### Question about aluminium electrolysis (little complex)

Posted:

**Fri Jul 10, 2020 1:39 pm**First of all, I'm not a native english speaker, some terms maybe a little off

Basically, my professor asked how much liters of oxigen are produced based on an aluminium electrolysis in an industrial process, this is the equation:

This electrolysis was done at 900Celsius, using graphite electrodes, the anodic atracts the oxigen and form that is discarted, the catodic atracts pure liquid Aluminium that is trasnferred to another place, here is an image about it:

https://imgur.com/a/rz8gDwd (unfortunately I don't have it in english, it's in portuguese)

But before the oxigen calculation, he asked how much mass of is produced based on a 8 hours production using 135000amp, we rounded faraday's constant to 96500. The calculation went like this:

Each 4 moles of need 12 moles of electrons;

m= 135000 x 28800 x 27 / 12 x 96500

m= ~1.04x1011 / 1158000

m= ~90652.8 grams of Aluminium = 90.6 Kilos of Aluminium

After we done this, he asked how many liters of oxigen are produced in this case at 2.5atm using the universal gas constant (0.082 Lâ‹…atmâ‹…K^âˆ’1â‹…mol^âˆ’1), but he also asked to do the calculation based on a 90.6 Kilos of aluminium production. My first thought was "well i'll just convert 90.6 kilos to moles, balance it and i'll have the results" I did it, but he said it's wrong, i don't know if I'm missing something, here's how I did:

First, converting to moles:

90600 / 27 = ~3355.5 Al moles

Balancing:
Calculation:

P x V = n x R x T

2.5 x V = 2516.625 x 0.082 x (900+273,15)

2.5V = ~242095.04

V = ~242095.04 / 2.5

V= ~96838.01 Liters of oxigen

Me and the boys did a debate and couldn't conclude if we're really wrong or the professor is. Asking for help on this question because he said the results and they are VERY far to mine and I don't even can thing how to reach them.

Basically, my professor asked how much liters of oxigen are produced based on an aluminium electrolysis in an industrial process, this is the equation:

Code: Select all

`2Al2O3 = 4Al + 3O2`

Code: Select all

`CO2`

https://imgur.com/a/rz8gDwd (unfortunately I don't have it in english, it's in portuguese)

But before the oxigen calculation, he asked how much mass of

Code: Select all

`Al`

Each 4 moles of

Code: Select all

`Al3+`

m= 135000 x 28800 x 27 / 12 x 96500

m= ~1.04x1011 / 1158000

m= ~90652.8 grams of Aluminium = 90.6 Kilos of Aluminium

After we done this, he asked how many liters of oxigen are produced in this case at 2.5atm using the universal gas constant (0.082 Lâ‹…atmâ‹…K^âˆ’1â‹…mol^âˆ’1), but he also asked to do the calculation based on a 90.6 Kilos of aluminium production. My first thought was "well i'll just convert 90.6 kilos to moles, balance it and i'll have the results" I did it, but he said it's wrong, i don't know if I'm missing something, here's how I did:

First, converting to moles:

90600 / 27 = ~3355.5 Al moles

Balancing:

Code: Select all

` 1677.75 Al2O3 = 3355.5 Al + 2516.625 O2`

P x V = n x R x T

2.5 x V = 2516.625 x 0.082 x (900+273,15)

2.5V = ~242095.04

V = ~242095.04 / 2.5

V= ~96838.01 Liters of oxigen

Me and the boys did a debate and couldn't conclude if we're really wrong or the professor is. Asking for help on this question because he said the results and they are VERY far to mine and I don't even can thing how to reach them.