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erik [133]
3 years ago
11

The refining of aluminum from bauxite ore (which contains 50.% Al2O3 by mass) proceeds by the overall reaction 2Al2O3 + 3C → 4Al

+ 3CO2. How much bauxite ore is required to give the 5.0  1013 grams of aluminum produced each year in the United States? (Assume 100% conversion.)
Chemistry
1 answer:
iragen [17]3 years ago
8 0

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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Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

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Mechanism C

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(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

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The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

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As:

K-1 [I]² = K1 [I2]:

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b) C = 0.50 J/(g°C)

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∴ m = 10.0 g

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Answer:

The answer to your question is letter B. 9

Explanation:

Unbalanced reaction

                     Al₂(SO₄)₃  +  Ca(OH)₂   ⇒    Al(OH)₃   +   CaSO₄

                     Reactants             Elements      Products

                          2                          Al                     1

                          3                           S                     1

                         14                           O                    7

                          1                           Ca                    1

                          2                           H                     3

Balanced reaction

                    Al₂(SO₄)₃  + 3Ca(OH)₂   ⇒   2Al(OH)₃   +   3CaSO₄

                     Reactants             Elements      Products

                          2                          Al                    2

                          3                           S                    3

                         18                           O                  18

                          3                           Ca                  3

                          6                            H                    6

The sum of the coefficients is 1 + 3+ 2+ 3 = 9

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3 years ago
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