The most common copper ore used in the production of copper is chalcopyrite. If a chalcopyrite ore contains 55.0% CuFeS2, how many grams of chalcopyrite ore are needed to produce 60.0 g of copper?

Answer is: 315,29 grams of copper ore.
m(chalcopyrite) = ?
ω(CuFeS₂) = 55,0 % = 0,55.
m(Cu) = 60,0 g.
mass percentage of copper in CuFeS₂:
ω(Cu) = Ar(Cu) ÷ Mr(CuFeS₂).
ω(Cu) = 63,55 ÷ 183,4.
ω(Cu) = 0,346 = 34,6 %.
mass percentage of copper in chalcopyrite:
ω(Cu) : ω₁(Cu) = 100% : ω(CuFeS₂).
ω₁(Cu) = 19,03 % = 0,1903.
m(chalcopyrite) = 60,0 g ÷ 0,1903.
m(chalcopyrite) = 315,29 g.

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Annainn Annainn · Answer 2 · 2018-12-14T14:36:18+01:00

The formation of Cu from CuFeS2 can be shown by the following equations:

4CuFeS2 + 7O2 → 4CuS + 2Fe2O3 + 4 SO2

4CuS + 4O2 → 4Cu + 4SO2

Based on the stoichiometry:

1 mole of CuFeS2 produces 1 mol of Cu

Now,

Mass of Cu to be produced = 60 g

# moles of Cu = 60 g/63.5 g.mol-1 = 0.945 moles

Therefore, moles of CuFeS2 required = 0.945 moles

Molar mass of CuFeS2 = 183.5 g.mol-1

Mass of CuFeS2 required = 0.945 * 183.5 = 173.41 g

It is given that the ore contains 55.0% CuFeS2

i.e.

100 g of chalcopyrite contains 55.0 g CuFeS2

The amount of ore corresponding to 173.41 g of CuFeS2 is

= 100 g chalcopyrite * 173.41 g CuFeS2/55 g CuFeS2 = 315.29 g

Ans: Around 315 g of the ore is required to produce 60 g Cu