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Roasting 100.0 g of a copper ore yields 71.8 g of pure copper. The ore contains only Cu2S and CuS along with 4.5% inert impurity (by mass). The relevant reactions are: Cu2S + O2 -> 2Cu + SO2 CuS + O2 -> Cu + SO2 Calculate the percentage by mass of Cu2S in the ore. (Atomic mass: Cu = 63.5, S = 32)

1. 50%
2. 55%
3. 60%
4. 65%

Correct answer: 50%

Solution

The ore has 95.5 g of active sulfide mixture. Using mole relationships from each reaction and the total Cu produced (1.13 mol), solving the two linear equations gives the mass of Cu2S as approximately 50 g, i.e. 50% of the ore.

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