Question

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today. In the first step, aluminum hydroxide reacts to form alumina and water: 2Al(OH)3(s) →Al2O3(s)+3H2OIn the second step, alumina and carbon react to form aluminum and carbon dioxide: Al2O3(s)+3C(s)→4Al(s)+3CO2(g) Suppose the yield of the first step is and the yield of the second step is. Calculate the mass of aluminum hydroxide required to make of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to significant digits.

Answer 1

Answer:

10 kg Al(OH)₃

Explanation:

There is some info missing. I think this is the original question.

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th-century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today.

In the first step, aluminum hydroxide reacts to form alumina (Al₂O₃) and water: 2 Al(OH)₃(s) → Al₂O₃(s) + 3H₂O(g). In the second step, alumina (Al₂O₃ and carbon react to form aluminum and carbon dioxide: 2Al₂O₃(s)+3C(s)→4Al(s)+3CO₂(g). Suppose the yield of the first step is 63% and the yield of the second step is 89%.

Calculate the mass of aluminum hydroxide required to make 2.0 kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.

Let's consider the 2 steps in the synthesis of Al.

Step 1: 2 Al(OH)₃(s) → Al₂O₃(s) + 3 H₂O(g)

Step 2: 2 Al₂O₃(s) + 3 C(s) → 4 Al(s) + 3 CO₂(g)

In Step 2, the percent yield of Al is 89% and the real yield is 2.0 kg. The theoretical yield is:

2.0 kg (R) × (100 kg (T) / 89 kg (R)) = 2.2 kg = 2.2 × 10³ g

In Step 2, the mass of Al is 4 × 26.98 g = 107.9 g and the mass of Al₂O₃ is 2 × 101.96 g = 203.92g. The mass of Al₂O₃ that produced 2.2 × 10³ g of Al is:

2.2 × 10³ g Al × (203.92g Al₂O₃ / 107.9 g Al) = 4.2 × 10³ g Al₂O₃

In Step 1, the percent yield of Al₂O₃ is 63% and the real yield is 4.2 × 10³ g. The theoretical yield is:

4.2 × 10³ g (R) × (100 g (T)/ 63 g (R)) = 6.7 × 10³ g

In Step 1, the mass of Al₂O₃ is 101.96 g and the mass of Al(OH)₃ is 2 × 78.00 g = 156.0 g. The mass of Al(OH)₃ that produced 6.7 × 10³ g of Al₂O₃ is:

6.7 × 10³ g Al₂O₃ × (156.0 g Al(OH)₃ / 101.96 g Al₂O₃) = 1.0 × 10⁴ g Al(OH)₃ = 10 kg Al(OH)₃