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
Answer:
b, their strong winds uproot many of the important plants
Explanation:
Answer:
E) None of the above
Explanation:
Metals react according to the activity series, thta is organized with the leats metals in in the bottom and the most reactive metals in the top. Most reactive metals displace less reactive metals in the compounds they are found.
in this series, Gold, Silver and Mercury are below Copper, this is, they are less reactive than Cu. this means none of them can displace it from its compounds.
Answer:
friend according to the question the answer is +3 ....
The given reaction is:
3Fe + 4H2O → Fe3O4 + 4H2
Given:
Mass of Fe = 354 g
Mass of H2O = 839 g
Calculation:
Step 1 : Find the limiting reagent
Molar mass of Fe = 56 g/mol
Molar mass of H2O = 18 g/mol
# moles of Fe = mass of Fe/molar mass Fe = 354/56 = 6.321 moles
# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles
Since moles of Fe is less than H2O; Fe is the limiting reagent.
Step 2: Calculate moles of Fe3O4 formed
As per reaction stoichiometry:
3 moles of Fe form 1 mole of Fe3O4
Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4
Step 4: calculate the mass of Fe3O4 formed
Molar mass of Fe3O4 = 232 g/mol
# moles = 2.107 moles
Mass of Fe3O4 = moles * molar mass
= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)
