The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum metal and carbon monoxide.If the yield of this reaction is 75%, what mass of aluminum metal can be produced from the reaction of 1.65 × 10⁶ g of aluminum oxide with 1.50 × 10⁶ g of carbon?

Question

Quizplus · Accepted Answer

First, we must determine the limiting reactant:

Al2O3 + 3C → 2Al + 3CO

1.65 × 10^16 g Al2O3 × (1 mol Al2O3 / 101.96 g Al2O3) × (2 mol Al / 1 mol Al2O3) × (26.98 g Al / 1 mol Al) = 8.74 × 10^15 g Al

1.50 × 10^16 g C × (1 mol C / 12.01 g C) × (2 mol Al / 3 mol C) × (26.98 g Al / 1 mol Al) = 2.54 × 10^15 g Al

Therefore, carbon is the limiting reactant. The maximum amount of Al that can be produced is:

2.54 × 10^15 g Al × (1 / 0.75) = 3.39 × 10^15 g Al or 3.39 × 10^12 kg Al or 6.6 × 10^15 lbs Al

Answer: D.

The Hall Process for the Production of Aluminum Involves the Reaction