An ore of iron contains FeS and some non-volatile impurity. Roasting if this ore converts all FeS into Fe2O3 and 4% loss in weight was observed. Mass percentage if FeS in ore will be

Engineering

Chemistry

Chemical Equation and Stoichiometric Calculation

Question

An ore of iron contains FeS and some non-volatile impurity. Roasting if this ore converts all FeS into Fe₂O₃ and 4% loss in weight was observed. Mass percentage if FeS in ore will be

66 %

44 %

30 %

60 %

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2FeS + $\frac{7}{2}$ O₂ → Fe₂O₃ + 2SO₂

_{$[\frac{\frac{ω}{56 + 32}}{2} = \frac{\frac{{ωFe}_{2} O_{3}}{2 \times 56 + 48}}{1}]$}

Two ωR₂O₃ = $\frac{(2 \times 56 + 48)}{(56 + 32) 2}$ w = 0.9 w
ω = 0.9ω = 4
0.1ω = 44%