Solving Kilowatt hours in electrolysis equation

I can't figure out how to do this problem

How many kilowatt-hours of electricity are used to produce 5.50 kg of magnesium in the electrolysis of molten MgCl2 with an applied emf of 5.00 V?

thanks will pick best answer

First let's see how many electrons we need. We produce 5500 grams of magnesium
with an average atomic weight of 24.3, so that is 5500/24.3, or 226.3 moles of Mg.
Each atom of Mg needs two electrons to reduce it to elementary magnesium so we
need 226.3 times 2, or 452.6 moles of electrons (also called faradays).
Each faraday contains Avogadro's number, or 6.02 * 10**23 electrons, so we need a
total of 452.6 times 6.02 * 10**23, or 2.724 * 10**26 electrons. Since one coulomb
contains 6.28 * 10**18 electrons, so we need (27.24 * 10**25) / (6.28 * 10**18) or
4.337 * 10**7 coulombs. If we have a constant emf of 5.0 volts, that is a total of
21.68 * 10**7 volt-coulombs, or joules of energy. (216,800,000 joules). Since 1 J
equals one watt-second and there are 3600 seconds in an hour, one kilowatt hour
equals 3,600,000 joules. In this example we use 216,800,000 joules so we use
216,800,000 divided by 3,600,000, or 60.22 kilowatt hours of energy.
Whew! Hope this answers your question

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