How Many Moles Of Al Are Necessary To Form 81.6 G Of A L B R 3 AlBr_3 A LB R 3 ​ From The Following Reaction? 2 A L ( S ) + 3 B R 2 ( L ) → 2 A L B R 3 ( S 2 \, Al(s) + 3 \, Br_2(l) \rightarrow 2 \, AlBr_3(s 2 A L ( S ) + 3 B R 2 ​ ( L ) → 2 A LB R 3 ​ ( S ]

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Introduction

In chemistry, understanding the stoichiometry of a reaction is crucial for determining the amount of reactants required to produce a specific amount of product. The reaction between aluminum (Al) and bromine (Br2) to form aluminum bromide (AlBr3) is a classic example of a single displacement reaction. In this article, we will explore the stoichiometry of this reaction and calculate the number of moles of Al necessary to form 81.6 g of AlBr3.

Stoichiometry of the Reaction

The balanced chemical equation for the reaction between Al and Br2 to form AlBr3 is:

2 Al(s) + 3 Br2(l) → 2 AlBr3(s)

From the equation, we can see that 2 moles of Al react with 3 moles of Br2 to produce 2 moles of AlBr3. This means that the mole ratio of Al to AlBr3 is 1:1, and the mole ratio of Br2 to AlBr3 is 3:2.

Molar Mass of AlBr3

To calculate the number of moles of AlBr3 produced, we need to know its molar mass. The molar mass of AlBr3 is the sum of the atomic masses of Al and three Br atoms.

Atomic mass of Al = 26.98 g/mol Atomic mass of Br = 79.90 g/mol

Molar mass of AlBr3 = 26.98 + (3 × 79.90) = 266.86 g/mol

Calculating the Number of Moles of AlBr3

Now that we know the molar mass of AlBr3, we can calculate the number of moles of AlBr3 produced from 81.6 g of AlBr3.

Number of moles of AlBr3 = mass of AlBr3 / molar mass of AlBr3 = 81.6 g / 266.86 g/mol = 0.307 mol

Calculating the Number of Moles of Al Required

Since the mole ratio of Al to AlBr3 is 1:1, the number of moles of Al required to form 0.307 mol of AlBr3 is also 0.307 mol.

Conclusion

In conclusion, to form 81.6 g of AlBr3 from the reaction between Al and Br2, we need 0.307 mol of Al. This calculation is based on the stoichiometry of the reaction and the molar mass of AlBr3.

Additional Information

  • The molar mass of Al is 26.98 g/mol.
  • The molar mass of Br2 is 159.80 g/mol.
  • The molar mass of Al is approximately 1/6th the molar mass of AlBr3.

References

  • Chemistry: An Atoms First Approach, by Steven S. Zumdahl
  • General Chemistry: Principles and Modern Applications, by Linus Pauling

Further Reading

  • Stoichiometry: A Guide to Balancing Chemical Equations
  • Molar Mass: A Guide to Calculating Molar Mass
  • Chemical Reactions: A Guide to Understanding Chemical Reactions

Introduction

In our previous article, we explored the stoichiometry of the reaction between aluminum (Al) and bromine (Br2) to form aluminum bromide (AlBr3). We also calculated the number of moles of Al required to form 81.6 g of AlBr3. In this article, we will answer some frequently asked questions (FAQs) about stoichiometry and molar mass.

Q: What is stoichiometry?

A: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It involves calculating the amounts of reactants and products required to produce a specific amount of product.

Q: What is molar mass?

A: Molar mass is the mass of one mole of a substance. It is calculated by summing the atomic masses of the atoms in a molecule.

Q: How do I calculate the molar mass of a compound?

A: To calculate the molar mass of a compound, you need to sum the atomic masses of the atoms in the compound. For example, the molar mass of water (H2O) is calculated as follows:

Atomic mass of H = 1.01 g/mol Atomic mass of O = 16.00 g/mol

Molar mass of H2O = (2 × 1.01) + 16.00 = 18.02 g/mol

Q: What is the difference between atomic mass and molar mass?

A: Atomic mass is the mass of a single atom of an element, while molar mass is the mass of one mole of a substance.

Q: How do I calculate the number of moles of a substance?

A: To calculate the number of moles of a substance, you need to divide the mass of the substance by its molar mass.

Number of moles = mass of substance / molar mass

Q: What is the mole ratio of reactants and products in a chemical reaction?

A: The mole ratio of reactants and products in a chemical reaction is the ratio of the number of moles of each reactant and product. It is calculated by comparing the coefficients of the reactants and products in the balanced chemical equation.

Q: How do I balance a chemical equation?

A: To balance a chemical equation, you need to ensure that the number of atoms of each element is the same on both the reactant and product sides. You can do this by adding coefficients to the reactants and products.

Q: What is the significance of stoichiometry in chemistry?

A: Stoichiometry is crucial in chemistry because it allows us to predict the amounts of reactants and products required to produce a specific amount of product. It is also essential in calculating the yields of chemical reactions and in designing chemical processes.

Q: Can you provide some examples of stoichiometry problems?

A: Here are a few examples of stoichiometry problems:

  • Calculate the number of moles of oxygen required to burn 1.5 g of methane (CH4).
  • Calculate the number of moles of hydrogen required to produce 2.5 g of ammonia (NH3).
  • Calculate the number of moles of carbon required to produce 3.5 g of carbon dioxide (CO2).

Conclusion

In conclusion, stoichiometry and molar mass are fundamental concepts in chemistry that are essential for understanding chemical reactions and calculating the amounts of reactants and products required to produce a specific amount of product. We hope that this article has provided you with a better understanding of these concepts and has answered some of your frequently asked questions.

Additional Information

  • Stoichiometry: A Guide to Balancing Chemical Equations
  • Molar Mass: A Guide to Calculating Molar Mass
  • Chemical Reactions: A Guide to Understanding Chemical Reactions

References

  • Chemistry: An Atoms First Approach, by Steven S. Zumdahl
  • General Chemistry: Principles and Modern Applications, by Linus Pauling

Further Reading

  • Stoichiometry: A Guide to Balancing Chemical Equations
  • Molar Mass: A Guide to Calculating Molar Mass
  • Chemical Reactions: A Guide to Understanding Chemical Reactions