copper refining traditionally involves \"roasting\" insoluble sulfides ores (CuS

Question

copper refining traditionally involves "roasting" insoluble sulfides ores (CuS) with oxygen. Unfortunately, the process produces large quantities of SO2(g),which is a major contributor to pollution and acid rain. An alternative process involves treating the sulfide ore with HNO3(aq), which dissolves the CuS without generating any SO2. The unbalanced chemical equation for the reaction is given below.

CuS(s) + NO3_(aq)----> Cu2+(aq) + NO(g) + HSO4_(aq) (not balanced)

What volume of concentrated nitric acid solution is required per kilogram of CuS? Assume that the concentrated nitric acid solution is 70% HNO3 by mass and has a density of 1.40g/ml.

Explanation / Answer

Oxidation Half (where S is oxidised)

CuS + 4 H2O --------> Cu2+ + HSO4- + 7H+ + 8e- ---- Eqn 1

Reduction Half ( N is getting reduced)

NO3- + 4 H+ + 3e- -----------> NO + 2 H2O ----------- eqn 2

Balancing electrons by multiplying eqn 1 by 3 & eqn 2 by 8

Final balanced eqn

3 CuS + 8 NO3- + 11H+ -------> 3Cu2+ + 3 HSO4- + 8 NO + 4 H2O

Moles of CuS = 1000/(63.5+32) = 10.471 moles

Moles of HNO3 required = 10.471*8/3 = 27.923 moles

Let volume of HNO3 = V mL

Mass of the solution = 1.40*V g

Mass of HNO3 = 1.40*V*70/100 g

Moles of HNO3 = 1.40*V*70/100*(1+48+14)

Hence

1.40*V*70/100*(1+48+14) = 27.923

=> V = 1795.06 mL = 1.795 L 1.8 L

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