If a piece of iron gains 10% of its weight due to partial rusting into `Fe_(2)O_(3)` the percentage of total iron that has rusted is:

Options

A. 23
B. 13
C. 23.3
D. 25.67

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Variables
Let the initial mass of iron be \( x \) grams. The mass of iron that rusts is \( y \) grams.

### Step 2: Understand the Reaction
The rusting of iron can be represented by the reaction:
\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \]
From this reaction, we can see that 4 moles of iron produce 2 moles of rust (Fe₂O₃).

### Step 3: Calculate the Mass Gain
According to the problem, the piece of iron gains 10% of its weight due to rusting. Therefore, the new mass after rusting is:
\[ \text{New mass} = x + 0.1x = 1.1x \]

### Step 4: Calculate the Mass of Iron Not Rusted
The mass of iron that has not rusted is:
\[ x - 56y \]
where \( 56y \) is the mass of iron that has rusted (since the molar mass of iron is 56 g/mol).

### Step 5: Set Up the Equation
We can set up the equation based on the mass before and after rusting:
\[ \text{Mass not rusted} + \text{Mass of rust formed} = \text{New mass} \]
The mass of rust formed can be calculated as follows:
- From the stoichiometry of the reaction, for every 4 moles of iron that rust, 2 moles of Fe₂O₃ are formed. Therefore, the mass of rust formed from \( y \) grams of iron is:
\[ \text{Mass of rust} = \frac{80y}{2} = 80y \]
Thus, the equation becomes:
\[ (x - 56y) + 80y = 1.1x \]

### Step 6: Simplify the Equation
Now we simplify the equation:
\[ x - 56y + 80y = 1.1x \]
\[ x + 24y = 1.1x \]
Subtract \( x \) from both sides:
\[ 24y = 0.1x \]
Now, we can express \( y \):
\[ y = \frac{0.1x}{24} \]

### Step 7: Calculate the Mass of Iron that has Rusted
The mass of iron that has rusted is:
\[ \text{Mass of rusted iron} = 56y = 56 \times \frac{0.1x}{24} = \frac{5.6x}{24} \]

### Step 8: Calculate the Percentage of Iron that has Rusted
To find the percentage of total iron that has rusted, we use the formula:
\[ \text{Percentage of iron rusted} = \left( \frac{\text{Mass of rusted iron}}{\text{Initial mass of iron}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage} = \left( \frac{\frac{5.6x}{24}}{x} \right) \times 100 \]
\[ = \left( \frac{5.6}{24} \right) \times 100 \]
\[ = 23.33\% \]

### Final Answer
The percentage of total iron that has rusted is approximately **23.33%**.

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