Moles of iron and copper
Observations and Questions listed below.
The nails become coated with a copper coloured substance as the reaction progresses. The solution changes from bright blue to a lighter blue colour.
Mass of the empty dry beaker = 105.52 g
Initial mass of the beaker + CuCl2 (s) = 113.02 g
Initial mass of the two iron nails = 3.82 g
Final mass of the dry beaker and copper = 106.41 g
Final mass of the two iron nails = 3.04 g
1.What mass of CuCl2 (s) is used?
2.What mass of iron is used in the reaction?
3.How many moles of iron are used?
4.What mass of Cu is produced in the reaction?
5.How many moles of Cu are produced?
6.Find the ratio of the moles of copper produced to moles of iron used.
7.Determine the number of atoms of iron reacting and the number of atoms of copper produced.
8.The blue colour of the solution gets lighter as the reaction proceeds. What does this indicate? Explain your answer.
9.Write a balanced equation for the reaction. This is a single replacement reaction.
10.How many moles, how many atoms, and what mass of Cu (s) are produced by the reaction when 100. grams of iron completely reacts, assuming there is enough CuCl2 (aq)?
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