Calculate the number of electrons required to deposit 41.5 g of Al

Explanation

To calculate the number of electrons required to deposit 41.5 g of aluminum (Al), we can follow these steps:

### Step 1: Write the half-reaction for aluminum deposition
The half-reaction for the reduction of aluminum ions to aluminum metal is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

This indicates that 3 moles of electrons are required to deposit 1 mole of aluminum.

...