How many amperes would be needed to produce 60 g of magnesium during the electrolysis of molten MgCl2 in 2 hours?
first, you need two electrons per atom.
atoms Mg= moleMg*AvagNumber= figure that out for 60 gMg
Then, each electron has this charge: 1.6E-19 Coulombs= e
amperes*time= number coulmbs
= 2*e*atoms electrons
amperes= above/120sec
Answer
Where
To determine the number of amperes required for the electrolysis of molten MgCl2, we need to use Faraday's laws of electrolysis.
The first step is to find the number of moles of magnesium (Mg) that will be produced. We know that the molar mass of magnesium is 24.31 g/mol. Therefore,
Number of moles of magnesium = Mass of magnesium / Molar mass of magnesium
= 60 g / 24.31 g/mol
= 2.47 mol
According to Faraday's first law, the amount of substance (in moles) produced during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The proportionality constant is called the Faraday constant, F, which is equal to 96,485 Coulombs/mol.
Next, we need to calculate the total charge (in Coulombs) required to produce 2.47 moles of Mg. We can use the equation:
Charge (Coulombs) = Number of moles of Mg x Faraday constant
= 2.47 mol x 96,485 C/mol
= 238,308 Coulombs
Finally, to determine the number of amperes (A) needed, we divide the total charge by the time (in seconds):
Amperes (A) = Charge (Coulombs) / Time (Seconds)
= 238,308 C / (2 hours x 3600 seconds/hour)
= 33.14 Amperes
Therefore, approximately 33.14 Amperes would be needed to produce 60 g of magnesium during the electrolysis of molten MgCl2 in 2 hours.