Question

pt 1: How much aluminum is produced from 3350 kg of Al2O3?

pt2: The oxygen produced in Part 1 is allowed to react with carbon to produce carbon monoxide. Write a balanced equation describing the reaction of alumina with carbon. You need not include the states of matter in the balanced equation.

pt3: How much CO is produced from the alumina in Part 1?

Answer 1

In the problem, molar mass of Alumina=101.96g/mol

Molar mass of aluminum = 26.981539 amu

Molar mass of Carbon monoxide = 28.01g/mol

All these values are taken from wikipedia.

Answer 2

Part 1: Aluminum Production from Al₂O₃

Balanced Equation (Alumina Electrolysis):

$$2{\text{Al}}_2{\text{O}}_3\to 4\text{Al}+3{\text{O}}_2$$

Given:

• Mass of Al₂O₃ = 3350 kg

• Molar mass of Al₂O₃ = 102 g/mol

• Molar mass of Al = 27 g/mol

• 
  

Steps:

1. Moles of Al₂O₃:

   $$\text{Moles}=\frac{3350\times 10^3\text{g}}{102\text{g/mol}}=32,843\text{mol}$$

2. Moles of Al produced (2:4 ratio):

   $$\text{Moles of Al}=32,843\times \frac{4}{2}=65,686\text{mol}$$

3. Mass of Al:

   $$\text{Mass}=65,686\text{mol}\times 27\text{g/mol}=1,773,522\text{g}=\ast \ast 1,774\text{kg}\ast \ast$$

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Part 2: Balanced Equation (O₂ + C → CO)

$$2\text{C}+{\text{O}}_2\to 2\text{CO}$$

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Part 3: CO Production from Alumina

From Part 1:

• Moles of O₂ produced (2:3 ratio):

  $${\text{Moles of O}}_2=32,843\times \frac{3}{2}=49,265\text{mol}$$

From Part 2 (1:2 ratio for O₂:CO):

1. Moles of CO:

   $$\text{Moles of CO}=49,265\times 2=98,530\text{mol}$$

2. Mass of CO (molar mass = 28 g/mol):

3. 
   

   $$\text{Mass}=98,530\text{mol}\times 28\text{g/mol}=2,758,840\text{g}=\ast \ast 2,759\text{kg}\ast \ast$$

4. 
   

Answers:

1. Aluminum produced: 1,774 kg

2. Balanced equation: $2\text{C}+{\text{O}}_2\to 2\text{CO}$

3. CO produced: 2,759 kg