Final answer:
Approximately 224.448 grams of calcium metal could be produced by the electrolysis of molten CaBr2 using a current of 30.0 amps for 10.0 hours, following Faraday's laws of electrolysis and using calculations involving charge, moles of electrons, and molar mass.
Step-by-step explanation:
To determine how many grams of Ca metal could be produced by the electrolysis of molten CaBr2 using a current of 30.0 amps for 10.0 hours, you need to use Faraday's laws of electrolysis. First, you calculate the total charge (Q) passed through the electrolyte using the formula Q = It, where I is the current in amps and t is the time in seconds.
Q = 30.0 A × 10.0 h × 3600 s/h = 1080000 C
The number of moles of electrons can be found by dividing the total charge by Faraday's constant (F), which is approximately 96485 C/mol.
moles of electrons = Q / F = 1080000 C / 96485 C/mol ≈ 11.20 mol
Since the reaction for producing Ca from CaBr2 involves the reduction of Ca2+ ions, 2 moles of electrons are needed to produce 1 mole of Ca. Thus, we need to divide the number of moles of electrons by 2 to get the moles of Ca produced.
moles of Ca = moles of electrons / 2 ≈ 11.20 mol / 2 = 5.60 mol
Finally, we calculate the mass of Ca produced using the molar mass of Ca, which is approximately 40.08 g/mol.
mass of Ca = moles of Ca × molar mass of Ca = 5.60 mol × 40.08 g/mol = 224.448 g
Therefore, approximately 224.448 grams of Ca metal could be produced by the electrolysis of molten CaBr2 under the given conditions.