Act 4B: Moles of Iron and Copper-Corrections
Purpose: To find the number of moles of copper produced by a reaction
of copper(II) chloride dihydrate and iron
Pre-Lab Questions
1) n = xg =34.0 g X mol
= 0.609 mol Fe
MM
55.8 g
2) # atoms = 2 mol Fe X 6.02 X 1023 atoms = 1.2 X 1024 atoms
mol Fe
3) MM(CuCl2•2H2O) = 170.5 g/mol
4) To remove impurities
5) Decant: Pouring out liquid while leaving the solid in the container.
Data: 2 decimal places!!!
2Fe + 3CuCl2•2H2O  2FeCl3 + 3Cu
+ 6H2O
2 mol 3 mol
2 mol 3 mol
6 mol
Calculations
1)
(a) (grams) (Fe)
(b) (grams)
(c) (grams) (Cu)
nails(before-after)
(beaker + CuCl2)-(beaker)
(beaker + Cu)-(beaker)
0.34
1.00
0.28
a. Mass of iron
0.34 g
b. Mass of copper(II)chloride dihydrate 1.00 g
c. Mass of copper
0.28 g
2a) n(Fe)used = 0.34 g X mol =
0.0061 mol
55.8 g
n(Cu)produced = 0.28 g X mol = 0.0044 mol
63.5 g
3a) atoms(Fe)used=0.0061 mol X 6.02 X 1023 atoms Fe = 3.7 X 1021 atoms Fe
mol
atoms(Cu)produced= 0.0044 mol X 6.02 X 1023 atoms Cu = 2.7 X 1021 atoms Cu
mole ratio
mol
4) n(Cu) = 3 mol Cu = 1.5 (In theory)Theoretical Yield
n(Fe) 2 mol Fe
n(Cu) = 0.0044 mol Cu = 0.72 (Actual Yield)
n(Fe) 0.0061 mol Fe
(Percent Yield: obtained
)
0.72 X100 = 48 %
calculated
1.5
5) If the beaker had a green/turquoise residue, some CuCl2•2H2O is present.
Synthesis Problems
=
2Fe + 3CuCl2•2H2O  2FeCl3 + 3Cu
+ 6H2O
2 mol 3 mol
2 mol 3 mol
6 mol
34.0 g
? mol
55.8 g/mol
Have MM
Mole Bridge Want
Road Map: xg(Fe)--- n(Fe) ------------- n(Cu)
Want
n(Cu) = 34.0 g X mol Fe X 3 mol Cu = 0.914 mol Cu
55.8 g 2 mol Fe
1)Eqn
Ratio
Data
MM
2)
2Fe + 3CuCl2•2H2O  2FeCl3 + 3Cu
+ 6H2O
Ratio 2 mol 3 mol
2 mol 3 mol
6 mol
Data ? mol
45.0 g
MM
63.5 g/mol
Road Map: Have MM
Mole Bridge Want
xg(Cu) ------ n(Cu) -------------- n(Fe)
Want
n(Fe) = 45.0 g X mol Cu X 2 mol Fe = 0.472 mol Fe
63.5 g 3 mol Cu
3) Road Map: Have MM
X A.N. Want
xg(Cu) ------ n(Cu) -------- # atoms Cu
45.0 g X mol Cu X 6.02 X 1023 atoms = 4.27 X 1023 atoms Cu
63.5 g
mol
4) Road Map:
Have X A.N. Want
n(Fe) ------- # atoms Fe
0.472 mol Fe X 6.02 X 1023 atoms = 2.84 X 1023 atoms Fe
mol
5)
2Fe + 3CuCl2•2H2O  2FeCl3 + 3Cu
+ 6H2O
Ratio 2 mol 3 mol
2 mol 3 mol
6 mol
Data 456 g
?g
Have MM
Mole Bridge
X MM
Road Map: xg(Fe) ----- n(Fe) ------------ n(Cu) ------ xg(Cu)
n(Cu) = 456 g X mol Fe X 3 mol Cu X 63.5 g = 778 g Cu
55.8 g 2 mol Fe mol Cu
Conclusion:
Did we meet the purpose?
Yes, we did produce pure copper that was weighed.
However, we noticed that the beaker still contained some copper(II)chloride
dihydrate residue and some orange residue of
iron(III) choride. These residues increased the beaker mass.