The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum m

Question

3 years ago

7

The Hall process for the production of aluminum involves the reaction of aluminum oxide with elemental carbon to give aluminum m

etal 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?

Chemistry

2 answers:

topjm [15] · Answer 1 · 2022-06-21T12:39:46+03:00

Answer:

<u><em>1.7 × 10³ kg of ore.</em></u>

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.

I am Lyosha [343] · Answer 2 · 2022-06-21T14:43:16+03:00

Mass of aluminium ore = 1.629.10³ kg

<h3>Further eplanation </h3>

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