Question

The ratio of ${Fe^{3+}}$ and ${Fe^{2+}}$ in ${Fe_{0.9} S_{1.0}}$ is

• A. 0.28
• B. 0.5
• C. 2
• D. 4

Answer 1

Answer: (A)

0.28

Answer 2

In ${Fe_{0.9}S_{1.0}}$, suppose Fe atoms = 90, then S- atoms = 100. Out of 90 Fe atoms, if ${Fe^{3+}}$ = x, then ${Fe^{2+}}$ = 90 - x. As ${Fe^{3+}}$ forms ${Fe_2S_3}$, S-atoms combine with x atoms of ${Fe^{3+} = \frac{3}{2} x = 1.5 \, x}$ As ${Fe^{2+}}$ forms FeS, S-atoms combine with 90 - x atoms of ${Fe^{2+} = (90 - x)}$ Total S-atoms = 1.5 x + (90 - x) =100 or x = 20 = ${Fe^{3+}}$ Ratio ${Fe^{3+} / Fe^{2+}} = \frac{20}{70}$ = 0.28