Question

calculate ratio of fe3+ and fe2+ in fe0.9s1.0

Answer 1

Answer: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

Explanation:

Answer 2

Answer: The ratio of $Fe^{3+}$ to $Fe^{2+}$ will be 0.28

Explanation:

In a neutral compound, the charge on anion and cation are equal or they balance each other.

We are given a chemical compound having formula of $Fe_{0.9}S_{1.0}$

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)

By balancing out the charge in compound, 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 ratio of $Fe^{3+}$ to $Fe^{2+}$ will be 0.28