Taking into account the reaction stoichiometry, 163.67 grams of C is needed to reduce Al₂O₃ to produce 491 grams of pure aluminum.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
2 Al₂O₃ + 3 C → 4 Al + 3 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂O₃: 2 moles
- C: 3 moles
- Al: 4 moles
- CO₂: 3 moles
The molar mass of the compounds is:
- Al₂O₃: 102 g/mole
- C: 12 g/mole
- Al: 27 g/mole
- CO₂: 44 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 2 moles ×102 g/mole= 204 grams
- C: 3 moles ×12 g/mole= 36 grams
- Al: 4 moles×27 g/mole= 108 grams
- CO₂: 3 moles×44 g/mole= 132 grams
<h3>Mass of carbon required</h3>
The following rule of three can be applied: if 108 grams of Al are produced by 36 grams of C, 491 grams of Al are produced by how much mass of C?
mass of C= (491 grams of Al× 36 grams of C)÷ 108 grams of Al
<u><em>mass of C= 163.67 grams</em></u>
Finally, 163.67 grams of C is required.
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