Respuesta :
Explanation:
It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.
[tex]24 \times 3600 sec[/tex]
Hence, total charge passed daily is calculated as follows.
[tex]150,000 \times 24 \times 3600 sec[/tex]
And, number of Faraday of charge is as follows.
[tex]\frac{150,000 \times 24 \times 3600 sec}{96500}[/tex]
= 134300.52 F
The oxidation state of aluminium in [tex]Al_{2}O_{3}[/tex] is +3.
[tex]Al^{3+} + 3e^{-} \rightarrow Al(s)[/tex]
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.
[tex]\frac{134300.52 F}{3} \times \frac{89}{100}[/tex]
= 39842.487 mol
or, = [tex]3.9842 \times 10^{4} mol[/tex]
Molar mass of Al = 27 g/mol
Therefore, mass in gram will be calculated as follows.
Mass in grams = [tex]3.9842 \times 10^{4} mol \times 27[/tex]
= [tex]107.57 \times 10^{4} g[/tex]
= 1075.7 kg/day
Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.
