How many coulombs are required to deposit `40.5` g of aluminium when the electrode reaction is :
`Al^(3+) + 3e^(-) to Al`

Explanation

To determine how many coulombs are required to deposit 40.5 g of aluminium using the reaction \( \text{Al}^{3+} + 3e^{-} \rightarrow \text{Al} \), we can follow these steps:

### Step 1: Determine the moles of aluminium
First, we need to calculate the number of moles of aluminium in 40.5 g. The molar mass of aluminium (Al) is approximately 27 g/mol.

\[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{40.5 \, \text{g}}{27 \, \text{g/mol}} = 1.5 \, \text{mol} \]
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