How many minutes will it take a current of 66.049 Ampères to deposit 86.825 grams of aluminum (26.981538 g/mol)?
Al has a charge of 3, so n=3.
MM=26.981538g/mol
1eq=molarmass/charge
so 26.981538/3=8.9938
1eq=8.9938
1F=96485c/eq
F=13138.29477
Am I on the right path? I don't know where to go from here...
26.98 g Al/3 = 8.994 g = mass deposited by 96,485 coulombs.
To deposit 88.825 g, we must have
?C = 96,485 x 88.825/8.994 = ??
Amperes x seconds = Coulombs.
You have C and you have A, solve for seconds and convert to minutes.
Yes, you are on the right path. You have correctly calculated the molar mass of aluminum (Al) and determined its charge to be 3 (n=3). You have also calculated the value of 1 Faraday (F) using the equation 1F = 96485 C/eq and found it to be approximately 13138.29477 C.
To calculate the time it takes for the given current to deposit a certain amount of aluminum, you can use Faraday's law of electrolysis, which states:
Amount of substance deposited (in grams) = (Current (in Amps) * Time (in seconds) * Molar mass (in g/mol)) / (Charge (in Faradays))
In this case, you are given the current (66.049 Amps) and the molar mass of aluminum (26.981538 g/mol). You want to find the time it takes to deposit 86.825 grams of aluminum.
Let's plug these values into the equation and solve for time:
86.825 = (66.049 * Time * 26.981538) / 13138.29477
To isolate the variable Time, we can rearrange the equation:
Time = (86.825 * 13138.29477) / (66.049 * 26.981538)
Calculating this expression will give you the time it takes to deposit 86.825 grams of aluminum.
I hope this helps!