`1.5gm` mixture of `SiO_(2)` and `Fe_(2)O_(3)` on very strong heating leave a residue weighting `1.46gm`. The reaction responsible for loss of weight is . `Fe_(2) O_(3) (s) rarr Fe_(3)O_(4) (s) + O_(2)(g)`
What is the percentage by mass of `Fe_(2)O_(3)` is original sample .

Options

A. `80%`
B. `20%`
C. `40%`
D. `60%`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the Reaction
The reaction given is:
\[ 3 \text{Fe}_2\text{O}_3 (s) \rightarrow 2 \text{Fe}_3\text{O}_4 (s) + \frac{3}{2} \text{O}_2 (g) \]
This indicates that when iron(III) oxide (\( \text{Fe}_2\text{O}_3 \)) is heated, it decomposes to iron(II,III) oxide (\( \text{Fe}_3\text{O}_4 \)) and oxygen gas (\( \text{O}_2 \)), resulting in a loss of mass due to the release of oxygen.

### Step 2: Calculate the Molar Masses
- Molar mass of \( \text{Fe}_2\text{O}_3 \):
- \( \text{Fe} = 55.85 \, \text{g/mol} \) (2 iron atoms)
- \( \text{O} = 16.00 \, \text{g/mol} \) (3 oxygen atoms)
- Total: \( 2 \times 55.85 + 3 \times 16.00 = 159.7 \, \text{g/mol} \) (approximately \( 160 \, \text{g/mol} \))

- Molar mass of \( \text{Fe}_3\text{O}_4 \):
- \( \text{Fe} = 55.85 \, \text{g/mol} \) (3 iron atoms)
- \( \text{O} = 16.00 \, \text{g/mol} \) (4 oxygen atoms)
- Total: \( 3 \times 55.85 + 4 \times 16.00 = 231.55 \, \text{g/mol} \) (approximately \( 232 \, \text{g/mol} \))

### Step 3: Determine the Mass Loss
From the balanced equation, we can see that:
- For every \( 3 \, \text{moles} \) of \( \text{Fe}_2\text{O}_3 \) that decompose, \( 1.5 \, \text{moles} \) of \( \text{O}_2 \) are produced.
- The loss of mass due to the reaction can be calculated as follows:
- Mass loss per \( 3 \, \text{moles} \) of \( \text{Fe}_2\text{O}_3 \):
\[ \text{Mass loss} = \text{Mass of } \text{Fe}_2\text{O}_3 - \text{Mass of } \text{Fe}_3\text{O}_4 = 3 \times 160 - 2 \times 232 = 480 - 464 = 16 \, \text{g} \]
- Therefore, the loss of mass per gram of \( \text{Fe}_2\text{O}_3 \) is:
\[ \text{Loss of mass per gram of } \text{Fe}_2\text{O}_3 = \frac{16 \, \text{g}}{480 \, \text{g}} = \frac{1}{30} \, \text{g} \]

### Step 4: Calculate the Mass of \( \text{Fe}_2\text{O}_3 \) in the Mixture
- The initial mass of the mixture is \( 1.5 \, \text{g} \) and the residue after heating is \( 1.46 \, \text{g} \).
- Therefore, the mass lost during the reaction is:
\[ \text{Mass lost} = 1.5 \, \text{g} - 1.46 \, \text{g} = 0.04 \, \text{g} \]

### Step 5: Relate Mass Loss to Mass of \( \text{Fe}_2\text{O}_3 \)
- Let \( x \) be the mass of \( \text{Fe}_2\text{O}_3 \) in the original mixture.
- The mass loss due to \( \text{Fe}_2\text{O}_3 \) is:
\[ \text{Mass loss} = \left(\frac{1}{30} \, \text{g}\right) \times x \]
- Setting this equal to the mass lost:
\[ \frac{1}{30} x = 0.04 \implies x = 0.04 \times 30 = 1.2 \, \text{g} \]

### Step 6: Calculate the Percentage of \( \text{Fe}_2\text{O}_3 \)
- The percentage by mass of \( \text{Fe}_2\text{O}_3 \) in the original sample is:
\[ \text{Percentage} = \left(\frac{x}{\text{Total mass}}\right) \times 100 = \left(\frac{1.2 \, \text{g}}{1.5 \, \text{g}}\right) \times 100 = 80\% \]

### Final Answer
The percentage by mass of \( \text{Fe}_2\text{O}_3 \) in the original sample is **80%**.

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