Question

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?

Answer 1

Copper Ore contains 2% Cu₂S

Therefore 2.0 tons of the ore = 2% of 2 tons = $\dfrac{2}{100}×2000kg$ = 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 = $\dfrac{40000×88.97}{159.2}$ = 22,354.3g

This means that 22.35 kg of Cu is recovered from 2.0 tons of the Copper ore