Moles of iron and copperObservations and Questions listed below.ObservationsThe nails become coated with a copper coloured substance as the...

The best way to approach this problem is to start by
writing a balanced equation to show the mole ratios of reactants and
products.

CuCl2  +  Fe -->  Cu  + 
FeCl2

But not all the reactants were used up, so what you
really have is:

CuCl2  + Fe  -->  Cu  +  FeCl2  + 
Fe  +  CuCl2

Now use the information given to determine
amounts of various materials.

Beaker + CuCl2 - beaker = 7.5
g of CuCl2 at beginning.

Nails at start - nails at end
means 3.82 - 3.04 of .78 g of iron were used up to form
FeCl2.

beaker + Cu - beaker = .89 g of Cu were
formed.

Convert known masses to
moles:

CuCl2:  7.5g/134.452 g/mole = .05578 moles at
start

Fe:  3.82/55.847g/mole = .0684 moles at
start

Fe: 3.04/55.847 g/mole = .0544 moles at
end

Fe used to react with Cu = .0684 - .0544 = .014
moles

Cu:  .89g/63.546 g/mole = .014 moles
produced

From the balanced equation, you know that one mole
of FeCl2 is produced for every mole of Fe consumed.

So if
.014 mole of Fe used, .014 moles FeCl2 made which equals .014 moles * 126.753 g/mole =
1.776 g of FeCl2

Since .014 moles of Fe were used up, that
means there were also .014 moles of CuCl2 used.  .05578 - .014 = .04178 moles of CuCl2
left.  .04178 * 134.452 = 5.617 g ofCuCl2 left.

So rewrite
equation adding in moles of each:

.05578 CuCl2 + .0684 Fe 
-->  .014 Cu  +  .014 FeCl2  +  .0544 Fe  + .042
CuCl2.

number of atoms is # of moles * 6.023X10^23
atoms/mole.

For both Cu & Fe:  .014 * 6.023 x 10^23
= 8.43 x 10^21 atoms.

Color gets lighter because CuCl2 is
being used up.

If you start with 100 g Fe, that is
100/55.847 g/mole = 1.79 moles.

You will need 1.79 moles of
CuCl2 = 1.79 * 134.452 = 240.75 g = 1.79 * 6.023 x 10^23 = 1.08 x 10^24 atoms of CuCl2,
Fe, & Cu