In the commercial preparation of aluminum,aluminum oxide `(Al_2O_3)` is electrolysed at `1000^(@)` C. How many coulombs of electricity are required to give 54kg of aluminum ? Assume following reaction takes place at cathode:
`Al^(3+)+3e^- \to Al`

Options

A. `17.3xx10^8`
B. `3.21xx10^7`
C. `1.82xx10^4`
D. `57.6xx10^7`

Answer

To find out how many coulombs of electricity are required to produce 54 kg of aluminum from aluminum oxide (Al₂O₃), we can follow these steps:

### Step 1: Determine the molar mass of aluminum (Al)
The molar mass of aluminum is approximately 27 g/mol.

### Step 2: Convert the mass of aluminum to grams
Given that we need to produce 54 kg of aluminum, we convert this to grams:
\[ 54 \text{ kg} = 54 \times 10^3 \text{ g} = 54000 \text{ g} \]

### Step 3: Calculate the number of moles of aluminum produced
Using the molar mass of aluminum, we can calculate the number of moles of aluminum:
\[ \text{Number of moles of Al} = \frac{\text{mass}}{\text{molar mass}} = \frac{54000 \text{ g}}{27 \text{ g/mol}} = 2000 \text{ mol} \]

### Step 4: Determine the number of electrons required
From the cathode reaction:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
Each mole of aluminum requires 3 moles of electrons. Therefore, for 2000 moles of aluminum:
\[ \text{Total moles of electrons} = 2000 \text{ mol} \times 3 = 6000 \text{ mol of electrons} \]

### Step 5: Calculate the total charge in coulombs
Using Faraday's constant, which is approximately \( 96500 \text{ C/mol} \), we can calculate the total charge:
\[ \text{Total charge (Q)} = \text{moles of electrons} \times \text{Faraday's constant} \]
\[ Q = 6000 \text{ mol} \times 96500 \text{ C/mol} = 57,600,000 \text{ C} \]

### Step 6: Final answer
Thus, the total charge required to produce 54 kg of aluminum is:
\[ Q = 57.6 \times 10^6 \text{ C} \]