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.)

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

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

what does the table look like