The mass of carbon anode consumed (giving only carbon dioxide) in the production of `270 kg` of aluminium metal from bauxite by the Hall process is

Options

A. 180 kg
B. 270 kg
C. 540 kg
D. 90 kg

Explanation

To solve the problem of determining the mass of carbon anode consumed in the production of 270 kg of aluminum metal from bauxite by the Hall process, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The Hall process involves the electrolysis of aluminum oxide (Al₂O₃). The reactions at the cathode and anode are as follows:
- At the cathode: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \)
- At the anode: \( 2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^- \)

Additionally, the carbon anode reacts with oxygen to form carbon dioxide:
- \( C + O_2 \rightarrow CO_2 \)

2. **Determine Moles of Aluminum**:
The molar mass of aluminum (Al) is 27 g/mol. To find the moles of aluminum produced from 270 kg:
\[ \text{Moles of Al} = \frac{270,000 \text{ g}}{27 \text{ g/mol}} = 10,000 \text{ moles} \]

3. **Relate Moles of Carbon to Moles of Aluminum**:
From the stoichiometry of the reaction, 3/2 moles of carbon (C) are required to produce 2 moles of aluminum (Al). Therefore, we can express the moles of carbon needed in terms of moles of aluminum:
\[ \text{Moles of C} = \frac{3}{4} \times \text{Moles of Al} \]
Substituting the moles of aluminum:
\[ \text{Moles of C} = \frac{3}{4} \times 10,000 = 7,500 \text{ moles} \]

4. **Calculate the Mass of Carbon**:
The molar mass of carbon (C) is 12 g/mol. To find the mass of carbon consumed:
\[ \text{Mass of C} = \text{Moles of C} \times \text{Molar Mass of C} = 7,500 \text{ moles} \times 12 \text{ g/mol} = 90,000 \text{ g} \]
Converting grams to kilograms:
\[ \text{Mass of C} = \frac{90,000 \text{ g}}{1000} = 90 \text{ kg} \]

### Final Answer:
The mass of carbon anode consumed is **90 kg**.

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