Chemistry,?

Question

Hi guys
I need to have a step by step solution way,I have the answers the weight percent of Cu2S in the Ore is 62.2% ,the weight Percent CuS in the Ore is 26.8 Percent.I need just Solution way

Copper metal can be prepared by roasting Copper Ore which can contain cuprite (Cu2S) and Copper (II) Sulfide.
Cu2S(s)+O2(g) then 2Cu(s)+SO2(g)
CuS(s)+O2 (g) then Cu(s)+SO2(g)
Suppose an Ore sample contains 11% impurity in addition to a mixture of CuS and Cu2S.Heating 100.0 gr of the mixture produces 75.4 gr of Copper metal with apurity of 89.5%.
What is the weight percent of CuS in the Ore?the weight percent of Cu2S?

Answer 1

62.2+26.8=89percent
100-89=11percent
Cu2S(s)+O2(g) then 2Cu(s)+SO2(g) (62.2percent)
CuS(s)+O2 (g) then Cu(s)+SO2(g) (26.8percent)
Burning uses oxygen gas and releases SO2 gas.

100(0.622)=62.2g of Cu2S
100(0.268)=26.8g of CuS

Atomic weights:
Cu=63.546 g per mole
S= 32.065 g per mole

Cu2S= 2(63.546)+32.065 = A
Cu2 = 2(63.546)
2(63.546) / A = B
CuS= 63.546+32.065 = C
Cu=63.546
63.546 / C = D

62.2B+26.8D = Z
Z / 0.895 = 75.4

62.2B / 75.4 = weight percent of Cu2S
26.8D / 75.4 = weight percent of CuS