Question

The overall cell reaction for aluminum production is $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}\left(\text { in } \mathrm{Na}_{3} \mathrm{AlF}_{6}\right)+3 \mathrm{C}(\mathrm{graphite}) \longrightarrow 4 \mathrm{Al}(t)+3 \mathrm{CO}_{2}(g) $$ (a) Assuming \(100 \%\) efficiency, how many metric tons (t) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) are consumed per metric ton of Al produced? (b) Assuming \(100 \%\) efficiency, how many metric tons of the graphite anode are consumed per metric ton of Al produced? (c) Actual conditions in an aluminum plant require \(1.89 \mathrm{t}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(0.45 \mathrm{t}\) of graphite per metric ton of Al. What is the percent yicld of \(\mathrm{Al}\) with respect to \(\mathrm{Al}_{2} \mathrm{O}_{3} ?\) (d) What is the percent yield of Al with respect to graphite? (e) What volume of \(\mathrm{CO}_{2}\) (in \(\mathrm{m}^{3}\) ) is produced per metric ton of Al at operating conditions of \(960 .{ }^{\circ} \mathrm{C}\) and exactly 1 atm?

Short answer

1.89 tons of \( \mathrm{Al}_{2} \mathrm{O}_{3} \) are consumed per ton of \( \mathrm{Al} \). 0.33 tons of graphite are consumed per ton of \( \mathrm{Al} \). Percent yield with respect to \( \mathrm{Al}_{2} \mathrm{O}_{3} \) is 100\%; with respect to graphite 73.33\%. Volume of \( \mathrm{CO}_{2} \) produced is 2.81 \( \mathrm{m}^{3} \).

Step-by-step solution

- Understand the balanced reaction

The balanced reaction given is: \[ 2 \mathrm{Al}_{2} \mathrm{O}_{3} + 3 \mathrm{C} \rightarrow 4 \mathrm{Al} + 3 \mathrm{CO}_{2} \] This equation will be used to find the relationships between the reactants and products.

- Find the molar masses

Calculate the molar masses of \( \mathrm{Al}_{2} \mathrm{O}_{3} \), \( \mathrm{C} \), and \( \mathrm{Al} \): \( \mathrm{Al}_{2} \mathrm{O}_{3} \): 2*(27) + 3*(16) = 102 \mathrm{g/mol} \( \mathrm{C} \): 12 \mathrm{g/mol} \( \mathrm{Al} \): 27 \mathrm{g/mol}

- Calculate the mass ratio for Al production from Al2O3

Using the equation, 2 moles of \( \mathrm{Al}_{2} \mathrm{O}_{3} \) produce 4 moles of \( \mathrm{Al} \). 1000 kg of \( \mathrm{Al} \) corresponds to 1000/27 moles = 37.04 moles of \( \mathrm{Al} \). The molar ratio \( \mathrm{Al}_{2} \mathrm{O}_{3} \) to \( \mathrm{Al} \) is 1:2, so for 37.04 moles of \( \mathrm{Al} \), we need 37.04/2 = 18.52 moles of \( \mathrm{Al}_{2} \mathrm{O}_{3} \). Convert this to mass: 18.52 moles * 102 g/mol = 1888.96 g \( \approx 1.89 \) metric tons.

- Graphite consumption per ton of Al produced

Using the balanced equation, 3 moles of \( \mathrm{C} \) are needed for 4 moles of \( \mathrm{Al} \). Thus, for 37.04 moles of \( \mathrm{Al} \), 37.04 * (3/4) = 27.78 moles of \( \mathrm{C} \) are needed. Convert this to mass: 27.78 moles * 12 g/mol = 333.36 g \( \approx 0.33 \) metric tons.

- Calculate percent yield for Al2O3

Given actual conditions require 1.89 metric tons, percent yield is calculated as: Yield = (Theoretical/Actual) * 100% \( = \frac{1.89} {1.89} * 100\% \) = 100%

- Calculate percent yield for graphite

Given actual conditions require 0.45 metric tons, percent yield is calculated as: Yield = (Theoretical/Actual) * 100% \( = \frac{0.33} {0.45} * 100\% \) = 73.33%

- Volume of CO2 produced per metric ton of Al

Using the ideal gas law \(PV = nRT\): Calculate moles of \( \mathrm{CO}_{2} \) produced: From the balanced reaction, 3 moles of \( \mathrm{CO}_{2} \) for 4 moles of \( \mathrm{Al} \), hence for 37.04 moles of \( \mathrm{Al} \), moles of \( \mathrm{CO}_{2} \) = 37.04 * (3/4) = 27.78 moles. Use \( R = 0.0821 \mathrm{L\cdot atm/(mol\cdot K)} \), \( T = 960+273 = 1233 \mathrm{K} \), \( P = 1 \mathrm{atm} \)\[ V = \frac{nRT}{P} = \frac{27.78*0.0821*1233}{1} \approx 2811.67 \mathrm{L} = 2.81 \mathrm{m}^{3} \]

Key concepts

Electrochemical Reactions

In the context of aluminum production, electrochemical reactions are utilized to extract aluminum metal from its oxide, aluminum oxide (\text{Al\textsubscript{2}O\textsubscript{3}}). This process involves using an electrolytic cell where aluminum oxide is dissolved in molten cryolite (\text{Na\textsubscript{3}AlF\textsubscript{6}}). Electrochemical reactions involve the transfer of electrons between two species. In the electrolytic cell, \text{Al\textsubscript{2}O\textsubscript{3}} undergoes reduction to form aluminum metal at the cathode, and carbon (graphite) is oxidized at the anode to produce carbon dioxide (\text{CO\textsubscript{2}}). The balanced reaction is written as: \[ 2 \text{Al\textsubscript{2}O\textsubscript{3}} + 3 \text{C} \rightarrow 4 \text{Al} + 3 \text{CO\textsubscript{2}} \]The balanced equation indicates that for every 2 moles of \text{Al\textsubscript{2}O\textsubscript{3}}, 4 moles of aluminum and 3 moles of carbon dioxide are produced. This stoichiometric relationship is crucial for further calculations regarding reactant use and product formation.

Molar Mass Calculations

To understand the quantities involved in the reaction, calculating molar masses is essential. Molar mass is the mass in grams of one mole of a substance, and it helps convert between mass and moles. For the compounds in this reaction:

• Aluminum oxide (\text{Al\textsubscript{2}O\textsubscript{3}}): 2*(27) + 3*(16) = 102 g/mol
• Carbon (graphite) (\text{C}): 12 g/mol
• Aluminum (\text{Al}): 27 g/mol

These values allow us to determine how much of each reactant is needed to produce a certain amount of product. For instance, to find out how much \text{Al\textsubscript{2}O\textsubscript{3}} is needed to produce one metric ton (1000 kg) of aluminum, we convert the mass of aluminum to moles: \[ \text{Moles of Al} = \frac{1000 \text{ kg}}{27 \text{ g/mol}} = 37.04 \text{ moles} \]Then, using the stoichiometric ratio from the balanced equation: \[ \text{Moles of Al\textsubscript{2}O\textsubscript{3}} required} = \frac{37.04}{2} = 18.52 \text{ moles} \]Multiplying by the molar mass of \text{Al\textsubscript{2}O\textsubscript{3}} to find the mass: \[ \text{Mass of Al\textsubscript{2}O\textsubscript{3}} = 18.52 \text{ moles} * 102 \text{ g/mol} = 1888.96 \text{ g} \text{(or 1.89 metric tons)} \]

Percent Yield Calculation

Percent yield measures the efficiency of a reaction by comparing the actual yield to the theoretical yield. It is calculated using the formula: \[ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} * 100 \]Given the actual conditions in the aluminum plant:

• 1.89 metric tons of \text{Al\textsubscript{2}O\textsubscript{3}} per metric ton of aluminum (theoretical efficiency is 100%)
• 0.45 metric tons of graphite consumed per metric ton of aluminum produced

To calculate the percent yield with respect to \text{Al\textsubscript{2}O\textsubscript{3}}:\[ \text{Percent Yield} = \frac{1.89 \text{ (actual)}}{1.89 \text{ (theoretical)}} * 100 = 100\text{%} \]For graphite, where the theoretical consumption is 0.33 metric tons, the percent yield is:\[ \text{Percent Yield} = \frac{0.33 \text{ (theoretical)}}{0.45 \text{ (actual)}} * 100 \text{%} = 73.33 \text{%} \]This means that only about 73.33% of the graphite is used efficiently compared to the theoretical prediction.

Ideal Gas Law Applications

The Ideal Gas Law is used to relate the volume, pressure, temperature, and number of moles of a gas. The law is given by the equation: \[ PV = nRT \]Where:

• \( P \) - Pressure (in atm)
• \( V \) - Volume (in liters or cubic meters)
• \( n \) - Number of moles
• \( R \) - Ideal gas constant (0.0821 L·atm/(mol·K))
• \( T \) - Temperature (in Kelvin)

For calculating the volume of \text{CO\textsubscript{2}} produced per metric ton of aluminum: 1. From the balanced reaction, 3 moles of \text{CO\textsubscript{2}} are produced for every 4 moles of aluminum. For 37.04 moles of \text{Al}, the moles of \text{CO\textsubscript{2}} produced are:\[ n = 37.04 \text{ moles} * \frac{3}{4} = 27.78 \text{ moles} \]2. Given the temperature (T) is 960°C, convert to Kelvin:\[ T = 960 + 273 = 1233 \text{ K} \]3. Using the Ideal Gas Law to find the volume (V): \[ V = \frac{nRT}{P} = \frac{27.78 * 0.0821 * 1233}{1} \backapprox 2811.67 \text{ L} \backapprox 2.81 \text{ m\textsuperscript{3}} \]Thus, approximately 2.81 cubic meters of \text{CO\textsubscript{2}} gas are produced per metric ton of aluminum at the given conditions.