Answer to Question #90601 in General Chemistry for karina

Question #90601
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 77.% and the yield of the second step is 67.%. Calculate the mass of aluminum hydroxide required to make 9.0kg of aluminum.

Be sure your answer has a unit symbol, if needed, and is rounded to 2 significant digits.
1
Expert's answer
2019-06-07T03:16:24-0400
"2Al(OH)_3 (s) \\rightarrow Al_2O_3 (s) + 3H_2O(g)"

"2Al_2O_3(s) + 3C(s) \\rightarrow 4Al(s) +3CO_2(g)"

Moles of aluminum:


"n= \\frac{m}{M} = \\frac {9000 g}{26.98\\frac{g}{mol}}= 333.58 mol"

"Yield \\% =\\frac{n_{actual}}{n_{thearetical}}\\times \\%"

"\\therefore n_{theoretical} = \\frac{n_{actual}\\times 100\\%}{yield \\%}= \\frac{333.58 mol\\times 100\\%}{67.0\\%}= 497.88 mol"

According to second equation mole ratio "n(Al_2O_3):n(Al)=2:4" then "n(Al_2O_3)= \\frac{n(Al)}{2} = \\frac{497.88 mol}{2} = 248.94 mol"



"n_{actual}(Al_2O_3)= 248.94 mol"

"n_{theoretical}(Al_2O_3)= \\frac{n_{actual}\\times 100\\%}{yield \\%} = \\frac{248.94 mol\\times 100\\%}{77.0\\%} = 323.3 mol"

According to the first equation mole ratio "n(Al(OH)_3 : n(Al_2O_3)= 2:1" ,then


"n(Al(OH)_3)= n(Al_2O_3)\\times 2 = 646.6 mol"

"m(Al(OH)_3)= n\\times M = 646.6 m\\times 78.00\\frac{g}{mol} = 50435 g = 50.435 kg = 50 kg"


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