So I did an experiment recently with copper(II) chloride and aluminum. As they reacted with each other the redox reaction made the aluminum a brown substance that is supposed to be copper and the color of the copper (II) chloride solution was supposed to be colorless, indicating that the reaction has reacted ''evenly''. Even though I made all my calculations right and put the right amount of aluminum foil in the solution, the copper(II) chloride didn't become colorless. Could there be any suggestions as to why it didn't become as expected? Im doing a lab report on this and I can't seem to figure out any possible suggestions... The calculations: V(CuCl2) = 50 ml = 50 cm3 = 0.05 dm3

c(CuCl2) = 0.20 mol/dm3

m(CuCl2) = ? g

n(CuCl2) = V(CuCl2) x c(CuCl2) = 0.05 dm3 x 0.20 mol/dm3 = 0.01 mol M(CuCl2 • 2 H2O) = 170.48 g/mol

m(CuCl2) = n(CuCl2) x M(CuCl2 • 2 H2O) = 0.01 mol x 170,48 g/mol = 1.7048 ≈ 1.705 g

(The 1,705 g is the copper chloride that was used to put in the distilled water to make the solution)

The amount went into 50 ml of distilled water. For every experiment, I was supposed to only use 10-20 ml so I used 15 ml for every try.

CuIICl2 + Al0 ⇸AlIICl2 + Cu0

3CuIICl + 2Al → 2AlCl3 + 3 Cu0

V(CuCl2) = 15 ml

n(CuCl2)= 0.01 mol

m(Al) = ? g

n(CuCl2) = Al x ⅔ = 0.0066666667 mol

n(Al) = 0.0066666667 mol

m(Al) = n x M = 0,0066666667 x 27 = 0.18 g

I even showed my teacher but i have no idea what could have gone wrong :/

  • 2
    $\begingroup$ This may be helpful: chemistry.stackexchange.com/a/128099/79678. You should also show your calculations: a trivial math mistake is all it takes to ruin a calculation. $\endgroup$
    – Ed V
    Aug 12, 2022 at 15:16
  • $\begingroup$ I edited them. Is there something that is wrong with the calculations? $\endgroup$ Aug 12, 2022 at 15:31
  • $\begingroup$ Well one gram and 0,705 gram as well if that makes sense not 1,7 kg. $\endgroup$ Aug 12, 2022 at 15:36
  • 3
    $\begingroup$ Could there be unreacted Al hidden inside a shell of Cu? $\endgroup$ Aug 12, 2022 at 15:57
  • 1
    $\begingroup$ Two mistakes. First : It is impossible to have a solution of $\ce{CuCl2}$ which has a concentration of $\ce{20 mol/L}$ . Look ! $\ce{20 mol CuCl2}$ weighs $\pu{2668 g}$. This cannot be dissolved in 1 liter. Second mistake : $\pu{15 mL}$ $\ce{CuCl2 0.2 M}$ does not contains $0.01$ mol $\ce{CuCl2}$. $\endgroup$
    – Maurice
    Aug 12, 2022 at 19:19


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