A quantity of electrical charge that brigns about the depositiion of `4.5g Al` from `Al^(3+)` at the cathode will also produce the following volume `(STP)` of `H_(2)(g)` from `H^(o+)` at the cathode.

Options

A. `44.8L`
B. `22.4L`
C. `11.2L`
D. `5.6L`

Explanation

To solve the problem of determining the volume of hydrogen gas produced at STP from the deposition of 4.5 g of aluminum, we will use Faraday's laws of electrolysis and the concept of equivalent weights.

### Step-by-Step Solution:

**Step 1: Determine the Equivalent Weights**
- The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n}} \]
where \( n \) is the number of electrons transferred per ion.

- For Hydrogen (\( H^+ \)):
- Molar Mass of \( H = 1 \, \text{g/mol} \)
- \( n = 2 \) (since \( 2H^+ + 2e^- \rightarrow H_2 \))
- Equivalent Weight of Hydrogen:
\[ \text{Equivalent Weight of } H = \frac{1}{2} = 1 \, \text{g/equiv} \]

- For Aluminum (\( Al^{3+} \)):
- Molar Mass of \( Al = 27 \, \text{g/mol} \)
- \( n = 3 \) (since \( Al^{3+} + 3e^- \rightarrow Al \))
- Equivalent Weight of Aluminum:
\[ \text{Equivalent Weight of } Al = \frac{27}{3} = 9 \, \text{g/equiv} \]

**Step 2: Use Faraday's Second Law of Electrolysis**
- According to Faraday's second law:
\[ \frac{\text{Weight of } H}{\text{Weight of } Al} = \frac{\text{Equivalent Weight of } H}{\text{Equivalent Weight of } Al} \]

- Given that the weight of Aluminum is \( 4.5 \, \text{g} \):
\[ \frac{W_H}{4.5} = \frac{1}{9} \]

**Step 3: Calculate the Weight of Hydrogen**
- Rearranging the equation to find \( W_H \):
\[ W_H = \frac{1}{9} \times 4.5 \]
\[ W_H = 0.5 \, \text{g} \]

**Step 4: Convert the Weight of Hydrogen to Volume at STP**
- The molar mass of \( H_2 \) is \( 2 \, \text{g/mol} \).
- At STP, \( 1 \, \text{mol} \) of gas occupies \( 22.4 \, \text{L} \).
- The number of moles of \( H_2 \) produced:
\[ \text{Moles of } H_2 = \frac{0.5 \, \text{g}}{2 \, \text{g/mol}} = 0.25 \, \text{mol} \]

- The volume of \( H_2 \) at STP:
\[ \text{Volume} = \text{Moles} \times 22.4 \, \text{L/mol} \]
\[ \text{Volume} = 0.25 \times 22.4 = 5.6 \, \text{L} \]

### Final Answer:
The volume of \( H_2(g) \) produced at STP is **5.6 liters**.