In an industrial plant, aluminium is produced by elecrolysis of alumina dissolved in cryolite. This takes a current of 20000A. If the current efficiency is 90%, how much Al will be produced per day?

To determine how much aluminum (Al) can be produced per day in an industrial plant using electrolysis, we need to consider several factors, including the current, the efficiency of the process, and the chemical reactions involved. Let's break it down step by step.

Understanding the Electrolysis Process

In the electrolysis of alumina (Al₂O₃) dissolved in cryolite (Na₃AlF₆), aluminum ions are reduced at the cathode, while oxygen ions are oxidized at the anode. The overall reaction can be simplified to:

• 2 Al₂O₃ (s) → 4 Al (l) + 3 O₂ (g)

This means that from 2 moles of alumina, we can produce 4 moles of aluminum. The molar mass of aluminum is approximately 27 g/mol.

Calculating the Theoretical Yield

First, we need to calculate the total charge (Q) that flows through the system in one day. The formula for charge is:

• Q = I × t

Where:

• I = current in amperes (A)
• t = time in seconds (s)

Given that the current (I) is 20,000 A and the time (t) in one day is 24 hours, we convert hours to seconds:

• t = 24 hours × 60 minutes/hour × 60 seconds/minute = 86,400 seconds

Now, substituting the values into the formula:

• Q = 20,000 A × 86,400 s = 1,728,000,000 C (coulombs)

Relating Charge to Aluminum Production

Next, we need to relate this charge to the amount of aluminum produced. The electrochemical equivalent of aluminum can be calculated using Faraday's laws of electrolysis. The number of moles of aluminum produced can be calculated using the formula:

• n = Q / (n × F)

Where:

• n = number of moles of aluminum produced
• F = Faraday's constant (approximately 96,485 C/mol)
• n = number of electrons transferred per mole of aluminum (3 for Al, since Al³⁺ + 3e⁻ → Al)

Substituting the values:

• n = 1,728,000,000 C / (3 × 96,485 C/mol) ≈ 5,9733.5 moles of Al

Calculating the Mass of Aluminum

Now, we can convert the number of moles of aluminum to grams:

• mass = n × molar mass

Substituting the values:

• mass = 5,9733.5 moles × 27 g/mol ≈ 161,200 g

Considering Current Efficiency

Since the current efficiency is 90%, we need to adjust the mass produced accordingly:

• Actual mass = theoretical mass × efficiency

Calculating the actual mass:

• Actual mass = 161,200 g × 0.90 ≈ 145,080 g

Final Conversion to Kilograms

Finally, converting grams to kilograms for a more practical measure:

• 145,080 g = 145.08 kg

Therefore, under the given conditions, approximately 145.08 kg of aluminum will be produced per day in the industrial plant.