A method for producing pure copper metal involves the reaction of copper(I) sulfide with oxygen gas to give copper metal and sulfur dioxide. Suppose the yield of this reaction is 87%. What mass of a copper ore consisting of 46% copper(I) sulfide must be mined in order to produce 1.0 *10³ kg (1.0 metric ton) of copper metal A) 1.3 * 10³ kg B) 1.4 * 10³ kg C) 1.5 * 10³ kg D) 3.2 * 10³ kg E) 8.0 * 10³ kg

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Quizplus · Accepted Answer

To produce 1.0 * 10^3 kg of copper with an 87% yield, more copper must be produced theoretically: 1.0 * 10^3 kg / 0.87 = 1.149 * 10^3 kg. The molar mass of Cu2S is 159.16 g/mol, and Cu is 63.55 g/mol. From Cu2S, 2 moles of Cu are produced, so 159.16 g of Cu2S produces 127.1 g of Cu. Therefore, 1.149 * 10^3 kg of Cu requires (1.149 * 10^3 kg * 159.16 g/mol) / 127.1 g/mol = 1.438 * 10^3 kg of Cu2S. Since the ore is 46% Cu2S, the mass of ore needed is 1.438 * 10^3 kg / 0.46 = 3.126 * 10^3 kg, approximately 3.2 * 10^3 kg.

A Method for Producing Pure Copper Metal Involves the Reaction