Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

The mass of carbon anode consumed (giving only \(\mathrm{CO}_{2}\) ) in the production of \(270 \mathrm{~kg}\) of aluminium metal from bauxite by the Hall process is (atomic mass of \(\mathrm{Al}=27\) ) (a) \(180 \mathrm{~kg}\) (b) \(270 \mathrm{~kg}\) (c) \(145 \mathrm{~kg}\) (d) \(90 \mathrm{~kg}\)

Step-by-Step Solution

Answer

The mass of carbon anode consumed is 90 kg.

Step 1: Understand the Chemical Reaction

In the Hall process, aluminum oxide ({ ext{Al}_2 ext{O}_3}) is electrolyzed to produce aluminum ( ext{Al}) and carbon dioxide ( ext{CO}_24). The balanced equation for the reaction is: 4{ ext{Al}_2 ext{O}_3} + 3 ext{C} ightarrow 4 ext{Al} + 3 ext{CO}_2}

Step 2: Calculate Moles of Aluminum Produced

To produce 270 kg of aluminum, convert the mass to grams:270 kg = 270,000 g.Next, calculate the moles of aluminum produced using its molar mass (27 g/mol):\[\text{Moles of } \text{Al} = \frac{270,000}{27} = 10,000 \text{ moles}\]

Step 3: Relate Moles of Aluminum to Moles of Carbon

According to the balanced equation, 4 moles of 4{ ext{Al}} are produced from 3 moles of ext{C}. Thus, the moles of carbon consumed can be calculated as:\[\text{Moles of } \text{C} = \frac{3}{4} imes 10,000 = 7,500 \text{ moles}\]

Step 4: Calculate Mass of Carbon Consumed

The molar mass of carbon is 12 g/mol. Calculate the mass of carbon consumed:\[\text{Mass of } \text{C} = 7,500 \times 12 = 90,000 \text{ g} = 90 \text{ kg}\]

Step 5: Select the Correct Answer

Compare the mass of carbon consumed calculated (90 kg) with the given options: (a) 180 kg, (b) 270 kg, (c) 145 kg, (d) 90 kg. The correct answer is (d) 90 kg.

Key Concepts

Electrolysis of Aluminum Oxide

Electrolysis is a chemical process that uses electricity to drive a non-spontaneous reaction. In the context of aluminum production, the Hall process employs electrolysis to extract aluminum from aluminum oxide (Al₂O₃). This oxide, also known as alumina, is dissolved in molten cryolite to lower its melting point, making it easier to perform electrolysis.

Inside an electrolytic cell, an electric current is passed through the molten alumina-cryolite mixture. This current causes the aluminum ions to migrate to the cathode, where they gain electrons to form metallic aluminum. Simultaneously, oxygen ions move toward the anode, where they release electrons and form oxygen gas. This oxygen then reacts with the carbon anodes in the cell to produce carbon dioxide gas.

The Hall process is highly energy-intensive due to the need for high temperatures and substantial electricity to maintain the electrolysis reaction. Despite the energy consumption, it remains the primary method for aluminum production worldwide due to its efficiency in converting alumina into aluminum.

Balanced Chemical Equation

A balanced chemical equation accurately represents the stoichiometry of a chemical reaction. It reflects the conservation of mass, where the number of atoms of each element is the same on both sides of the equation. In the Hall process, the reaction can be represented as:

• 4Al₂O₃ + 3C → 4Al + 3CO₂

When balancing a chemical equation, we ensure that the quantities of reactants and products correspond to their stoichiometric coefficients.

In the given equation:

• 4Al₂O₃ indicates that four formula units of aluminum oxide are decomposed.
• 3C represents three atoms of carbon (the anode material).
• 4Al signals the formation of four aluminum atoms.
• 3CO₂ reveals that three molecules of carbon dioxide are produced.

Balancing ensures that each step in the calculation corresponds properly to the theoretical yields described by the equation.

Stoichiometry in Chemical Reactions

Stoichiometry is the method we use to calculate the quantities of reactants and products in chemical reactions. It allows us to make accurate predictions about how much of each substance is consumed or produced. In our balanced equation, stoichiometry plays a critical role when determining how much carbon is consumed to produce a specific amount of aluminum.

The stoichiometric coefficients from the balanced equation help us understand the proportional relationship:

• To produce 4 moles of aluminum, 3 moles of carbon are needed.

Suppose we start with 10,000 moles of aluminum, as in the example. Stoichiometry tells us how to calculate the number of moles of carbon consumed:

• Moles of C = (3/4) × 10,000 = 7,500 moles

Through stoichiometry, we can efficiently use reactants and optimize the production process.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is a vital concept for converting between mass and moles in chemical equations. Molar mass calculations allow us to determine how much of a substance is being used or produced in a reaction.

For aluminum, the molar mass is 27 g/mol. We use this to convert the mass of aluminum produced into moles:

• Moles of Al = 270,000 g / 27 g/mol = 10,000 moles

For carbon, the molar mass is 12 g/mol. From the moles of carbon calculated using stoichiometry, we then determine the mass:

• Mass of C = 7,500 moles × 12 g/mol = 90,000 g = 90 kg

Molar mass calculations provide the necessary link between theoretical stoichiometric relationships and real-world chemical quantities.