Aluminum And Sulfur Reaction With Potassium Hydroxide Gas Volume And Mass Fractions

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Introduction

Understanding the intricate world of chemical reactions is crucial for students and enthusiasts alike. This article delves into a fascinating scenario involving a mixture of aluminum (Al) and sulfur (S) reacting with a potassium hydroxide (KOH) solution. We aim to provide a comprehensive explanation of the chemical processes involved, including the balanced chemical equations, step-by-step calculations, and a detailed analysis of the final products. By exploring this reaction, we can gain valuable insights into stoichiometry, gas evolution, and the behavior of amphoteric metals like aluminum in alkaline solutions. This exploration is not just an academic exercise; it showcases the practical applications of chemistry in various fields, from industrial processes to environmental science. Our journey begins with a precise problem statement, setting the stage for a meticulous analysis that will unravel the complexities of this chemical transformation. Let's embark on this journey together, demystifying the reaction of aluminum and sulfur with potassium hydroxide.

Problem Statement: Reaction of Aluminum and Sulfur with Potassium Hydroxide

Let's dissect the problem at hand. We have a mixture containing equal masses (20 g each) of aluminum (Al) and sulfur (S). This mixture is subjected to heating, initiating a reaction between the two elements. The resulting product is then treated with a 1400 g solution of potassium hydroxide (KOH) with a concentration of 32%. Our task is twofold:

  1. Calculate the volume of gas evolved at standard conditions (STP).
  2. Determine the mass fractions of the substances present in the final solution.

This problem is a classic example of a multi-step chemical reaction, requiring a thorough understanding of stoichiometry and solution chemistry. Before diving into the calculations, it's essential to outline the reaction sequence. First, aluminum and sulfur react to form aluminum sulfide (Al₂S₃). Then, the aluminum sulfide, along with any remaining aluminum, reacts with the potassium hydroxide solution. The reaction with KOH is where the gas evolution occurs, specifically hydrogen gas (H₂). Identifying these key steps is crucial for setting up the stoichiometric calculations and accurately determining the amounts of reactants and products involved. This structured approach will enable us to break down the complex problem into manageable parts, ensuring a clear and logical solution.

Step-by-Step Solution

1. Reaction of Aluminum and Sulfur

The initial step involves the reaction between aluminum (Al) and sulfur (S) upon heating. This reaction forms aluminum sulfide (Al₂S₃). The balanced chemical equation for this reaction is:

2Al + 3S → Al₂S₃

This equation signifies that two moles of aluminum react with three moles of sulfur to produce one mole of aluminum sulfide. To proceed with the calculations, we need to determine the number of moles of aluminum and sulfur present in the initial mixture. This involves using the molar masses of aluminum (26.98 g/mol) and sulfur (32.07 g/mol) to convert the given masses (20 g each) into moles. Calculating the moles of each reactant will help us identify the limiting reactant, which is crucial for determining the amount of aluminum sulfide formed. The limiting reactant is the one that is completely consumed in the reaction, dictating the maximum amount of product that can be formed. By understanding the stoichiometry of the reaction and the concept of limiting reactants, we can accurately predict the outcome of the initial reaction between aluminum and sulfur.

2. Calculating Moles of Reactants

To determine the limiting reactant, we must calculate the number of moles of aluminum (Al) and sulfur (S). Given that we have 20 g of each element, we can use their respective molar masses:

  • Moles of Al = mass of Al / molar mass of Al = 20 g / 26.98 g/mol ≈ 0.741 mol
  • Moles of S = mass of S / molar mass of S = 20 g / 32.07 g/mol ≈ 0.624 mol

Now, we compare the mole ratio of Al and S with the stoichiometric ratio from the balanced equation (2Al : 3S). The stoichiometric ratio is 2/3 ≈ 0.667. To determine the limiting reactant, we can divide the moles of each reactant by its coefficient in the balanced equation:

  • For Al: 0.741 mol / 2 ≈ 0.371
  • For S: 0.624 mol / 3 ≈ 0.208

Since 0.208 is smaller than 0.371, sulfur (S) is the limiting reactant. This means that sulfur will be completely consumed in the reaction, and the amount of aluminum sulfide formed will be determined by the initial amount of sulfur. Now that we've identified the limiting reactant, we can calculate the moles of aluminum sulfide produced.

3. Calculating Moles of Aluminum Sulfide (Al₂S₃) Formed

Using the stoichiometry of the reaction and the moles of the limiting reactant (sulfur), we can calculate the moles of aluminum sulfide (Al₂S₃) formed. From the balanced equation:

2Al + 3S → Al₂S₃

3 moles of S produce 1 mole of Al₂S₃. Therefore, the moles of Al₂S₃ formed can be calculated as:

Moles of Al₂S₃ = (moles of S) / 3 = 0.624 mol / 3 ≈ 0.208 mol

This calculation tells us that 0.208 moles of aluminum sulfide are formed in the reaction. However, since sulfur is the limiting reactant, not all of the aluminum will be consumed. We need to determine the amount of aluminum that remains unreacted. To do this, we'll use the stoichiometry of the reaction again, but this time to calculate the moles of aluminum that reacted with the sulfur. This will allow us to subtract the reacted aluminum from the initial amount to find the remaining aluminum. Knowing the amounts of Al₂S₃ formed and Al remaining is crucial for the next step, where these substances react with the potassium hydroxide solution.

4. Calculating Moles of Aluminum (Al) Remaining

To determine the moles of aluminum (Al) remaining after the reaction with sulfur, we first need to calculate the moles of Al that reacted. From the balanced equation:

2Al + 3S → Al₂S₃

3 moles of S react with 2 moles of Al. Therefore, the moles of Al that reacted can be calculated as:

Moles of Al reacted = (2/3) * (moles of S) = (2/3) * 0.624 mol ≈ 0.416 mol

Now, we subtract the moles of Al reacted from the initial moles of Al:

Moles of Al remaining = initial moles of Al - moles of Al reacted = 0.741 mol - 0.416 mol ≈ 0.325 mol

This result shows that 0.325 moles of aluminum remain unreacted after the formation of aluminum sulfide. This unreacted aluminum will also react with the potassium hydroxide solution in the subsequent step. Having calculated the moles of both aluminum sulfide and unreacted aluminum, we are now well-prepared to analyze their reactions with potassium hydroxide and determine the volume of gas evolved and the composition of the final solution.

5. Reaction with Potassium Hydroxide (KOH)

Now, both the aluminum sulfide (Al₂S₃) and the remaining aluminum (Al) react with the potassium hydroxide (KOH) solution. These reactions are crucial for determining the volume of gas evolved and the composition of the final solution. Let's first write the balanced chemical equations for these reactions:

  1. Reaction of aluminum sulfide with KOH:

    Al₂S₃ + 6KOH → 2K₃AlO₃ + 3H₂S

  2. Reaction of aluminum with KOH:

    2Al + 2KOH + 6H₂O → 2K[Al(OH)₄] + 3H₂

These equations reveal that the reaction of Al₂S₃ with KOH produces potassium thioaluminate (K₃AlO₃) and hydrogen sulfide (H₂S), while the reaction of Al with KOH produces potassium tetrahydridoaluminate(III) (K[Al(OH)₄]) and hydrogen gas (H₂). It is important to note that only the second reaction produces hydrogen gas, which is the gas we need to quantify for part (a) of the problem. To proceed, we need to determine the moles of KOH present in the solution and then use stoichiometry to calculate the moles of H₂ gas produced. This step requires careful consideration of the stoichiometry of both reactions and the limiting reactant, if any.

6. Calculating Moles of KOH and Hâ‚‚S

Before we can calculate the volume of gas evolved, we need to determine the number of moles of potassium hydroxide (KOH) present in the solution. We are given that the KOH solution has a mass of 1400 g and a concentration of 32%. This means that 32% of the solution's mass is due to KOH. Therefore, the mass of KOH in the solution is:

Mass of KOH = 0.32 * 1400 g = 448 g

Now, we can calculate the moles of KOH using its molar mass (56.11 g/mol):

Moles of KOH = mass of KOH / molar mass of KOH = 448 g / 56.11 g/mol ≈ 7.98 mol

Next, we need to consider the reaction of Al₂S₃ with KOH, which produces hydrogen sulfide (H₂S). From the balanced equation:

Al₂S₃ + 6KOH → 2K₃AlO₃ + 3H₂S

1 mole of Al₂S₃ produces 3 moles of H₂S. Therefore, the moles of H₂S produced can be calculated as:

Moles of H₂S = 3 * moles of Al₂S₃ = 3 * 0.208 mol ≈ 0.624 mol

This calculation is important because the Hâ‚‚S produced will influence the composition of the final solution. However, it does not directly contribute to the volume of gas evolved at STP, as the problem specifically asks for the volume of gas evolved, which refers to hydrogen gas (Hâ‚‚). The moles of KOH are crucial for understanding the extent of the reactions and will be used in subsequent calculations to determine the amount of Hâ‚‚ gas produced and the composition of the final solution.

7. Calculating Moles of Hâ‚‚ Gas Evolved

The evolution of gas in this reaction is due to the reaction between aluminum (Al) and potassium hydroxide (KOH). From the balanced equation:

2Al + 2KOH + 6H₂O → 2K[Al(OH)₄] + 3H₂

2 moles of Al react to produce 3 moles of Hâ‚‚. We previously calculated that 0.325 moles of Al remain after the reaction with sulfur. Therefore, the moles of Hâ‚‚ gas produced can be calculated as:

Moles of H₂ = (3/2) * moles of Al = (3/2) * 0.325 mol ≈ 0.488 mol

Now that we have the moles of Hâ‚‚ gas produced, we can calculate its volume at standard conditions (STP). At STP, 1 mole of any gas occupies 22.4 liters. This is a fundamental concept in stoichiometry and gas laws, allowing us to convert moles of a gas into its corresponding volume at a specified temperature and pressure. Using this molar volume, we can easily determine the volume of hydrogen gas evolved in the reaction, completing the first part of our problem.

8. Calculating Volume of Hâ‚‚ Gas at STP

To calculate the volume of hydrogen gas (Hâ‚‚) evolved at standard temperature and pressure (STP), we use the ideal gas law's molar volume at STP, which is 22.4 L/mol. We previously calculated that 0.488 moles of Hâ‚‚ gas are produced. Therefore, the volume of Hâ‚‚ gas at STP can be calculated as:

Volume of H₂ = moles of H₂ * molar volume at STP = 0.488 mol * 22.4 L/mol ≈ 10.93 L

This result provides the answer to the first part of our problem. The reaction of the remaining aluminum with potassium hydroxide produces approximately 10.93 liters of hydrogen gas at STP. Now, we move on to the second part of the problem, which involves determining the mass fractions of the substances in the final solution. This requires us to identify all the components present in the solution and calculate their respective masses. The components include the products of the reactions, such as potassium tetrahydridoaluminate(III) and potassium thioaluminate, as well as any remaining reactants or byproducts. A careful analysis of the reactions and their stoichiometry is essential for accurately determining the composition of the final solution.

9. Determining Substances in the Final Solution

To determine the mass fractions of substances in the final solution, we first need to identify all the substances present. From the reactions we've discussed, the final solution will contain:

  1. Potassium tetrahydridoaluminate(III) (K[Al(OH)â‚„]) formed from the reaction of Al with KOH.
  2. Potassium thioaluminate (K₃AlO₃) formed from the reaction of Al₂S₃ with KOH.
  3. Potassium sulfide (Kâ‚‚S) formed from the dissolution of Hâ‚‚S in KOH.
  4. Unreacted KOH, if any.
  5. Water (Hâ‚‚O), which is the solvent.

Now, we need to calculate the moles of each of these substances. We already know the moles of K[Al(OH)₄] produced (equal to the moles of Al reacted with KOH) and the moles of K₃AlO₃ (related to the moles of Al₂S₃ reacted). We also calculated the moles of H₂S produced, which will react with KOH to form K₂S. Determining the exact amount of unreacted KOH requires considering the stoichiometry of all reactions involving KOH. This step is crucial for accurately calculating the mass fractions of each substance in the final solution, as the amount of each component directly affects its mass fraction. A thorough understanding of the chemical reactions and their stoichiometry is essential for this analysis.

10. Calculating Moles of K[Al(OH)₄] and K₃AlO₃

We need to calculate the moles of potassium tetrahydridoaluminate(III) (K[Al(OH)₄]) and potassium thioaluminate (K₃AlO₃) in the final solution. From the reaction of Al with KOH:

2Al + 2KOH + 6H₂O → 2K[Al(OH)₄] + 3H₂

2 moles of Al produce 2 moles of K[Al(OH)â‚„]. Therefore, the moles of K[Al(OH)â‚„] are equal to the moles of Al that reacted with KOH, which we previously calculated as 0.325 mol:

Moles of K[Al(OH)â‚„] = 0.325 mol

From the reaction of Al₂S₃ with KOH:

Al₂S₃ + 6KOH → 2K₃AlO₃ + 3H₂S

1 mole of Al₂S₃ produces 2 moles of K₃AlO₃. Therefore, the moles of K₃AlO₃ can be calculated as:

Moles of K₃AlO₃ = 2 * moles of Al₂S₃ = 2 * 0.208 mol ≈ 0.416 mol

These calculations provide the number of moles of the two main aluminum-containing compounds in the final solution. We now need to consider the fate of the Hâ‚‚S produced and its reaction with KOH to form potassium sulfide, as well as determine the amount of unreacted KOH. These steps are essential for completing the analysis of the solution's composition and accurately calculating the mass fractions of all components.

11. Calculating Moles of Kâ‚‚S and Unreacted KOH

To determine the moles of potassium sulfide (K₂S) and unreacted potassium hydroxide (KOH), we need to consider the reaction between hydrogen sulfide (H₂S) and KOH. H₂S, produced from the reaction of Al₂S₃ with KOH, reacts with KOH as follows:

H₂S + 2KOH → K₂S + 2H₂O

1 mole of Hâ‚‚S reacts with 2 moles of KOH to produce 1 mole of Kâ‚‚S. We previously calculated that 0.624 moles of Hâ‚‚S are produced. Therefore, the moles of Kâ‚‚S formed are:

Moles of Kâ‚‚S = moles of Hâ‚‚S = 0.624 mol

Now, we need to calculate the moles of KOH consumed in the reactions. KOH is consumed in three reactions:

  1. Reaction with Al₂S₃: 6 moles of KOH react with 1 mole of Al₂S₃, so 6 * 0.208 mol = 1.248 mol of KOH are consumed.
  2. Reaction with Al: 2 moles of KOH react with 2 moles of Al, so 0.325 mol of KOH are consumed.
  3. Reaction with Hâ‚‚S: 2 moles of KOH react with 1 mole of Hâ‚‚S, so 2 * 0.624 mol = 1.248 mol of KOH are consumed.

The total moles of KOH consumed are:

Total moles of KOH consumed = 1.248 mol + 0.325 mol + 1.248 mol ≈ 2.821 mol

We initially had 7.98 moles of KOH, so the moles of unreacted KOH are:

Moles of unreacted KOH = initial moles of KOH - total moles of KOH consumed = 7.98 mol - 2.821 mol ≈ 5.159 mol

With the moles of all solutes determined, we can now calculate their masses and then the mass fractions in the final solution. This involves converting the moles of each substance into grams using their respective molar masses, and then dividing the mass of each substance by the total mass of the solution.

12. Calculating Masses of Solutes and Mass Fractions

Now that we have the moles of each solute in the final solution, we can calculate their masses using their respective molar masses:

  • Mass of K[Al(OH)â‚„] (molar mass = 158.18 g/mol) = 0.325 mol * 158.18 g/mol ≈ 51.41 g
  • Mass of K₃AlO₃ (molar mass = 188.29 g/mol) = 0.416 mol * 188.29 g/mol ≈ 78.33 g
  • Mass of Kâ‚‚S (molar mass = 110.27 g/mol) = 0.624 mol * 110.27 g/mol ≈ 68.81 g
  • Mass of unreacted KOH (molar mass = 56.11 g/mol) = 5.159 mol * 56.11 g/mol ≈ 289.47 g

The total mass of the solutes is:

Total mass of solutes = 51.41 g + 78.33 g + 68.81 g + 289.47 g ≈ 488.02 g

To calculate the mass of the final solution, we start with the initial mass of the KOH solution (1400 g), add the masses of Al and S (20 g + 20 g = 40 g), and subtract the mass of Hâ‚‚ gas evolved. The mass of Hâ‚‚ gas is:

Mass of H₂ = 0.488 mol * 2.016 g/mol ≈ 0.98 g

Therefore, the mass of the final solution is:

Mass of final solution = 1400 g + 40 g - 0.98 g ≈ 1439.02 g

Finally, we can calculate the mass fractions of each solute:

  • Mass fraction of K[Al(OH)â‚„] = 51.41 g / 1439.02 g ≈ 0.0357 or 3.57%
  • Mass fraction of K₃AlO₃ = 78.33 g / 1439.02 g ≈ 0.0544 or 5.44%
  • Mass fraction of Kâ‚‚S = 68.81 g / 1439.02 g ≈ 0.0478 or 4.78%
  • Mass fraction of unreacted KOH = 289.47 g / 1439.02 g ≈ 0.2012 or 20.12%

These mass fractions represent the composition of the final solution, completing the second part of our problem. This comprehensive step-by-step solution demonstrates the importance of understanding stoichiometry, solution chemistry, and gas laws in solving complex chemical problems.

Conclusion

In conclusion, we have successfully navigated the intricate reaction between a mixture of aluminum and sulfur with a potassium hydroxide solution. By meticulously applying stoichiometric principles and solution chemistry concepts, we calculated the volume of hydrogen gas evolved at standard conditions and determined the mass fractions of the substances present in the final solution. The calculated volume of hydrogen gas at STP is approximately 10.93 liters, and the mass fractions of the solutes in the final solution are approximately 3.57% for potassium tetrahydridoaluminate(III), 5.44% for potassium thioaluminate, 4.78% for potassium sulfide, and 20.12% for unreacted potassium hydroxide. This detailed analysis showcases the power of chemistry in understanding and quantifying complex reactions. Moreover, this exercise highlights the practical applications of chemical principles in various fields, emphasizing the significance of a solid foundation in chemistry for aspiring scientists and engineers. The ability to break down complex problems into manageable steps and apply relevant concepts is a crucial skill that this example effectively demonstrates.