Question

Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell reaction is 2Al2O3 + 3C → 4A1 + 3CO2

Answer 1

The daily aluminum production of a 150,000 A aluminum cell is approximately 8,640,000,000 mol.

Step-by-step explanation:

To calculate the daily aluminum production, we need to determine the number of moles of aluminum produced per second and then convert it to daily production. First, let's calculate the number of moles of aluminum produced per second:

2 moles of Al = 4 moles of Al2O3 (according to the balanced equation)

So, 1 mole of Al = 2 moles of Al2O3

Now, let's calculate the number of moles of Al2O3 consumed per second:

150,000 A × 1 mol Al2O3/3 A × 2 mol Al/4 mol Al2O3 = 100,000 mol Al/s

Finally, let's convert it to daily production:

100,000 mol Al/s × 60 s/min × 60 min/h × 24 h/day = 8,640,000,000 mol Al/day

Therefore, the daily aluminum production of a 150,000 A aluminum cell is approximately 8,640,000,000 mol.

Answer 2

Step-by-step explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

`24 * 3600 sec`

Hence, total charge passed daily is calculated as follows.

`150,000 * 24 * 3600 sec`

And, number of Faraday of charge is as follows.

`(150,000 * 24 * 3600 sec)/(96500)`

= 134300.52 F

The oxidation state of aluminium in
`Al_(2)O_(3)` is +3.

`Al^(3+) + 3e^(-) \rightarrow Al(s)`

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

`(134300.52 F)/(3) * (89)/(100)`

= 39842.487 mol

or, =
`3.9842 * 10^(4) mol`

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

Mass in grams =
`3.9842 * 10^(4) mol * 27`

=
`107.57 * 10^(4) g`

= 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.