Answer:
18.9 x 10¹³ grams of Bauxite Ore
Explanation:
Al₂O₃ = 50% of Bauxite Ore
Al₂O₃ = 0.5 (Bauxite Ore)--------------------------------------- (1)
Overall reaction:
2Al₂O₃ + 3C → 4Al + 3CO₂--------------------------------------- (2)
[ Al= 27 , O=16, C=12]
From  (2), 2 moles of  Aluminium oxide (Al₂O₃) gives  4 moles of Aluminium
In terms of grams, we can say:
Al₂O₃ = [2(27) +3(16)]
     = 54 +48
     =102grams
2 moles of Al₂O₃ = 2 x102grams
              =204grams
4 moles of Al = 4 x 27
           =108 grams
So from (2):
204 grams of  Al₂O₃  = 108 grams of  Aluminium
x grams of  Al₂O₃     =  5.0 x 10¹³grams of Aluminium
Calculating for x:
x =  (204 x  5.0 x 10¹³)/ 108
 = 9.44 x 10¹³ grams
So 9.44 x 10¹³ grams of pure bauxite (Bauxite) is required.
However the to calculate the quantity of raw bauxite, we use (1):
Bauxite ore = Pure Bauxite/0.5
          =  9.44 x 10¹³ grams/0.5
          =  18.88 x 10¹³ grams
          ≈ 18.9 x 10¹³ grams
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