CIEM 1211 Lab Manual-Revised 95/2017 The Copper Lab Name Instructor Section Part

Question

CIEM 1211 Lab Manual-Revised 95/2017 The Copper Lab Name Instructor Section Partner Experimental Data (3 pts, for blanks in experimental section) Mass of copper(II) chloride CuCl Mass of aluminum Mass of aluminum remaining after reaction Mass of aluminum that reacted Mass of filter paper and watchglass Mass of filter paper, watchglass, and copper Mass o 880 1.42 63 o.454 2-422, Toy f copper recovered Show calculations with each blank except for the molar mass of Cu (Unless otherwise noted, 6 pts. for blanks including calculations in this section) Copper Molar mass of CuCl Mole of CuCl: dissolved Balanced equation for the dissociation of CuClz in water (4 pts.) Mole of copper(II) ions (Cu") initially in solution Balanced net ionic equation for the reaction of copper(II) chloride with aluminum (4pts.) Mole of copper metal (Cu') expected to be formed 57

Explanation / Answer

a) Moles of CuCl2 dissolved.

For this part, we must convert the grams of CuCl2 given in experimental data (3.023g) to moles, using the given molar mass of CuCl2 (134.542 g/mol). Use the formula:

MM = g/n

Where MM = molar mass, n = moles, g = grams. Solve for n:

n = g/MM

And substitute the values:

n = (3.023g) / (134.542 g/mol)

n = 0.02247 moles CuCl2

b) Balanced equation for the dissociation of CuCl2 in water.

For this, simply split the compound into its ions with their respective charges and coefficients. CuCl2 will dissolve into 1 copper(II) ion and 2 chloride ions in solution.

CuCl2 = Cu2+ + 2Cl-

c) Moles of Cu2+ ions initially in solution.

As we can see in the equation above, we get 1 mole of Cu2+ per every mole of CuCl2 (in other words, their moles will be the same, it would be different with Cl- because each mole of CuCl2 produces 2 moles of Cl-, so in that case we would have twice the moles of Cl- in solution).

n = 0.02247 moles Cu2+

d) Balanced net-ionic equation between CuCl2 and Al.

In this reaction, Cu(II) will be reduced to Cu0 and Al0 will be oxidized to Al3+:

First write the complete equation that takes place:

CuCl2 (aq) + Al(s) = Cu(s) + AlCl3(aq)

Split the (aq) species into ions:

Cu2+ + 2Cl- + Al(s) = Cu(s) + Al3+ + 3Cl-

For the net ionic equation, eliminate the species that appear on both sides of the equation (Cl- in this case).

Cu2+ + Al(s) = Cu(s) + Al3+

Balance charges (the smallest charge that we can have on both sides is 6+, so multiply all Cu species by 3 and all Al species by 2).

3Cu2+ + 2Al(s) = 3Cu(s) + 2Al3+

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