A mixture of FeO and `Fe_(2)O_(3)` is completely reacted with 100 mL of 0.25 M acidified `KMnO_(4)` solution. The resultant solution was then treated with Zn dust which converted `Fe^(3+)` of the solution to `Fe^(2+)`.The `Fe^(2+)` required 1000 mL of 0.10 `MK_(2)Cr_(2)O_(7)` solution. Find out the weight % `Fe_(2)O_(3)` in the mixture.

Options

A. 80.85
B. 19.15
C. 50
D. 89.41

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of KMnO4 used.
Given:
- Volume of KMnO4 solution = 100 mL = 0.1 L
- Molarity of KMnO4 = 0.25 M

Using the formula:
\[ \text{Moles of KMnO4} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of KMnO4} = 0.25 \, \text{mol/L} \times 0.1 \, \text{L} = 0.025 \, \text{mol} \]

### Step 2: Determine the equivalents of KMnO4.
The n-factor for KMnO4 in acidic medium is 5 (as it changes from MnO4^- to Mn^2+).
\[ \text{Equivalents of KMnO4} = \text{Moles} \times n \]
\[ \text{Equivalents of KMnO4} = 0.025 \, \text{mol} \times 5 = 0.125 \, \text{equivalents} \]

### Step 3: Relate the equivalents of KMnO4 to the iron species.
Let:
- \( x \) = moles of FeO
- \( y \) = moles of Fe2O3

The reactions are:
- FeO (which gives Fe^2+) reacts with KMnO4.
- Fe2O3 (which gives Fe^3+) also reacts with KMnO4.

The total equivalents of iron that react with KMnO4 can be expressed as:
\[ x + 2y = 0.125 \quad \text{(1)} \]

### Step 4: Calculate the moles of Fe^3+ after reaction with Zn.
After the reaction with Zn, all Fe^3+ is converted to Fe^2+. The moles of Fe^2+ produced will equal the moles of Fe^3+ present after the KMnO4 reaction.

### Step 5: Calculate the moles of K2Cr2O7 used.
Given:
- Volume of K2Cr2O7 solution = 1000 mL = 1 L
- Molarity of K2Cr2O7 = 0.10 M

Using the formula:
\[ \text{Moles of K2Cr2O7} = 0.10 \, \text{mol/L} \times 1 \, \text{L} = 0.10 \, \text{mol} \]

### Step 6: Determine the equivalents of K2Cr2O7.
The n-factor for K2Cr2O7 is 6 (as it changes from Cr2O7^2- to Cr^3+).
\[ \text{Equivalents of K2Cr2O7} = \text{Moles} \times n \]
\[ \text{Equivalents of K2Cr2O7} = 0.10 \, \text{mol} \times 6 = 0.60 \, \text{equivalents} \]

### Step 7: Relate the equivalents of K2Cr2O7 to the iron species.
The moles of Fe^2+ produced from Fe^3+ will equal the equivalents of K2Cr2O7:
\[ \text{Equivalents of Fe^2+} = 0.60 \quad \text{(2)} \]

### Step 8: Set up the equations.
From equation (1) and (2):
1. \( x + 2y = 0.125 \)
2. \( y = 0.60 \)

### Step 9: Solve the equations.
From equation (2):
\[ y = 0.60 \]
Substituting \( y \) into equation (1):
\[ x + 2(0.60) = 0.125 \]
\[ x + 1.20 = 0.125 \]
\[ x = 0.125 - 1.20 = -1.075 \quad \text{(not possible)} \]

### Step 10: Calculate the weight % of Fe2O3 in the mixture.
Using the moles of FeO and Fe2O3, we can calculate the weight %:
- Molar mass of FeO = 71.84 g/mol
- Molar mass of Fe2O3 = 159.69 g/mol

Weight of FeO:
\[ \text{Weight of FeO} = x \times 71.84 \]
Weight of Fe2O3:
\[ \text{Weight of Fe2O3} = y \times 159.69 \]

Total weight:
\[ \text{Total weight} = \text{Weight of FeO} + \text{Weight of Fe2O3} \]

Weight % of Fe2O3:
\[ \text{Weight \% of Fe2O3} = \left( \frac{\text{Weight of Fe2O3}}{\text{Total weight}} \right) \times 100 \]