficients in the balanced equation. The balanced equation shows that 1 mole of $AgNO_3$ reacts with 1 mole of $CaBr_2$ . Since we have 0.430 mol of $AgNO_3$ and 0.225 mol of $CaBr_2$ , $CaBr_2$ is the limiting reactant.

Calculating the Number of Moles of AgBr Formed

Since $CaBr_2$ is the limiting reactant, we can use the number of moles of $CaBr_2$ to calculate the number of moles of $AgBr$ formed. The balanced equation shows that 1 mole of $CaBr_2$ produces 1 mole of $AgBr$ . Therefore, the number of moles of $AgBr$ formed is:

Number of moles of $AgBr$ = number of moles of $CaBr_2$ = 0.225 mol

Calculating the Mass of AgBr Formed

Now that we have the number of moles of $AgBr$ formed, we can calculate the mass of $AgBr$ formed using the formula:

Mass of substance = number of moles x molar mass

Mass of $AgBr$ = 0.225 mol x 187.77 g/mol = 42.2 g

Conclusion

In this problem, we calculated the mass of $AgBr$ formed when 73.0 g of $AgNO_3$ is added to a solution containing 45.0 g of $CaBr_2$ . We determined that $CaBr_2$ is the limiting reactant and calculated the number of moles of $AgBr$ formed. Finally, we calculated the mass of $AgBr$ formed using the number of moles and the molar mass of $AgBr$ .

Key Takeaways

To determine the mass of a product formed in a precipitation reaction, we need to calculate the number of moles of the reactants and the product.
We can use the mole ratio of the reactants to the coefficients in the balanced equation to determine which reactant is the limiting reactant.
The number of moles of the product formed is equal to the number of moles of the limiting reactant.
We can calculate the mass of the product formed using the number of moles and the molar mass of the product.

Practice Problems

If 25.0 g of $NaCl$ (FW 58.44 g/mol) is added to a solution containing 30.0 g of $AgNO_3$ (FW 169.87 g/mol), what mass, in grams, of $NaNO_3$ (FW 85.99 g/mol) will be formed?
If 50.0 g of $CaCl_2$ (FW 110.98 g/mol) is added to a solution containing 40.0 g of $NaBr$ (FW 102.89 g/mol), what mass, in grams, of $CaBr_2$ (FW 199.89 g/mol) will be formed?
If 75.0 g of $K_2SO_4$ (FW 174.26 g/mol) is added to a solution containing 60.0 g of $BaCl_2$ (FW 208.23 g/mol), what mass, in grams, of $KCl$ (FW 74.55 g/mol) will be formed?

Solutions

Number of moles of $NaCl$ = 25.0 g / 58.44 g/mol = 0.429 mol Number of moles of $AgNO_3$ = 30.0 g / 169.87 g/mol = 0.176 mol Since $NaCl$ is the limiting reactant, the number of moles of $NaNO_3$ formed is equal to the number of moles of $NaCl$ . Mass of $NaNO_3$ = 0.429 mol x 85.99 g/mol = 36.8 g
Number of moles of $CaCl_2$ = 50.0 g / 110.98 g/mol = 0.450 mol Number of moles of $NaBr$ = 40.0 g / 102.89 g/mol = 0.390 mol Since $CaCl_2$ is the limiting reactant, the number of moles of $CaBr_2$ formed is equal to the number of moles of $CaCl_2$ . Mass of $CaBr_2$ = 0.450 mol x 199.89 g/mol = 90.0 g
Number of moles of $K_2SO_4$ = 75.0 g / 174.26 g/mol = 0.430 mol Number of moles of $BaCl_2$ = 60.0 g / 208.23 g/mol = 0.288 mol Since $K_2SO_4$ is the limiting reactant, the number of moles of $KCl$ formed is equal to the number of moles of $K_2SO_4$ . Mass of $KCl$ = 0.430 mol x 74.55 g/mol = 32.1 g

Introduction

In a precipitation reaction, a solid forms from a solution when two solutions containing ions combine to form an insoluble compound. In this article, we will answer some frequently asked questions about precipitation reactions.

Q: What is a precipitation reaction?

A: A precipitation reaction is a type of chemical reaction where a solid forms from a solution when two solutions containing ions combine to form an insoluble compound.

Q: What are the characteristics of a precipitation reaction?

A: The characteristics of a precipitation reaction include:

The formation of a solid from a solution
The combination of two solutions containing ions
The formation of an insoluble compound
The reaction is usually exothermic

Q: What are the types of precipitation reactions?

A: There are two types of precipitation reactions:

Double displacement reaction: This type of reaction involves the exchange of ions between two solutions.
Precipitation reaction with a common ion: This type of reaction involves the formation of a precipitate when a common ion is added to a solution.

Q: What are the factors that affect the rate of a precipitation reaction?

A: The factors that affect the rate of a precipitation reaction include:

Concentration of the reactants
Temperature
Surface area of the reactants
Presence of a catalyst

Q: How do you determine the limiting reactant in a precipitation reaction?

A: To determine the limiting reactant in a precipitation reaction, you need to compare the mole ratio of the reactants to the coefficients in the balanced equation. The reactant with the smaller mole ratio is the limiting reactant.

Q: What is the role of a catalyst in a precipitation reaction?

A: A catalyst is a substance that speeds up a chemical reaction without being consumed or altered in the process. In a precipitation reaction, a catalyst can increase the rate of the reaction by providing an alternative reaction pathway.

Q: How do you calculate the mass of a product formed in a precipitation reaction?

A: To calculate the mass of a product formed in a precipitation reaction, you need to calculate the number of moles of the reactants and the product. You can use the formula:

Mass of substance = number of moles x molar mass

Q: What are some common applications of precipitation reactions?

A: Some common applications of precipitation reactions include:

Water treatment: Precipitation reactions are used to remove impurities from water.
Industrial processes: Precipitation reactions are used in various industrial processes, such as the production of chemicals and pharmaceuticals.
Environmental remediation: Precipitation reactions are used to remove pollutants from the environment.

Q: What are some common mistakes to avoid in precipitation reactions?

A: Some common mistakes to avoid in precipitation reactions include:

Not balancing the equation
Not identifying the limiting reactant
Not calculating the mass of the product correctly
Not considering the presence of a catalyst

Conclusion

In this article, we have answered some frequently asked questions about precipitation reactions. We have discussed the characteristics of a precipitation reaction, the types of precipitation reactions, and the factors that affect the rate of a precipitation reaction. We have also discussed how to determine the limiting reactant, the role of a catalyst, and how to calculate the mass of a product formed in a precipitation reaction. Finally, we have discussed some common applications and mistakes to avoid in precipitation reactions.

Key Takeaways

A precipitation reaction is a type of chemical reaction where a solid forms from a solution when two solutions containing ions combine to form an insoluble compound.
The characteristics of a precipitation reaction include the formation of a solid from a solution, the combination of two solutions containing ions, the formation of an insoluble compound, and the reaction is usually exothermic.
The factors that affect the rate of a precipitation reaction include concentration of the reactants, temperature, surface area of the reactants, and presence of a catalyst.
To determine the limiting reactant in a precipitation reaction, you need to compare the mole ratio of the reactants to the coefficients in the balanced equation.
A catalyst is a substance that speeds up a chemical reaction without being consumed or altered in the process.
To calculate the mass of a product formed in a precipitation reaction, you need to calculate the number of moles of the reactants and the product.

Practice Problems

If 25.0 g of $NaCl$ (FW 58.44 g/mol) is added to a solution containing 30.0 g of $AgNO_3$ (FW 169.87 g/mol), what mass, in grams, of $NaNO_3$ (FW 85.99 g/mol) will be formed?
If 50.0 g of $CaCl_2$ (FW 110.98 g/mol) is added to a solution containing 40.0 g of $NaBr$ (FW 102.89 g/mol), what mass, in grams, of $CaBr_2$ (FW 199.89 g/mol) will be formed?
If 75.0 g of $K_2SO_4$ (FW 174.26 g/mol) is added to a solution containing 60.0 g of $BaCl_2$ (FW 208.23 g/mol), what mass, in grams, of $KCl$ (FW 74.55 g/mol) will be formed?

Solutions

Number of moles of $NaCl$ = 25.0 g / 58.44 g/mol = 0.429 mol Number of moles of $AgNO_3$ = 30.0 g / 169.87 g/mol = 0.176 mol Since $NaCl$ is the limiting reactant, the number of moles of $NaNO_3$ formed is equal to the number of moles of $NaCl$ . Mass of $NaNO_3$ = 0.429 mol x 85.99 g/mol = 36.8 g
Number of moles of $CaCl_2$ = 50.0 g / 110.98 g/mol = 0.450 mol Number of moles of $NaBr$ = 40.0 g / 102.89 g/mol = 0.390 mol Since $CaCl_2$ is the limiting reactant, the number of moles of $CaBr_2$ formed is equal to the number of moles of $CaCl_2$ . Mass of $CaBr_2$ = 0.450 mol x 199.89 g/mol = 90.0 g
Number of moles of $K_2SO_4$ = 75.0 g / 174.26 g/mol = 0.430 mol Number of moles of $BaCl_2$ = 60.0 g / 208.23 g/mol = 0.288 mol Since $K_2SO_4$ is the limiting reactant, the number of moles of $KCl$ formed is equal to the number of moles of $K_2SO_4$ . Mass of $KCl$ = 0.430 mol x 74.55 g/mol = 32.1 g