An electrolytic cell produces aluminum from Al2O3 at the rate of ten kilograms a day.?

An electrolytic cell produces aluminum from Al2O3 at the rate of ten kilograms a dayAssuming a yield of 100%a) how many moles of electrons must pass through the cell in one day?c) how many moles of oxygen (O2) are being produced simultaneously? I understand that for part a, you start with 5kg, and that you use dimensional analysis, but what I'm stuck on is the stoichometry for the electrons (3 for Al, 4 for O, 7 for both, or 12 for canceling out?) and if you go from kg to g (molar mass) of Al2O3.Part c is a mystery to me at the moment.

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Dimensional Analysis? This is a question of ElectrochemistryFrom Faraday's first law in this domain, Mass of Aluminium produced Equivelant mass of source( Al2O3 ) / Charge on a mole of electrons x Amount of electricity consumed( quantity of charge )Eqmass of Al2O3 Molecular mass / noof electrons transferred during all redox reactions involved per mole of Al2O3 Al2O3 - 2Al(3+) + 3O(2-)So, 6 electrons have been transferredSo, mass of Al 10000 grams 102/(6 grams x 96500 Coulomb ) x Q CoulombSo, Q comes out to be about 56765160 Using Faraday's first law again, calculate the amount of O2 produced yourself

nochi is much better :)

i don't like conairi buy my Straighteners and curlers at sephora.

i bought a conair curlerstraightener (2 in 1) at walmart :) mine works really well and curls hair really nicely.D My friend has the same and she really likes too :) i recommend you to buy a conair one.

Dimensional Analysis? This is a question of ElectrochemistryFrom Faraday's first law in this domain, Mass of Aluminium produced Equivelant mass of source( Al2O3 ) / Charge on a mole of electrons x Amount of electricity consumed( quantity of charge )Eqmass of Al2O3 Molecular mass / noof electrons transferred during all redox reactions involved per mole of Al2O3 Al2O3 - 2Al(3+) + 3O(2-)So, 6 electrons have been transferredSo, mass of Al 10000 grams 102/(6 grams x 96500 Coulomb ) x Q CoulombSo, Q comes out to be about 56765160 Using Faraday's first law again, calculate the amount of O2 produced yourself