Abstract
Calcium oxide favors greatly the reduction of metal sulfides by H2 at 800°C. Copper sulfides and copper-iron sulfides are reduced rapidly to metals: Cu2S+H2+CaO→2Cu+CaS+H2O CuS+H2+CaO→Cu+CaS+H2O Cu5FeS4+4H2+4CaO→5Cu+ Fe+4CaS+4H2O
The reduction of chalcopyrite takes place according to any of the following reactions depending on the amount of CaO added: Cu2S·Fe2S3+H2+CaO→Cu2S·2FeS+CaS+H2O Cu2S·Fe2S3+3H2+3CaO→Cu2S+2Fe+3CaS+3H2O Cu2S·Fe2S3+4H2+4CaO→2Cu+2Fe+4CaS+4H2O
Gravity and magnetic methods were not successful in separating the reaction products. However, leaching by dilute HCl was successful in eliminating CaS and iron (as FeS as well as metal) leaving behind Cu2S concentrate or metallic copper, respectively. Reduction of molybdenite is also influenced by the presence of CaO, but that of pentlandite is not.
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FATHI HABASHI and RAYMOND DUGADALE, formerly with the Research Division of the Anaconda Company, Tucson, Arizona
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Habashi, F., Dugdale, R. The reduction of sulfide minerals by hydrogen in the presence of lime. Metall Trans 4, 1865–1871 (1973). https://doi.org/10.1007/BF02665414
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DOI: https://doi.org/10.1007/BF02665414