Question

The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum metal and carbon monoxide. If the yield of this reaction is 82% and aluminum ore is 71% by mass aluminum oxide, what mass of aluminum ore must be mined in order to produce 1.0 x 103 kg (1 metric ton) of aluminum metal by the Hall process?

Answer 1

Answer:

• 1.7 × 10³ kg of ore.

Explanation:

Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.

Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:

•  X         ×          71%                ×       82%     =     1.0 × 10³ kg

         ↑                      ↑                              ↑                     ↑

(mass of ore)    (% of Al in the ore)        (yield)        ( Al metal to obtain)

You must just simplify, solve and compute:

• 0.71 × 0.82 × X = 1,000

• X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg

Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.

Answer 2

Mass of aluminium ore = 1.629.10³ kg

Further eplanation

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

The Hall process:

Al₂O₃ + 3C ---> 2 Al + 3CO

To produce 1.0 x 10³ kg (1 metric ton) of aluminum metal(Al) and the yield of this reaction is 82% then

theoretical yield = An actual yield : percent yield

theoretical yield = 10³ : 0.82(82%)

theoretical yield = 1.219 .10³ kg = 1.219 .10⁶ gr mass of Al

atom mass Al = 27

mole Al = 1.219 .10⁶ : 27

mole Al = 4.5.10⁴

From reaction  :

mole Al₂O₃ : mole Al = 1 : 2 , so

mole Al₂O₃ = 0.5 x mole Al

mole Al₂O₃ = 0.5 x 4.5.10⁴

mole Al₂O₃ = 2.25.10⁴

Molecular mass  Al₂O₃ = 102 gr/mole

mass Al₂O₃ = mole x Molecular mass

mass Al₂O₃ = 2.25.10⁴ x 102

mass Al₂O₃ = 2.295.10⁶ gr

Because aluminum ore is 71% by mass aluminum oxide,then :

mass of aluminum ore = 0.71 (71%) x 2.295.10⁶ gr

mass of aluminum ore  = 1.629.10⁶ gr = 1.629.10³ kg

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Keywords: mole,The Hall process,mass of aluminum ore