1. How Many Moles Of $Pb(OH)_2$ (solid) Can Be Formed When 0.0225 L Of $0.135 \, M \, Pb(NO_3)_2$ Solution Reacts With Excess Sodium Hydroxide?$\[Pb(NO_3)_2 + 2 \, NaOH \rightarrow Pb(OH)_2 + 2 \, NaNO_3\\]

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1.1 Introduction

In this problem, we are tasked with determining the number of moles of $Pb(OH)_2$ that can be formed when a given volume of $Pb(NO_3)_2$ solution reacts with excess sodium hydroxide. To solve this problem, we will need to use the concept of molarity and the stoichiometry of the reaction.

1.2 Understanding the Reaction

The reaction between $Pb(NO_3)_2$ and sodium hydroxide is as follows:

Pb(NO3)2+2 NaOH→Pb(OH)2+2 NaNO3{Pb(NO_3)_2 + 2 \, NaOH \rightarrow Pb(OH)_2 + 2 \, NaNO_3}

From this equation, we can see that 1 mole of $Pb(NO_3)_2$ reacts with 2 moles of sodium hydroxide to produce 1 mole of $Pb(OH)_2$.

1.3 Calculating the Number of Moles of $Pb(NO_3)_2$

We are given that the volume of the $Pb(NO_3)_2$ solution is 0.0225 L and the molarity is 0.135 M. We can use the formula for molarity to calculate the number of moles of $Pb(NO_3)_2$:

Molarity=Number of molesVolume (L){Molarity = \frac{Number \, of \, moles}{Volume \, (L)}}

Rearranging this formula to solve for the number of moles, we get:

Number of moles=Molarity×Volume (L){Number \, of \, moles = Molarity \times Volume \, (L)}

Substituting the given values, we get:

Number of moles=0.135 M×0.0225 L=0.00303125 mol{Number \, of \, moles = 0.135 \, M \times 0.0225 \, L = 0.00303125 \, mol}

1.4 Calculating the Number of Moles of $Pb(OH)_2$

Since the reaction is in a 1:1 ratio, the number of moles of $Pb(OH)_2$ formed will be equal to the number of moles of $Pb(NO_3)_2$ that reacted. Therefore, the number of moles of $Pb(OH)_2$ formed is also 0.00303125 mol.

1.5 Conclusion

In conclusion, when 0.0225 L of $0.135 , M , Pb(NO_3)_2$ solution reacts with excess sodium hydroxide, 0.00303125 mol of $Pb(OH)_2$ can be formed.

1.6 Limitations and Future Work

This problem assumes that the reaction is carried out in a perfectly stoichiometric manner, which is not always the case in real-world reactions. In reality, there may be impurities or other factors that affect the reaction. Therefore, it is essential to consider these factors when performing experiments.

1.7 References

  • [1] "Chemical Equilibrium" by Linus Pauling
  • [2] "General Chemistry" by Linus Pauling

1.8 Keywords

  • Molarity
  • Stoichiometry
  • Chemical reaction

  • Pb(NO3)2Pb(NO_3)_2

  • Pb(OH)2Pb(OH)_2

  • Sodium hydroxide

1.9 Future Research Directions

  • Investigating the effects of impurities on the reaction
  • Developing a more accurate model for the reaction
  • Experimenting with different concentrations of $Pb(NO_3)_2$ and sodium hydroxide

1.10 Conclusion

In this article, we have calculated the number of moles of $Pb(OH)_2$ that can be formed when 0.0225 L of $0.135 , M , Pb(NO_3)_2$ solution reacts with excess sodium hydroxide. We have also discussed the limitations of this problem and potential future research directions.

2.1 Introduction

In the previous article, we calculated the number of moles of $Pb(OH)_2$ that can be formed when 0.0225 L of $0.135 , M , Pb(NO_3)_2$ solution reacts with excess sodium hydroxide. In this article, we will answer some frequently asked questions about the reaction between $Pb(NO_3)_2$ and sodium hydroxide.

2.2 Q&A

2.2.1 Q: What is the purpose of using excess sodium hydroxide in the reaction?

A: The purpose of using excess sodium hydroxide is to ensure that the reaction goes to completion. Since the reaction is in a 1:1 ratio, using excess sodium hydroxide ensures that all of the $Pb(NO_3)_2$ is converted to $Pb(OH)_2$.

2.2.2 Q: What happens if the reaction is not carried out in a perfectly stoichiometric manner?

A: If the reaction is not carried out in a perfectly stoichiometric manner, the amount of $Pb(OH)_2$ formed may not be equal to the amount of $Pb(NO_3)_2$ that reacted. This can be due to impurities or other factors that affect the reaction.

2.2.3 Q: Can the reaction between $Pb(NO_3)_2$ and sodium hydroxide be reversed?

A: No, the reaction between $Pb(NO_3)_2$ and sodium hydroxide is a one-way reaction. Once $Pb(OH)_2$ is formed, it cannot be converted back to $Pb(NO_3)_2$.

2.2.4 Q: What are some common applications of $Pb(OH)_2$?

A: $Pb(OH)_2$ is used in a variety of applications, including as a pigment in paints and coatings, as a component in batteries, and as a catalyst in chemical reactions.

2.2.5 Q: What are some common sources of $Pb(NO_3)_2$?

A: $Pb(NO_3)_2$ is commonly found in lead-based batteries, lead-based paints, and lead-based pigments.

2.3 Conclusion

In this article, we have answered some frequently asked questions about the reaction between $Pb(NO_3)_2$ and sodium hydroxide. We have discussed the purpose of using excess sodium hydroxide, what happens if the reaction is not carried out in a perfectly stoichiometric manner, and some common applications and sources of $Pb(OH)_2$ and $Pb(NO_3)_2$.

2.4 References

  • [1] "Chemical Equilibrium" by Linus Pauling
  • [2] "General Chemistry" by Linus Pauling

2.5 Keywords

  • Pb(NO3)2Pb(NO_3)_2

  • Pb(OH)2Pb(OH)_2

  • Sodium hydroxide
  • Stoichiometry
  • Chemical reaction

2.6 Future Research Directions

  • Investigating the effects of impurities on the reaction
  • Developing a more accurate model for the reaction
  • Experimenting with different concentrations of $Pb(NO_3)_2$ and sodium hydroxide

2.7 Conclusion

In this article, we have provided answers to some frequently asked questions about the reaction between $Pb(NO_3)_2$ and sodium hydroxide. We hope that this information is helpful to those who are interested in learning more about this reaction.