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 solve the question of how many moles of ferric alum \((NH_4)_2SO_4 \cdot Fe_2(SO_4)_3 \cdot 24H_2O\) can be made from a sample of iron (Fe) containing \(0.0056 \, g\), we will follow these steps:

### Step 1: Calculate the number of moles of Fe
We can calculate the number of moles of iron using the formula:

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

Given:
- Mass of Fe = \(0.0056 \, g\)
- Molar mass of Fe = \(56 \, g/mol\)

Substituting the values:

\[ \text{Number of moles of Fe} = \frac{0.0056 \, g}{56 \, g/mol} = 0.0001 \, moles \]

### Step 2: Determine the stoichiometry of Fe in ferric alum
From the formula of ferric alum \((NH_4)_2SO_4 \cdot Fe_2(SO_4)_3 \cdot 24H_2O\), we can see that there are 2 moles of Fe in 1 mole of ferric alum.

### Step 3: Calculate the number of moles of ferric alum produced
Using the stoichiometric relationship, we can find the number of moles of ferric alum that can be produced from the moles of Fe calculated in Step 1.

From the stoichiometry:
- 2 moles of Fe produce 1 mole of ferric alum.

Thus, the relationship can be expressed as:

\[ \text{Number of moles of ferric alum} = \frac{1}{2} \times \text{Number of moles of Fe} \]

Substituting the value from Step 1:

\[ \text{Number of moles of ferric alum} = \frac{1}{2} \times 0.0001 \, moles = 0.00005 \, moles \]

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

\[ 0.00005 \, moles \text{ or } 5 \times 10^{-5} \, moles \]