When electricity is passed through a solution of `AlCl_(3)` and `13.5g` of `Al` is deposited, the number of Faraday of electricity passed must be ………………….F.

Explanation

To solve the problem of how many Faraday of electricity are required to deposit 13.5 g of aluminum from an AlCl₃ solution, we can follow these steps:

### Step 1: Write the reduction reaction for aluminum
The reduction reaction for aluminum can be expressed as:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This indicates that 3 moles of electrons are required to reduce 1 mole of aluminum ion to aluminum metal.

### Step 2: Determine the molar mass of aluminum
The molar mass of aluminum (Al) is 27 g/mol. This means that 27 grams of aluminum corresponds to 1 mole of aluminum.

### Step 3: Calculate the number of moles of aluminum deposited
To find the number of moles of aluminum deposited when 13.5 g is produced, we can use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles of Al} = \frac{13.5 \, \text{g}}{27 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 4: Calculate the number of moles of electrons required
Since it takes 3 moles of electrons to deposit 1 mole of aluminum, the total number of moles of electrons required for 0.5 moles of aluminum is:
\[ \text{Moles of electrons} = 0.5 \, \text{mol Al} \times 3 \, \text{mol e}^- / \text{mol Al} = 1.5 \, \text{mol e}^- \]

### Step 5: Convert moles of electrons to Faraday
1 Faraday (F) is defined as the charge of 1 mole of electrons, which is approximately 96500 coulombs. Therefore, the number of Faraday required is equal to the number of moles of electrons:
\[ \text{Faraday} = 1.5 \, \text{mol e}^- = 1.5 \, F \]

### Final Answer
The number of Faraday of electricity passed must be **1.5 F**.

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