What would the result be if the solution was not basic when it was heated to for

Question

What would the result be if the solution was not basic when it was heated to form CuO? The [Cu(H_2 O)_6)^2+ solution would be directly converted back to Cu(s), in a one step reaction, causing a very high recovery of Cu(s) Too much of the [Cu(H_2 O)_6)^2+ would be converted to CuO(s) causing an increased recovery Cu(s) Some of the (Cu(H_2 O)_6]^2+ would not be converted to Cu(OH)_2 and would remain in solution, causing a lower recovery Cu(s) The acidity level of the solution is irrelevant What would be the result if a slight blue solution was poured off in the last step (part V) of the reaction? If the solution decanted was even slightly blue, then not all of the [Cu(H_2 O)_6)^2+ ions have been reduced by Zn and the yield of the copper would be reduced If the solution decanted was even slightly blue, then Zn has been dissolved in solution and the yield of the copper would be reduced This is the anticipated result, and no yield changes are anticipated If the solution decanted was even slightly blue, this is a sign that not all of the Cu(OH)_2 molecules have been transformed to CuO(s) What would be the result if, in the final step, all of the water boiled away and the Cu was exposed to the Bunsen burner flame for a long period of time? The Cu would oxidize to (Cu(H_2 O)_6]^2+ resulting in no copper recovery The Cu would oxidize to CuO causing the percent copper recovery to appear much too high The Cu would evaporate resulting in a very low percent recovery The Cu would oxidize to CuO causing the percent copper recovery to appear much too low

Explanation / Answer

1)

Correct option is 1. Some of the [Cu(H2O)6]2+ would not be converted to Cu(OH)2 and would remain in solution, causing a lower recovery Cu(s)

Explanation - When all Cu (H2O)6]2+ converted to Cu(OH) 2 ,then solution became basic. Before it solution starts out acidic because of excess nitric acid from the previous step, so the first OH^- added goes into neutralizing the acid; once the acid is neutralized, the next OH^- added goes to forming the blue Cu(OH)2 precipitate. Only after that OH^- is added, solution became basic.

2)

Correct option is If the solution decanted was slightly blue, then not all of Cu(H2O )62+ion have been reduced by Zn and yield of copper is reduced.

Explanation – If the solution decanted was slightly blue, then some of Cu is wasted and yield of copper is reduced.

[Cu(H2 O)6]2+ (aq) + Zn (s) --> Cu (s) + Zn2+ (aq) + 6 H2 O

3) Correct option is Cu is oxidize to CuO, causing the % copper recovery is much high

Explanation -In final step, if all the water boiled away and Cu is exposed to Bunsen burner Cu is converted to CuO.

Cu + O2 -------> CuO

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