How much electricity in terms of Faraday is required to produce `40.0 g` of `Al` from molter `Al_(2)O_(3)`?

How much electricity in terms of Faraday is required to produce `40.0

1.	How much electricity in terms of Faraday is required to produce `40.0 g` of `Al` from molter `Al_(2)O_(3)`?
Answer» Eq. of `Al = (i.t)/(96500)` `(6e+Al_(2)^(3+) rarr 2Al)` `:.` Eq. `wt.` of `Al = (27)/(6//2)` or `(40)/(27//3) = (i.t)/(96500)` or i.t `= (120)/(27) xx 96500 = 4.44 F`

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How much electricity in terms of Faraday is required to produce `40.0 g` of `Al` from molter `Al_(2)O_(3)`?

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