Answer:
There is 29.8 kilowatt hours needed.
Explanation:
Step 1: Data given
Mass of Magnesium = 3.00 kg
Molar mass of magnesium = 24.31 g/mol
Applied emf = 4.50 V (= 4.50 J/C)
Step 2: Calculate moles of Magnesium
Moles = Mass Mg / Molar mass Mg
Moles Mg = 3000 grams / 24.31 g/mol
Moles Mg = 123.4 moles
Step 3: Calculate how many electrons are needed to produce the magnesium.
The ionic equation for the reduction of Mg^2+ :
Mg^2+ + 2e^- → Mg
Every mole of Mg requires 2 mol of electrons.
For 123.4 mol of Mg, we have 246.8 mol of electrons.
Step 3: Find how many coulombs are involved.
The Faraday constant = 96500 couloumbs
1 mole of electrons is 96500 coulombs.
246.8 mol of electrons need 2.38 *10^7 Coulombs
Step 4: Calculate kilowatt-hours of electricity needed
2.38 * 10^7 C * 4.5 J/C = 10.7 * 10^7 J
10.7 * 10^7 J * ( 1 kW-h-/ 3.6*10^6 J ) = 29.8 kWh
There is 29.8 kilowatt hours needed.
