The Cu^2+ ions in this experiment are produced by the reaction of 1.0 g of coppe

Question

The Cu^2+ ions in this experiment are produced by the reaction of 1.0 g of copper turnings with excess nitric acid. How mans moles of Cu^2+ are produced? Why isn't hydrochloric acid used in a direct reaction with copper to prepare the CuCl_2 solution? How many grams of metallic copper are required to react with the number of moles of Cu^2+ calculated in Problem 1 to form the CuCI? The overall reaction can be taken to be: Cu^2+ (aq) + 2 Cl^- (aq) + Cu(s) rightarrow 2 CuCl(s) What is the maximum mass of CuCI that can be prepared from the reaction sequence of this experiment, using 1.0 g of Cu turnings to prepare the Cu^2+ solution? A sample of the compound prepared in this experiment, weighing 0.1021 g, is dissolved in HNO_3, and diluted to a volume of 100 mL. A 10-mL aliquot of that solution is mixed with 10 mL 6 M NH_3. The |Cu(NH_3)_4^2+| in the resulting solution is found to be 5.16 times 10^-3 M. How many moles of Cu were in the original sample, which had been effectively diluted to a volume of 200 mL? How many grams of Cu were in the sample? How many grams of Cl were in the sample? How many moles? What is the formula of the copper chloride compound?

Explanation / Answer

Q1.

1 g of Copper + HNO3 ...

mol of Cu = mass/MW = 1/63.546 = 0.015736 mol of copper... 0.015736 mol of Cu2+

Q2.

HCl will not dissociate CuCl2, since chlorine atoms are present

Q3.

Cu+2 + 2Cl- + Cu(s) -> 2 CuCl(s)

so

1 mol of Cu+2 --> 2 mol of CuCl

0.015736 mol --> 2*0.015736 = 0.031472 mol of CuCl

Q4

max mass

mas = mol*MW = 0.031472 *98.999 = 3.1156 g of CuCl

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