I need help with determing the mass of a reaction containing Cu and Cl2 with iron nails Given the information: mass of the empty dry beaker:...

Balanced equation:

CuCl2  +
Fe  -->  FeCl2  +  Cu  + Fe

For every mole of Fe
used, one mole of CuCl2 is also used.

1.  Since
from #3 there were 0.014 moles of Fe used, then there were 0.014 moles of CuCl2 * 134.45
g/mol = 1.88 g of CuCl2 used.

2.  mass of iron.  3.82 g -
3.04 g = .78 g of Fe used.

3. Moles of iron used: 
.78/55.85 = 0.014 moles of Fe

4.  Mass of Cu produced: 
106.41-105.52 = 0.89 g of Cu

5. moles of Cu:  0.89/63.55 =
0.014 moles of Cu

6. mole ratio:  0.014: 0.014 = 1:1 ratio
as expected from the balanced chemical equation.

7.# of
atoms of Fe &Cu.   0.014 moles * 6.023x10^23 atoms/mole = 8.43 x 10^21 atoms of
each.

8. Blue color fades because the CuCl2 concentration
is decreasing.

9. Balanced equation is at
top.

10. moles, mass, atoms if use 100 g of Fe.  100/55.85
= 1.79 moles of Fe used.  So produce 1.79 moles of Cu, which is 1.79*63.55 = 113.75 g of
Cu.  1.79 * 6.023x 10^23 = 1.078 x 10^24 atoms of Cu.