Question

The ratio of fe 3 and fe 2 ions in fe0.9s1 is

Answer 1

Answer:

0.28 is the answer

Explanation:

Say Fe3+ =x & Fe2+ = 0.9-x

Total Charge in different oxidation state = +2 is 3x + 2(0.9-x)= 2

= 3x +1.8 - 2x= 2

= x = 2-1.8

= x = 0.2

Fe3+ = 0.2

Fe 2+ = 0.9-0.2 =0.7

hence 0.2 / 0.7 = 0.28 Ans

Answer 2

Answer:

The ratio of Fe³⁺ and Fe²⁺ ions in the $Fe_{0.9}S_{1}$  sample will be 0.28.

Explanation:

Assume that the charge in Fe²⁺ is x

The charge on Fe³⁺ $=0.9-x$

The given composition is $Fe_{0.9}S_{1}$

We know that the charge on Sulphur S is -2

The overall charge on the compound is equal to zero

     $(+3)x +(0.9-x)(+2)+(-2)=0$

               $1.8+x=2$

             $x=2-1.8=0.2$

The amount of Fe³⁺ present$=0.9-0.2=0.7$

The ratio of Fe³⁺ and Fe²⁺ ions:

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

Therefore the ratio of Fe ions in two different oxidation state will be 0.28.