2 Al2O3 = 4 Al + 3O2
In the Hall-Héroult electrolytic cell, the time required for the cell to operate at 5550 A to produce 20.0 kg of aluminium is __________h.
First, we need to calculate the number of moles of aluminum produced using the given mass of aluminum.
The molar mass of aluminum (Al) is 26.98 g/mol.
Number of moles of Al = Mass of Al / Molar mass of Al
Number of moles of Al = 20000 g / 26.98 g/mol
Number of moles of Al = 740.125 mol
From the balanced equation, we know that 2 moles of Al2O3 produce 4 moles of Al. Therefore, the number of moles of Al2O3 consumed can be calculated as follows:
Number of moles of Al2O3 = (Number of moles of Al / 4) * 2
Number of moles of Al2O3 = (740.125 mol / 4) * 2
Number of moles of Al2O3 = 370.0625 mol
Now, let's calculate the number of moles of O2 produced. From the balanced equation, we know that 2 moles of Al2O3 produce 3 moles of O2. Therefore, the number of moles of O2 produced can be calculated as follows:
Number of moles of O2 = (Number of moles of Al2O3 / 2) * 3
Number of moles of O2 = (370.0625 mol / 2) * 3
Number of moles of O2 = 555.09375 mol
Next, we need to calculate the total charge required to produce these moles of aluminum, using Faraday's law:
Q = (Number of moles of Al * Faraday's Constant) + (Number of moles of O2 * Faraday's Constant)
Q = (740.125 mol * 3 * 96500 C/mol) + (555.09375 mol * 2 * 96500 C/mol)
Q = 215426250 C + 107593359375 C
Q = 107808785625 C
Now, let's calculate the time required using the given current (5550 A):
T = Q / I
T = 107808785625 C / 5550 A
T ≈ 19442856.41 s
Finally, let's convert the time from seconds to hours:
T = 19442856.41 s * 1 hr / 3600 s
T ≈ 5406.35 hr
Therefore, the time required for the cell to operate at 5550 A to produce 20.0 kg of aluminum is approximately 5406.35 hours.