Question

The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today. In the first step, aluminum hydroxide reacts to form alumina Al2O3 and water: 2AlOH3(s)→ Al2O3(s)+ 3H2O(g) In the second step, alumina Al2O3 and carbon react to form aluminum and carbon dioxide: 2Al2O3(s)+ 3C(s)→ 4Al(s)+ 3CO2(g) Suppose the yield of the first step is 73.% and the yield of the second step is 71.%. Calculate the mass of aluminum hydroxide required to make 1.0kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.

Answer 1

First Step:  2AlOH3(s)→ Al2O3(s)+ 3H2O(g)

Second Step:  2Al2O3(s)+ 3C(s)→ 4Al(s)+ 3CO2(g)

Mass of aluminium produced = 1.0 kg

Molar mass of Aluminium = 27 g/mol

Number of moles of Aluminium = 1.0 * 10^3/27 = 37.03 moles

Let the number of moles of Al(OH)3 needed be x

x moles of Al(OH)3 will produce 0.73*x/2 or 0.365*x moles of Al2O3 (Reason: 2 mole of Al(OH)3 produces one mole of Al2O3 and there is a 73% yield)

0.365*x moles of Al2O3, will produce 0.365 * x * 2 * 0.71 moles of Al, which is 0.5183*x moles of Al(OH)3

Moles of Al(OH)3 needed = 37.03/0.5183 = 71.4 moles

Mass of Al(OH)3 needed = number of moles * molar mass = 71.4 mol * 78 g/mol = 5569.2 g or 5.6 kg (in two significant digits)

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