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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