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 L⋅atm⋅K^−1⋅mol^−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 L⋅atm⋅K^−1⋅mol^−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.