Bauxite, the principal source of aluminum oxide, contains about 55% Al2O3(s) by

Question

Bauxite, the principal source of aluminum oxide, contains about 55% Al2O3(s) by mass.

(a) How much bauxite is required to produce the 2.0 million metric tons of aluminum metal pro-duced each year by electrolysis?

(b) Estimate the amount of energy needed to pro-duce this much aluminum per year using Hall-Héroult process if the operating voltage of a typi-cal aluminum generating cell is 5.0 V.

(c) What fraction of the 13 × 1015 kJ of electrical energy generated annually in the United States does this represent?

Explanation / Answer

Mass percent of Al2O3 (s) is 55 %

Mass of Al = 2.0 million metric ton

We calculate mol of Al by using molar mass of Al and by using mol ratio of Al2O3 to Al we get moles of Al2O3 and by using we get its mass.

Once we get mass of Al2O3 then by using its percent we get mass of Bauxite.

Calculation of Al in g

Mass of Al in g = 2.0 E 6 ton * (1000 kg / 1 ton ) * ( 1000 g / 1 kg)

= 2.0E12 g Al

Mol Al = 2.0E12 g Al * 1 mol Al/ 26.982 g

= 7.41 E10 mol

1 mol Al2O3 contains 2 mol Al

Mol of Al2O3 required to produce 7.41 E10 mol Al

= 7.41 E10 mol Al * 1 mol Al2O3 / 2 mol Al

= 3.71 E10 mol Al2O3

Mass of Al2O3

= 3.71 E10 mol Al2O3 * 101.961 g /mol

= 3.78 E12 g

Mass of Bauxite

= 3.78 E12 g * 100 Bauxite / 55 g bauxite

= 6.87 E 12 g bauxite

Mass of bauxite required = 6.87 E12 g

b). Volt =5.0 V

We know E = V * Q

We know volt, we calculate Q

1 mol Al extraction needs 3 mol electrons and

1 mol electrons needs 96500 C charge

Charge needed = 7.41 E10 mol * (3 mol e/ 1 mol Al) * (96500 C / 1 mol e)

= 2.15E16 C

E = 5.0 V * 2.15 E16 C

= 1.07 E17 V.C

= 1.07E17 J

c). For this we convert calculated energy into kJ

= 1.07 E17 J * 1kJ / 1000 J

= 1.07E14 kJ

Fraction of energy = 1.07E14 / 13E15 kJ

= 0.0082

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