Answer to Question #114839 in General Chemistry for maria

Question #114839
Copper metal can be prepared by roasting copper ore, which can contain cuprite (Cu2S) and copper(II) sulfide.
Cu2S(s)+O2(g)-->2 Cu(s) +SO2(g)
CuS(g)+O2(g)-->Cu(s)+ SO2(g)
Suppose an ore sample contains 11.0% impurity in addition to a mixture of CuS and Cu2S . Heating 100.0 g of the mixture produces 76.3 g of copper metal with a purity of 89.3%. What is the weight percent of CuS in the ore? The weight percent of Cu2S ?
Weight Percent of CuS=____
Weight percent of Cu2S=_____
1
Expert's answer
2020-05-08T14:09:35-0400


1.find the mass of impurities

"\\frac{11}{100}\\times100=11"

100-11=89 g

2.let Cu2S - x g, CuS=89-x g

Cu from Cu2S - 127x/159g

Total amount of Cu:

"\\frac{127x}{159}+\\frac{63.5(89.0-x)}{95.5}=\\frac{76.3\\times89.3}{100}"


"\\frac{127x}{159}+\\frac{63.5(89.0-x)}{95.5}=68.1359"


x=66.95 - m Cu2S


"\\omega (Cu2S)=\\frac{66.95}{100}=0.6695" or 66.95%

"m (CuS)=89-66.95=22.05" or 22.05%


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