This is example one from the notes about
electroplating.  So, the question is... "What
mass of gold could be electroplated from
a solution of gold (III) cyanide using a
current of 30.0A for five
minutes so that should have been 5.00
minutes.  Now... all we need to do is
understand the half reaction. It does not
matter that the compound is gold (III)
cyanide, it only matters that the ion
is the Au3+ ion, and during
electroplating the ion gets reduced out
of solution, so this would require three
electrons to balance that half reaction,
and will produce Au(s).  So... it's always
the ion, plus however many electrons are
needed to neutralize it, going to the
neutral metal.  So... that's what we need,  is
that half reaction.  So... the current from
the equation is 30A, the time is
5.00 minutes, but time in
minutes is not what we need to go into
Faraday's Law for electroplating.  We need time in seconds, so we'll convert that to
seconds immediately.  Number of moles in Faraday's equation it may be labeled
n sub(script) e,  it's the number of moles of electrons in this reduction
half-reaction, and it's just an integer
value.  It's an exact number.  Faraday's
constant from your data booklet, 96485
coulombs/mole,  and mass is our
unknown.  So,  the equation m = (I t M) / (n F)
m = (I t M) / (n F), so it's an equation in six parts,
with a constant, and four of the other five
quantities (known)... you have to
figure out the remainder. So... the
important thing with a Faraday's law (calculation) is
knowing what to do with it when you copy
it from your formula sheet.  I is pretty
straight forward... it's the current in Amps.  If you don't change time to seconds
you're not going to get this right,  because
you remember the unit an ampere is a
coulomb per second, so we need time in
seconds.  The next thing you have to know
is molar mass of what???  It's the molar
mass of gold...  gold and only gold...  it does
not matter what compound it comes from.  It's only gold ion that gets converted.
It's the molar mass of gold that goes in
here.
(The) number of moles of electrons is 3... that's an exact number, and (Coulomb's...  oops!)  Faraday's
Constant 96485 coulombs per mole.  So... in this
calculation, the amp, which is a Coulomb
per second, allows me to cancel seconds
and coulombs from the denominator. 
 Moles will cancel, and I'm left with units of grams.
Your answer 6.13 g,  3 Sig. Figs.
