A certain copper ore contains 2 % Cu2S. The metallurgical process for recovering elemental
Cu from this ore gives 70 % of the theoretical yield. How much Cu is recovered from 2.0 tons of this ore?
Copper Ore contains 2% Cu2S
Therefore 2.0 tons of the ore = 2% of 2 tons = "\\dfrac{2}{100}\u00d72000kg" = 40kg = 40,000g
"Cu_2S_{(s)} \\to 2Cu_{(s)} + S"
from the reaction above, 2 moles of Copper can be gotten from 1 mole of Copper Sulfide.
This means that 127.1g of Copper can be gotten from 159.2g of Copper Sulfide
Since the actual yield is 70% of the theoretical yield, this means for 159.2g of Copper Sulfide, 88.97g (70% of 127.1g) of Copper is recovered.
88.97g of Cu = 159.2g of Copper Sulfide
xg of Cu = 40,000g
x = "\\dfrac{40000\u00d788.97}{159.2}" = 22,354.3g
This means that 22.35 kg of Cu is recovered from 2.0 tons of the Copper ore
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