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“The first that probably occurs is that in which two equivalents of the metallic sulphide react on one of the metallic sulphate with reduction to the metal, metallic sulphide, and sulphurous acid, as shown by the following equation in the case of lead:

 2PbS + PbSO4 = 2Pb + PbS + 2SO2.

“The metal so formed, in the presence of air, is oxidized, and in this state reacts on a further portion of the metallic sulphide produced, with an increased formation of metal and evolution of sulphurous acid, according to the following equation, in the case of lead:

 2PbO + PbS = Pb + SO2.

“The metal so produced in this reaction is wholly reoxidized by the oxygen of the air current, and being free to react on still further portions of the metallic sulphide, repeats the reaction, and becomes an important factor in the desulphurizing of the undecomposed portion of the material. As the desulphurization proceeds, and the sulphate of metal accumulates, reactions are set up between the metallic sulphide and different multiple proportions of the metallic sulphate, with the formation of metal, metallic oxide, and evolution of sulphurous acid, as follows: