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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]

Asked by deepakmardi22 21st November 2020, 10:09 AM
Answered by Expert
Answer:
Let weight of CuFeS2 = x, Weigght of FeS2 = y
fraction numerator x cross times 64 over denominator left parenthesis 64 plus 56 plus 32 cross times 2 right parenthesis end fraction equals 0.06
fraction numerator x cross times 64 over denominator left parenthesis 64 plus 56 plus 32 cross times 2 right parenthesis end fraction plus fraction numerator y cross times 32 cross times 2 over denominator left parenthesis 56 plus 32 cross times 2 right parenthesis end fraction equals 0.35
x cross times 0.348 equals 0.06
x cross times 0.348 plus y cross times 0.533 equals 0.35
x equals 0.172 comma space y equals 0.334

W left parenthesis C u F e S subscript 2 right parenthesis equals 0.172 comma space W left parenthesis F e S subscript 2 right parenthesis equals 0.344 comma space W left parenthesis g a n g u e right parenthesis equals 0.484

E x c e s s space o f space a i r equals 20 percent sign
X left parenthesis O subscript 2 space i n space a i r right parenthesis equals 0.21
X left parenthesis N 2 space i n space a i r right parenthesis equals 0.79

4 C u F e S subscript 2 plus 9 O subscript 2 space rightwards arrow 2 C u subscript 2 S space plus 2 F e subscript 2 O subscript 3 plus 6 S O subscript 2
4 F e S subscript 2 plus 11 O subscript 2 rightwards arrow 2 F e subscript 2 O subscript 3 plus 8 S O subscript 2
n left parenthesis C u F e S subscript 2 right parenthesis equals 1000000 cross times fraction numerator 0.172 over denominator left parenthesis 64 plus 56 plus 32 cross times 2 right parenthesis end fraction equals 935 m o l
n left parenthesis F e S subscript 2 right parenthesis equals 1000000 cross times fraction numerator 0.344 over denominator left parenthesis 56 plus 32 cross times 2 right parenthesis end fraction equals 2867 m o l
n left parenthesis O subscript 2 right parenthesis equals fraction numerator 935 over denominator 4 cross times 9 end fraction plus fraction numerator 2867 over denominator 4 cross times 11 end fraction equals 9988 space m o l
V left parenthesis a i r right parenthesis equals fraction numerator 9988 over denominator 0.21 end fraction cross times 1.20 cross times fraction numerator 22.4 over denominator 1000 end fraction equals 1278 space m cubed

V left parenthesis O subscript 2 right parenthesis equals 9988 cross times 0.2 cross times fraction numerator 22.4 over denominator 1000 end fraction equals 44.7 space m cubed
V left parenthesis N subscript 2 right parenthesis equals 1278 cross times 0.79 equals 1010 space m cubed
V left parenthesis S O subscript 2 right parenthesis equals left parenthesis fraction numerator 935 over denominator 4 cross times 6 end fraction plus 2867 cross times 2 right parenthesis cross times fraction numerator 22.4 over denominator 1000 end fraction equals 160 space m cubed
V left parenthesis f l u e right parenthesis equals 44.7 plus 1010 plus 160 equals 1215 space m cubed
X left parenthesis N subscript 2 right parenthesis equals 0.831 comma space X left parenthesis S O subscript 2 right parenthesis equals 0.132 comma space X left parenthesis O subscript 2 right parenthesis equals 0.037
Answered by Expert 25th November 2020, 12:19 PM
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