If 500.0 G Of Potassium Bromide Reacts With Lead (II) Nitrate, What Mass Of Lead (II) Bromide Is Formed?$ \text{Chemical Reaction \quad \text{Pb(NO}_3\text{)}_2 + \text{2 KBr} \rightarrow \text{PbBr}_2 + \text{2 KNO}_3 }$

Understanding the Chemical Reaction

The given chemical reaction is: $\text{Pb(NO}_3\text{)}_2 + \text{2 KBr} \rightarrow \text{PbBr}_2 + \text{2 KNO}_3$. This is a double displacement reaction where lead (II) nitrate reacts with potassium bromide to form lead (II) bromide and potassium nitrate.

Calculating the Molar Mass of Potassium Bromide

To solve this problem, we need to calculate the molar mass of potassium bromide (KBr). The atomic mass of potassium (K) is 39.1 g/mol, and the atomic mass of bromine (Br) is 79.9 g/mol. Therefore, the molar mass of KBr is:

$$\text{Molar Mass of KBr} = 39.1 \text{ g/mol} + 79.9 \text{ g/mol} = 119.0 \text{ g/mol}$$

Calculating the Number of Moles of Potassium Bromide

Given that we have 500.0 g of potassium bromide, we can calculate the number of moles of KBr using the formula:

$$\text{Number of Moles} = \frac{\text{Mass of KBr}}{\text{Molar Mass of KBr}}$$

$$\text{Number of Moles of KBr} = \frac{500.0 \text{ g}}{119.0 \text{ g/mol}} = 4.20 \text{ mol}$$

Determining the Molar Ratio of Potassium Bromide to Lead (II) Bromide

From the balanced chemical equation, we can see that 2 moles of KBr react with 1 mole of Pb(NO3)2 to form 1 mole of PbBr2. Therefore, the molar ratio of KBr to PbBr2 is 2:1.

Calculating the Number of Moles of Lead (II) Bromide

Since the molar ratio of KBr to PbBr2 is 2:1, we can calculate the number of moles of PbBr2 formed by dividing the number of moles of KBr by 2:

$$\text{Number of Moles of PbBr2} = \frac{\text{Number of Moles of KBr}}{2}$$

$$\text{Number of Moles of PbBr2} = \frac{4.20 \text{ mol}}{2} = 2.10 \text{ mol}$$

Calculating the Molar Mass of Lead (II) Bromide

The atomic mass of lead (Pb) is 207.2 g/mol, and the atomic mass of bromine (Br) is 79.9 g/mol. Therefore, the molar mass of PbBr2 is:

$$\text{Molar Mass of PbBr2} = 207.2 \text{ g/mol} + 2 \times 79.9 \text{ g/mol} = 366.0 \text{ g/mol}$$

Calculating the Mass of Lead (II) Bromide

Finally, we can calculate the mass of PbBr2 formed by multiplying the number of moles of PbBr2 by its molar mass:

$$\text{Mass of PbBr2} = \text{Number of Moles of PbBr2} \times \text{Molar Mass of PbBr2}$$

$$\text{Mass of PbBr2} = 2.10 \text{ mol} \times 366.0 \text{ g/mol} = 768.6 \text{ g}$$

Therefore, the mass of lead (II) bromide formed when 500.0 g of potassium bromide reacts with lead (II) nitrate is 768.6 g.

Conclusion

In this problem, we calculated the mass of lead (II) bromide formed when 500.0 g of potassium bromide reacts with lead (II) nitrate. We first calculated the number of moles of potassium bromide, then determined the molar ratio of potassium bromide to lead (II) bromide, and finally calculated the mass of lead (II) bromide formed. The result is 768.6 g of lead (II) bromide.

Limitations and Assumptions

This problem assumes that the reaction is carried out in a perfectly stoichiometric manner, meaning that the reactants are present in the exact proportions required by the balanced chemical equation. In reality, the reaction may not be perfectly stoichiometric, and the actual yield of lead (II) bromide may be different from the calculated value.

Future Directions

This problem can be extended to explore the effects of different reaction conditions on the yield of lead (II) bromide. For example, we could investigate how changes in temperature, pressure, or concentration of the reactants affect the reaction rate and yield of lead (II) bromide.

References

• Chemical Reaction: $\text{Pb(NO}_3\text{)}_2 + \text{2 KBr} \rightarrow \text{PbBr}_2 + \text{2 KNO}_3$
• Molar Mass of KBr: 119.0 g/mol
• Molar Mass of PbBr2: 366.0 g/mol

Note: The references provided are the chemical reaction and the molar masses of KBr and PbBr2, which are used to solve the problem.

Frequently Asked Questions

Q: What is the chemical reaction between potassium bromide and lead (II) nitrate?

A: The chemical reaction between potassium bromide (KBr) and lead (II) nitrate (Pb(NO3)2) is: $\text{Pb(NO}_3\text{)}_2 + \text{2 KBr} \rightarrow \text{PbBr}_2 + \text{2 KNO}_3$. This is a double displacement reaction where lead (II) nitrate reacts with potassium bromide to form lead (II) bromide and potassium nitrate.

Q: How do I calculate the molar mass of potassium bromide?

A: To calculate the molar mass of potassium bromide (KBr), you need to add the atomic masses of potassium (K) and bromine (Br). The atomic mass of potassium is 39.1 g/mol, and the atomic mass of bromine is 79.9 g/mol. Therefore, the molar mass of KBr is: $\text{Molar Mass of KBr} = 39.1 \text{ g/mol} + 79.9 \text{ g/mol} = 119.0 \text{ g/mol}$

Q: How do I calculate the number of moles of potassium bromide?

A: To calculate the number of moles of potassium bromide (KBr), you need to divide the mass of KBr by its molar mass. Given that we have 500.0 g of KBr, we can calculate the number of moles of KBr using the formula: $\text{Number of Moles} = \frac{\text{Mass of KBr}}{\text{Molar Mass of KBr}}$. Therefore, the number of moles of KBr is: $\text{Number of Moles of KBr} = \frac{500.0 \text{ g}}{119.0 \text{ g/mol}} = 4.20 \text{ mol}$

Q: What is the molar ratio of potassium bromide to lead (II) bromide?

A: From the balanced chemical equation, we can see that 2 moles of KBr react with 1 mole of Pb(NO3)2 to form 1 mole of PbBr2. Therefore, the molar ratio of KBr to PbBr2 is 2:1.

Q: How do I calculate the number of moles of lead (II) bromide?

A: Since the molar ratio of KBr to PbBr2 is 2:1, we can calculate the number of moles of PbBr2 formed by dividing the number of moles of KBr by 2. Therefore, the number of moles of PbBr2 is: $\text{Number of Moles of PbBr2} = \frac{\text{Number of Moles of KBr}}{2} = \frac{4.20 \text{ mol}}{2} = 2.10 \text{ mol}$

Q: What is the molar mass of lead (II) bromide?

A: The atomic mass of lead (Pb) is 207.2 g/mol, and the atomic mass of bromine (Br) is 79.9 g/mol. Therefore, the molar mass of PbBr2 is: $\text{Molar Mass of PbBr2} = 207.2 \text{ g/mol} + 2 \times 79.9 \text{ g/mol} = 366.0 \text{ g/mol}$

Q: How do I calculate the mass of lead (II) bromide?

A: Finally, we can calculate the mass of PbBr2 formed by multiplying the number of moles of PbBr2 by its molar mass. Therefore, the mass of PbBr2 is: $\text{Mass of PbBr2} = \text{Number of Moles of PbBr2} \times \text{Molar Mass of PbBr2} = 2.10 \text{ mol} \times 366.0 \text{ g/mol} = 768.6 \text{ g}$

Q: What is the mass of lead (II) bromide formed when 500.0 g of potassium bromide reacts with lead (II) nitrate?

A: The mass of lead (II) bromide formed when 500.0 g of potassium bromide reacts with lead (II) nitrate is 768.6 g.

Q: What are the limitations and assumptions of this problem?

A: This problem assumes that the reaction is carried out in a perfectly stoichiometric manner, meaning that the reactants are present in the exact proportions required by the balanced chemical equation. In reality, the reaction may not be perfectly stoichiometric, and the actual yield of lead (II) bromide may be different from the calculated value.

Q: What are the future directions of this problem?

A: This problem can be extended to explore the effects of different reaction conditions on the yield of lead (II) bromide. For example, we could investigate how changes in temperature, pressure, or concentration of the reactants affect the reaction rate and yield of lead (II) bromide.

Q: What are the references used in this problem?

A: The references used in this problem are the chemical reaction and the molar masses of KBr and PbBr2, which are used to solve the problem.