How many kilowatt-hours of electricity are used to produce 3.00 kg of magnesium in theelectrolysis of molten mgcl2 with an applied emf of 4.50 v?
a) 0.0336
b) 0.0298
c) 7.4
d) 29.8
e) 14.9

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.

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