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 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.

Answer 1

$$2Al(OH)_3 (s) \rightarrow Al_2O_3 (s) + 3H_2O(g)$$

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

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

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

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