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 \(Al^{3+}\), we can follow these steps:

### Step 1: Determine the number of moles of \(Al^{3+}\)

To find the number of moles, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of aluminum (Al) is approximately 27 g/mol.

\[ \text{Number of moles of } Al^{3+} = \frac{4.5 \times 10^{-5} \text{ g}}{27 \text{ g/mol}} \]

Calculating this gives:

\[ \text{Number of moles of } Al^{3+} = \frac{4.5 \times 10^{-5}}{27} \approx 1.67 \times 10^{-6} \text{ moles} \]

### Step 2: Determine the number of electrons required for reduction

Each mole of \(Al^{3+}\) requires 3 moles of electrons to be reduced to aluminum (Al). Therefore, we can calculate the total moles of electrons required:

\[ \text{Moles of electrons} = 3 \times \text{Number of moles of } Al^{3+} \]

Substituting the value we found:

\[ \text{Moles of electrons} = 3 \times 1.67 \times 10^{-6} \approx 5.01 \times 10^{-6} \text{ moles of electrons} \]

### Step 3: Convert moles of electrons to number of electrons

To convert moles of electrons to the actual number of electrons, we multiply by Avogadro's number (\(6.022 \times 10^{23} \text{ electrons/mol}\)):

\[ \text{Number of electrons} = \text{Moles of electrons} \times 6.022 \times 10^{23} \]

Substituting the value we calculated:

\[ \text{Number of electrons} = 5.01 \times 10^{-6} \times 6.022 \times 10^{23} \]

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 \(Al^{3+}\) is approximately \(3.01 \times 10^{18}\) electrons.

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