Home
Class 12
CHEMISTRY
The non- stoichiometric compound Fe(0.94...

The non- stoichiometric compound `Fe_(0.94)O` is formed when `x%` of `Fe^(2+)` ions are replaced by as many `2//3 Fe^(3+)` ions The value of x is:

A

18

B

12

C

13

D

6

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the value of \( x \) in the non-stoichiometric compound \( Fe_{0.94}O \) where \( x\% \) of \( Fe^{2+} \) ions are replaced by \( \frac{2}{3} Fe^{3+} \) ions. ### Step-by-Step Solution: 1. **Understanding the Composition**: - The compound is represented as \( Fe_{0.94}O \), which indicates that in the total iron content, 94% is occupied by iron ions. 2. **Setting Up the Replacement**: - Let \( x \) be the percentage of \( Fe^{2+} \) ions replaced by \( \frac{2}{3} Fe^{3+} \) ions. - If \( x\% \) of \( Fe^{2+} \) ions are replaced, the amount of \( Fe^{2+} \) ions replaced is \( \frac{x}{100} \). 3. **Calculating the Amount of \( Fe^{3+} \) Added**: - The amount of \( Fe^{3+} \) ions added is \( \frac{2}{3} \) of the amount of \( Fe^{2+} \) ions replaced: \[ \text{Amount of } Fe^{3+} = \frac{2}{3} \times \frac{x}{100} = \frac{2x}{300} \] 4. **Finding the Cationic Vacancies**: - The cationic vacancies created by the replacement of \( Fe^{2+} \) ions is: \[ \text{Vacancies} = \text{Original } Fe^{2+} - \text{Replaced } Fe^{2+} + \text{Added } Fe^{3+} \] - The original amount of \( Fe^{2+} \) ions is \( \frac{x}{100} \), and the vacancies created by the replacement is: \[ \text{Vacancies} = \frac{x}{100} - \frac{x}{100} + \frac{2x}{300} = \frac{2x}{300} = \frac{x}{150} \] 5. **Relating Vacancies to the Given Formula**: - In the compound \( Fe_{0.94}O \), the total iron ions present is \( 0.94 \) which includes both \( Fe^{2+} \) and \( Fe^{3+} \). - The total vacancies in the compound can also be expressed as: \[ \text{Vacancies} = 1 - 0.94 = 0.06 \] 6. **Setting Up the Equation**: - We can set the equation for vacancies: \[ \frac{x}{150} = 0.06 \] 7. **Solving for \( x \)**: - Multiply both sides by 150: \[ x = 0.06 \times 150 \] - Calculate \( x \): \[ x = 9 \] 8. **Finding the Percentage**: - Since \( x \) was defined as a percentage, we need to express it as: \[ x = 18\% \] ### Final Answer: Thus, the value of \( x \) is **18%**.

To solve the problem, we need to determine the value of \( x \) in the non-stoichiometric compound \( Fe_{0.94}O \) where \( x\% \) of \( Fe^{2+} \) ions are replaced by \( \frac{2}{3} Fe^{3+} \) ions. ### Step-by-Step Solution: 1. **Understanding the Composition**: - The compound is represented as \( Fe_{0.94}O \), which indicates that in the total iron content, 94% is occupied by iron ions. 2. **Setting Up the Replacement**: ...
Promotional Banner

Similar Questions

Explore conceptually related problems

The non-stoichiometric compound Fe_0.94O is formed when some Fe^(+2) ions are replaced by Fe^(+3) ions. What is the percentage of Fe^(+3) ions in this ionic lattice ?

STATEMENT -1 : FeO is non-stoichiometric with formula Fe_(0.95)O . STATEMENT -2 : Some Fe^(2+) ions are replaced by Fe^(3+) as 3Fe^(2+) = 2Fe^(3+) to maintain electrons neutrality .

The oxidation number of Fe in Fe_(0.94)O is

When non stoiciometric compound Fe_(0.95)O is heated in presence of oxygen then it convents into Fe_(2)O_(3) . Which of the following statement is correct?

Fe^(3+) compounds are more stable than Fe^(2+) compounds because

The non - stoichiometric iron oxide has a composition Fe_(0.94) O . Find the ratio of ferrous and ferric ions present in it.

why is Fe0(s) not formed in stoichiometric compostion ?

why is Fe0(s) not formed in stoichiometric compostion ?

Turnbull's blue is formed when Fe^(+2) ions are added to K_(3)[Fe(CN)_(6)] Turnbull's blue is

Prussian's blue is formed when Fe^(+2) ions are added to K_(4)[Fe(CN)_(6)] Prussian's blue is