ELECTRON TRANSFER REACTIONS DATA Part A NAgen 70.243 a. Wt. of 150-ml beaker and
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ELECTRON TRANSFER REACTIONS DATA Part A NAgen 70.243 a. Wt. of 150-ml beaker and copper(II) chloride dihydrate b. Wt., of 150-ml beaker c. Wt. of copper (II) chloride dihydrate (a b) d. Wt of aluminum foil e. In the space that follows, write a brief description of everything you observed when you added the aluminum to your copper(II)chloride solution. ciclale ci to the w Ces t an to fi 22 and bubble Turning the iuminum to darvi rect brown color. TH began to d is inter cute the aluminum, Smouse as rise as well and the color o the m ik Hure is grayand what do you observe in your tests for copper ions and aluminum ions in the initial filtrate obtained from the collection of your copper metal. Aluminum ion test: ci r-e ci and Clark e Copper ion test Clear to whi Per cipitate time Rer Cip He an Clear 2nd Com Heid time clear liquid Wt of copper and vial G wt. of empty vial Owt of copper recovered (g h)Explanation / Answer
Given-
Wt. of beaker = 73.75g
Wt. of beaker + copper (II) chloride hydrate = 76.24 g
Wt. of copper (II) chloride hydrate = 76.24g -73.75g = 2.49 g
Wt. of aluminum foil = 0.42g
a) The balanced molecular equation for reaction-
3 CuCl2.2H2O (aq) + 2 Al (s) ---> 3 Cu (s) + 2 AlCl3 (aq) + 6 H2O (l)
b) Number of grams of Al necessary to react with entire sample of copper (II) chloride hydrate-
Molar mass of CuCl2 ·2H2O = 63.546 + 35.453*2 + 2*(1.00794*2 + 15.9994) = 170.48256 g/mol
Molar mass of copper (II) chloride hydrate = 170.48 g/mol
Molar mass of Al = 26.981g/mol
Step -1 Calculate the moles of copper (II) chloride hydrate
Moles of copper (II) chloride hydrate = Wt. of copper (II) chloride hydrate/ Molar mass of copper (II) chloride hydrate
Moles of copper (II) chloride hydrate =2.49 g/170.48 g/mol = 0.0146 moles
Step -2 calculate the moles Al
Since the stoichiometric ratio between Cu and Al is 2/3
So, multiply 0.0146 moles by 2/3 to find the moles of Al
Moles of Al = 0.0146 moles x 2/3 = 0.0097 moles
Step -3 calculate the g of Al
Number of g of Al = Moles of Al x Molar mass of Al
Number of g of Al = 0.0097 moles x 26.981g/mol = 0.262 g
Number of grams of Al necessary to react with entire sample of copper (II) chloride hydrate = 0.26 g
c) Only 0.26 g of Al from aluminum foil (0.42g) react with entire sample of copper (II) chloride hydrate, so copper (II) chloride hydrate is a limiting reagent. The limiting reagent is the one that is totally consumed. In this reaction copper (II) chloride hydrate is totally consumed.
d) Al is excess and copper (II) chloride hydrate is limiting, so the blue solution, caused by Cu2+(aq), becoming colorless. Unreacted Al (silver grey) can be seen. The excess Al has first to be removed before the Cu can be determined. We can remove Al by adding hydrochloric acid.
2Al(s) + 6HCl --------> 2AlCl3 (aq) + 3H2 (g).
This soluble salt will be part of the filtrate with the excess acid and copper will be collected in the filter paper.
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