The extraction of aluminum metal from the aluminum hydroxide found in bauxite by

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 (Al203) and water: 2 Al(OH)3(s) ? Al2O3(s) + 3H20(g) In the second step, alumina (Al203 and carbon react to form aluminum and carbon dioxide: 2A1,03(5) + 3Ca) ? 4 Al(s) + 3CO2(g) Suppose the yield of the first step is 63% and the yield of the second step is 89.%. Calculate the mass of aluminum hydroxide required to make 2.0 kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits. x10

Explanation / Answer

mass of aluminium = 2 kg, atomic weight of Al= 27 g/mole, 1 kg= 1000gm

moles of Al= mass/atomic weight= 2*1000 gm/27 =74.1 moles

from the reaction -2, 2Al2O3+3(c) ----------->4Al(s)+ 3CO2 (g)

4 moles of Al requires 2 moles of Al2O3 for 100% completion

74.1 moles of Al requires 74.1*2/4 =37.05 moles of Al2O3 for 100% completion. but the reaction is only 89% complete. Hence Al2O3 required will be more and this moles are =37.05/0.89=41.63

from the 1st reaction, 2Al(OH)3 ---------->Al2O3+ 3H2O

1 mole of Al2O3 requires 2 mole of Al(OH)3 for 100% completion

41. 63 moles of Al2O3 requires 41.63*2= 83.26 moles of Al(OH)3

but the reaction is only 63% complete. Hence moles of Al(OH)3 required= 83.26/0.63 =132 moles of Al(OH)3

molar mass of Al(OH)3= 78 g/mole, mass of Al(OH)3= moles* molar mass= 132*78 gm =10296 gm

=10296 gm/1000 = 10.296 kg

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