The number of electrons required to reduce `4.5 xx 10^(–5)` g of `Al^(+3)` is :

Options

A. `1.03 xx 10^(18)`
B. `3.01 xx 10^(18)`
C. `4.95 xx 10^(26)`
D. `7.31 xx 10^(20)`

Answer

To solve the problem of determining the number of electrons required to reduce \(4.5 \times 10^{-5}\) g of \( \text{Al}^{3+} \), we can follow these steps:

### Step 1: Write the Reduction Reaction
The reduction of aluminum ion can be represented as:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This equation indicates that one mole of \( \text{Al}^{3+} \) requires 3 moles of electrons for reduction.

### Step 2: Determine the Molar Mass of Aluminum
The molar mass of aluminum (Al) is approximately 27 g/mol.

### Step 3: Calculate the Number of Moles of Aluminum
To find the number of moles of aluminum in \(4.5 \times 10^{-5}\) g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of Al} = \frac{4.5 \times 10^{-5} \text{ g}}{27 \text{ g/mol}} = 1.6667 \times 10^{-6} \text{ moles} \]

### Step 4: Calculate the Number of Electrons Required
Since each mole of \( \text{Al}^{3+} \) requires 3 moles of electrons, the total number of moles of electrons required can be calculated as:
\[ \text{Number of moles of electrons} = 3 \times \text{Number of moles of Al} \]
Substituting the value from Step 3:
\[ \text{Number of moles of electrons} = 3 \times 1.6667 \times 10^{-6} = 5.0001 \times 10^{-6} \text{ moles} \]

### Step 5: Convert Moles of Electrons to Number of Electrons
Using Avogadro's number (\(6.022 \times 10^{23}\) electrons/mol), we can find the total number of electrons:
\[ \text{Number of electrons} = \text{Number of moles of electrons} \times \text{Avogadro's number} \]
Substituting the values:
\[ \text{Number of electrons} = 5.0001 \times 10^{-6} \text{ moles} \times 6.022 \times 10^{23} \text{ electrons/mol} \]
Calculating this gives:
\[ \text{Number of electrons} \approx 3.01 \times 10^{18} \text{ electrons} \]

### Final Answer
The number of electrons required to reduce \(4.5 \times 10^{-5}\) g of \( \text{Al}^{3+} \) is approximately \(3.01 \times 10^{18}\) electrons.

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