Answer to Question #145808 in Physical Chemistry for deepak

Question

A copper concentrate (CuFeS2 + FeS2 + gangue) of composition (wt%) Cu = 6%, S = 35% is
roasted with (20+RN*)% excess air. Copper in the roasted product is as Cu2S and all the iron is
converted to Fe2O3. Assuming that the gangue does not contain any Cu, Fe and S, Calculate for 1
ton of concentrate i) amount of minerals and gangue ii) volume of air supplied for roasting at STP.
iii) Volume and composition of the evolved gas (flue) after roasting [atomic weight of Cu=64, Fe=
56, S=32, O=16, N=14]

Answer 1

(i)Let $\omega$(CuFeS₂)=c and $\omega$(FeS₂)=f

c*64/(64+56+32*2) = 0.06

c*32*2/(64+56+32*2)+f*32*2/(56+32*2) = 0.35

c*0.348 = 0.06

c*0.348+f*0.533 = 0.35

c=0.172; f=0.334

$\omega$(CuFeS₂)=0.172; $\omega$(FeS₂)=0.344; $\omega$(gangue)=0.484

(ii) excess of air=20%; $\chi$ [mole fraction](O₂ in air)=0.21; $\chi$ (N₂ in air)=0.79

4CuFeS₂+9O₂=2Cu₂S+2Fe₂O₃+6SO₂

4FeS₂+11O₂=2Fe₂O₃+8SO₂

n(CuFeS₂)=1000000*0.172/(64+56+32*2)=935 mol

n(FeS₂)=1000000*0.344/(56+32*2)=2867 mol

n(O₂, theor)=935/4*9+2867/4*11=9988 mol

V(air)=9988/0.21*1.20*22.4/1000=1278 m³

(iii)

V(O₂)=9988*0.2*22.4/1000=44.7 m³

V(N₂)=1278*0.79=1010 m³

V(SO₂)=(935/4*6+2867*2)*22.4/1000=160 m³

V(flue)=44.7+1010+160=1215 m³

$\chi$ (N₂)=0.831; $\chi$ (SO₂)=0.132; $\chi$ (O₂)=0.037.