What Mass Of $H_2$ Forms When 35.25 G Of Al Reacts With Excess Hydrochloric Acid?$[ \begin Array}{c} 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \ \text{Al 26.98 \text{ G/mol} \ \text{H}_2: 2.02 \text{

Introduction

Chemical reactions involving metals and acids are fundamental concepts in chemistry. In this scenario, we are dealing with the reaction between aluminum (Al) and hydrochloric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas (H2). The given reaction is:

$$2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2$$

The molar mass of aluminum (Al) is 26.98 g/mol, and the molar mass of hydrogen gas (H2) is 2.02 g/mol. We are asked to find the mass of H2 that forms when 35.25 g of Al reacts with excess hydrochloric acid.

Step 1: Calculate the Number of Moles of Al

To find the mass of H2 produced, we first need to calculate the number of moles of Al that reacts. We can use the formula:

$$\text{moles} = \frac{\text{mass}}{\text{molar mass}}$$

Substituting the given values, we get:

$$\text{moles of Al} = \frac{35.25 \text{ g}}{26.98 \text{ g/mol}}$$

Step 2: Calculate the Number of Moles of H2 Produced

From the balanced equation, we can see that 2 moles of Al produce 3 moles of H2. Therefore, the mole ratio of Al to H2 is 2:3. We can use this ratio to calculate the number of moles of H2 produced:

$$\text{moles of H}_2 = \frac{3}{2} \times \text{moles of Al}$$

Step 3: Calculate the Mass of H2 Produced

Now that we have the number of moles of H2 produced, we can calculate the mass of H2 produced using the formula:

$$\text{mass} = \text{moles} \times \text{molar mass}$$

Substituting the values, we get:

$$\text{mass of H}_2 = 2.02 \text{ g/mol} \times \text{moles of H}_2$$

Calculations

Step 1: Calculate the Number of Moles of Al

$$\text{moles of Al} = \frac{35.25 \text{ g}}{26.98 \text{ g/mol}} = 1.308 \text{ mol}$$

Step 2: Calculate the Number of Moles of H2 Produced

$$\text{moles of H}_2 = \frac{3}{2} \times 1.308 \text{ mol} = 1.962 \text{ mol}$$

Step 3: Calculate the Mass of H2 Produced

$$\text{mass of H}_2 = 2.02 \text{ g/mol} \times 1.962 \text{ mol} = 3.97 \text{ g}$$

Conclusion

Therefore, when 35.25 g of Al reacts with excess hydrochloric acid, the mass of H2 produced is 3.97 g.

Discussion

This problem involves a simple chemical reaction between a metal (Al) and an acid (HCl) to produce a gas (H2). The balanced equation provides the mole ratio of Al to H2, which is used to calculate the number of moles of H2 produced. Finally, the mass of H2 produced is calculated using the number of moles and the molar mass of H2.

Limitations

This problem assumes that the reaction is carried out in excess HCl, which means that the amount of HCl is much larger than the amount of Al. This is a common assumption in many chemical reactions, but it may not always be the case in real-world scenarios.

Future Work

This problem can be extended to more complex reactions involving multiple reactants and products. Additionally, the reaction can be carried out under different conditions, such as varying temperatures or pressures, to study the effects on the reaction rate and product yield.

References

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

Note: The references provided are for general chemistry textbooks and are not specific to this problem. However, they provide a good starting point for further reading and research.

Introduction

In our previous article, we explored the chemical reaction between aluminum (Al) and hydrochloric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas (H2). We calculated the mass of H2 produced when 35.25 g of Al reacts with excess HCl. In this Q&A article, we will address some common questions related to this problem.

Q1: What is the balanced equation for the reaction between Al and HCl?

A1: The balanced equation for the reaction between Al and HCl is:

$$2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2$$

Q2: What is the mole ratio of Al to H2 in the reaction?

A2: The mole ratio of Al to H2 in the reaction is 2:3. This means that 2 moles of Al produce 3 moles of H2.

Q3: How do I calculate the number of moles of Al that reacts?

A3: To calculate the number of moles of Al that reacts, you can use the formula:

$$\text{moles} = \frac{\text{mass}}{\text{molar mass}}$$

Substituting the given values, you get:

$$\text{moles of Al} = \frac{35.25 \text{ g}}{26.98 \text{ g/mol}} = 1.308 \text{ mol}$$

Q4: How do I calculate the number of moles of H2 produced?

A4: To calculate the number of moles of H2 produced, you can use the mole ratio of Al to H2. Since 2 moles of Al produce 3 moles of H2, you can multiply the number of moles of Al by 3/2 to get the number of moles of H2:

$$\text{moles of H}_2 = \frac{3}{2} \times 1.308 \text{ mol} = 1.962 \text{ mol}$$

Q5: How do I calculate the mass of H2 produced?

A5: To calculate the mass of H2 produced, you can multiply the number of moles of H2 by the molar mass of H2:

$$\text{mass of H}_2 = 2.02 \text{ g/mol} \times 1.962 \text{ mol} = 3.97 \text{ g}$$

Q6: What are the limitations of this problem?

A6: This problem assumes that the reaction is carried out in excess HCl, which means that the amount of HCl is much larger than the amount of Al. This is a common assumption in many chemical reactions, but it may not always be the case in real-world scenarios.

Q7: Can this problem be extended to more complex reactions?

A7: Yes, this problem can be extended to more complex reactions involving multiple reactants and products. Additionally, the reaction can be carried out under different conditions, such as varying temperatures or pressures, to study the effects on the reaction rate and product yield.

Q8: What are some common applications of this reaction?

A8: This reaction is commonly used in the production of aluminum chloride, which is used as a catalyst in various industrial processes. Additionally, the reaction can be used to produce hydrogen gas, which is used as a fuel in various applications.

Conclusion

In this Q&A article, we addressed some common questions related to the chemical reaction between aluminum (Al) and hydrochloric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas (H2). We provided step-by-step calculations for the number of moles of Al that reacts, the number of moles of H2 produced, and the mass of H2 produced. We also discussed the limitations of this problem and some common applications of this reaction.

Discussion

This problem is a fundamental concept in chemistry, and it is essential to understand the principles of chemical reactions and stoichiometry. The balanced equation, mole ratio, and molar mass are all critical concepts in this problem. By understanding these concepts, you can apply them to more complex reactions and real-world scenarios.

Future Work

This problem can be extended to more complex reactions involving multiple reactants and products. Additionally, the reaction can be carried out under different conditions, such as varying temperatures or pressures, to study the effects on the reaction rate and product yield.

References

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

Note: The references provided are for general chemistry textbooks and are not specific to this problem. However, they provide a good starting point for further reading and research.