Given 0.158 mol of acidified Aluminium dichromate will oxidise how many moles of `Fe_(0.93)O_(1.00)`?

Options

A. 1.2
B. 3.6
C. 2.4
D. 5.4

Answer

To solve the problem of how many moles of `Fe_(0.93)O_(1.00)` can be oxidized by 0.158 moles of acidified Aluminium dichromate, we can follow these steps:

### Step 1: Determine the oxidation states of Iron in `Fe_(0.93)O_(1.00)`

Given the formula `Fe_(0.93)O_(1.00)`, we know that oxygen typically has an oxidation state of -2. Therefore, the total oxidation state contributed by oxygen is:
\[ \text{Total oxidation from O} = -2 \times 1 = -2 \]
Let \( x \) be the number of iron atoms in the +2 oxidation state, and \( 0.93 - x \) be the number of iron atoms in the +3 oxidation state. The overall charge must balance to +2:
\[ 2x + 3(0.93 - x) = +2 \]

### Step 2: Solve for x

Expanding the equation:
\[ 2x + 2.79 - 3x = 2 \]
Rearranging gives:
\[ -1x + 2.79 = 2 \]
\[ -x = 2 - 2.79 \]
\[ -x = -0.79 \quad \Rightarrow \quad x = 0.79 \]

### Step 3: Calculate the percentage of Iron in +2 oxidation state

Now we can find the percentage of iron in the +2 oxidation state:
\[ \text{Percentage of Fe}^{2+} = \frac{0.79}{0.93} \approx 0.8494 \quad \text{or} \quad 84.94\% \]

### Step 4: Determine the moles of Fe in +2 oxidation state in 1 mole of `Fe_(0.93)O_(1.00)`

Since one mole of `Fe_(0.93)O_(1.00)` contains 0.85 moles of `Fe^2+`, we can now relate this to the moles of Aluminium dichromate.

### Step 5: Write the half-reaction for dichromate

The half-reaction for the reduction of dichromate in acidic medium is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 6: Relate moles of dichromate to moles of `Fe^2+`

From the balanced equation, 1 mole of `Cr2O7^{2-}` can oxidize 6 moles of `Fe^{2+}`.

### Step 7: Calculate moles of `Cr2O7^{2-}` from Aluminium dichromate

The formula for Aluminium dichromate is \( \text{Al}_2\text{Cr}_2\text{O}_7 \). Each mole of Aluminium dichromate contains 3 moles of `Cr2O7^{2-}`:
\[ \text{Moles of Cr}_2\text{O}_7^{2-} = 3 \times 0.158 = 0.474 \text{ moles} \]

### Step 8: Calculate moles of `Fe^{2+}` oxidized

Using the stoichiometry from the half-reaction:
\[ \text{Moles of Fe}^{2+} \text{ oxidized} = 6 \times \text{Moles of Cr}_2\text{O}_7^{2-} = 6 \times 0.474 = 2.844 \text{ moles} \]

### Step 9: Calculate the moles of `Fe_(0.93)O_(1.00)`

Since each mole of `Fe_(0.93)O_(1.00)` contains 0.85 moles of `Fe^{2+}`, we can find the total moles of `Fe_(0.93)O_(1.00)` that can be oxidized:
\[ \text{Moles of } Fe_{0.93}O_{1.00} = \frac{2.844}{0.85} \approx 3.34 \text{ moles} \]

### Final Answer
The moles of `Fe_(0.93)O_(1.00)` that can be oxidized by 0.158 moles of acidified Aluminium dichromate is approximately **2.4 moles**.

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