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
Atomic radius atomic number ionization energy and then electronegativity
Solar energy will not run out (unless the sun burns out), but the supply of fossil fuels we use is slowly depleting. Burning fossil fuel also releases greenhouse gas into the air, which is bad for the atmosphere.
Hey there!
The best way to balance chemical equations is to first start by balancing polyatomic ions such as OH and SO₄.
Next, balance other elements, but save elements that are by themselves for last, such as H₂ or Fe. Once you balance everything else you can do the ones by themselves, it's much easier.
Hope this helps!
