Question

The ratio of Fe3+ and Fe2+ in Fe0.9S1.0 is-
1- 0.28
2- 0.5
3- 2
4- 4

Answer 1

Answer:

Explanation:

Fe3+= x and Fe2+=.93-x. Using charge balance: 3*x+2*(0.93-x) = 2*1. ( 1 for oxygen). x = .14. % of Fe3+= (.14/.93)* 100. =15%. This conversation is already

Answer 2

Answer: The correct answer is Option 1.

Explanation:

In a neutral compound, the charge on anion and cation is equal. So, for a given compound $Fe_{0.9}S_{1.0}$

The charge on sulfur atom is -2.

Let the number of $Fe^{3+}$ ions be x, so the number of $Fe^{2+}$ ions will be (0.9 - x)

Balancing out the charge, we get:

$3x+2(0.9-x)=2\times 1\\\\x=2-1.8\\\\x=0.2$

Thus, the number of $Fe^{2+}$ ions will be (0.9 - 0.2) = 0.7

Taking the ratio of $Fe^{3+}$ and $Fe^{2+}$, we get:

$\frac{Fe^{3+}}{Fe^{2+}}=\frac{0.2}{0.7}=0.28$

Hence, the correct answer is Option 1.