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

Explanation:

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

Hence, total charge passed daily is calculated as follows.

And, number of Faraday of charge is as follows.

                     = 134300.52 F

The oxidation state of aluminium in is +3.

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.

                   = 39842.487 mol

or,               =

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

            Mass in grams =

                                     =

                                     = 1075.7 kg/day

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

where's the diagram?

b.

the accuracy of the data.