The refining of aluminum from bauxite ore (which contains 50.% Al₂O₃ by mass) proceeds by the overall reaction 2Al₂O₃ + 3C $\rightarrow$ 4Al + 3CO₂. How much bauxite ore is required to give the 5.0*10¹³ g of aluminum produced each year in the United States? (Assume 100% conversion.)

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 2 moles of Al2O3 are required to produce 4 moles of Al. Therefore, the molar ratio of Al2O3 to Al is 2:4 or 1:2. The molar mass of Al2O3 is 102 g/mol, and the molar mass of Al is 27 g/mol. Therefore, the mass ratio of Al2O3 to Al is 102 g/mol / 27 g/mol = 3.8. To produce 5.0 × 10^13 g of Al, we need 3.8 × 5.0 × 10^13 g = 1.9 × 10^14 g of Al2O3. Since bauxite ore contains 50% Al2O3 by mass, we need 1.9 × 10^14 g / 0.50 = 3.8 × 10^14 g of bauxite ore.

The Refining of Aluminum from Bauxite Ore (Which Contains 50 →\rightarrow→

The Refining of Aluminum from Bauxite Ore (Which Contains 50 $\rightarrow$