How many moles of ferric alum
`(NH_(4))_(2) SO_(4) Fe_(2) (SO_(4))_(3). 24 H_(2)O` can be made from the sample of `Fe` containing `0.0056 g` of it?

Options

A. `10^(-4) mol`
B. `0.5 xx 10^(-4) mol`
C. `0.33 xx 10^(-4) mol`
D. `2 xx 10^(-4) mol`

Explanation

To determine how many moles of ferric alum \((NH_{4})_{2}SO_{4}Fe_{2}(SO_{4})_{3} \cdot 24 H_{2}O\) can be made from a sample of iron (Fe) weighing \(0.0056 \, g\), we can follow these steps:

### Step 1: Calculate the number of moles of Fe
To find the number of moles of iron, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molar mass}} \]

The molar mass of iron (Fe) is approximately \(56 \, g/mol\).

\[ \text{Number of moles of Fe} = \frac{0.0056 \, g}{56 \, g/mol} \]

Calculating this gives:

\[ \text{Number of moles of Fe} = \frac{0.0056}{56} = 1 \times 10^{-4} \, moles \]

### Step 2: Determine the relationship between moles of Fe and moles of ferric alum
From the formula of ferric alum, we can see that:

\[ (NH_{4})_{2}SO_{4}Fe_{2}(SO_{4})_{3} \cdot 24 H_{2}O \]

indicates that 1 mole of ferric alum contains 2 moles of Fe. Therefore, the relationship is:

\[ 1 \, \text{mole of alum} \rightarrow 2 \, \text{moles of Fe} \]

### Step 3: Calculate the moles of ferric alum that can be produced
Since 2 moles of Fe are required to produce 1 mole of ferric alum, we can set up the following relationship:

\[ \text{Moles of alum} = \frac{1}{2} \times \text{Moles of Fe} \]

Substituting the moles of Fe we calculated:

\[ \text{Moles of alum} = \frac{1}{2} \times (1 \times 10^{-4}) = 0.5 \times 10^{-4} \, moles \]

### Final Answer
Thus, the number of moles of ferric alum that can be produced from \(0.0056 \, g\) of Fe is:

\[ 0.5 \times 10^{-4} \, moles \]