Solve Al2O3 to Produce 20g Al from 86% Bauxite Ore

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Discussion Overview

The discussion revolves around calculating the mass of bauxite ore required to produce a specific amount of aluminum metal, given the composition of the ore and the relevant chemical reactions. Participants explore different methods and calculations related to this stoichiometric problem.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents their calculation method for determining the mass of bauxite ore needed to produce 20 grams of aluminum, leading to a result of 119 grams, which they believe is incorrect based on the answer key.
  • Another participant suggests a different approach using ratios of aluminum to alumina and alumina to bauxite, implying a more systematic method for solving the problem.
  • A third participant references a similar problem with different parameters, applying their method to arrive at a mass of bauxite ore required for a large-scale aluminum production scenario, suggesting their approach is valid.
  • One participant asserts that 44 grams is the correct answer to the original question, acknowledging the original poster's reasoning but finding it difficult to follow.

Areas of Agreement / Disagreement

There is no consensus on the correct method or answer, as participants present differing calculations and reasoning. Some agree on the final answer being 44 grams, while others question the correctness of the original poster's approach.

Contextual Notes

Participants express uncertainty regarding the validity of their methods and the assumptions made in their calculations. There are also references to different compositions of bauxite ore in related problems, which may affect the calculations.

Who May Find This Useful

This discussion may be useful for students or individuals interested in stoichiometry, particularly in the context of aluminum production from bauxite ore, as well as those looking for different approaches to solving similar chemistry problems.

kuahji
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A bauxite ore 86.0 percent Al2O3 by weight. How many grams of the ore are required to produce 20.0 grams of aluminum metal? (Formula masses: Al2O3=102 & Al=27.0)

A) 12.3g, B) 32.5g, C) 37.8g, D) 43.9g

So this is how I did it on the test but got it marked wrong.
I reasoned take 1/.86=1.16 then 1.16(102g)=119g is the actual weight of the bauxite ore.

Then with the Al 2(27)x=20 x=.37 so that means you're going to need 37% of 1 119g of bauxite ore. The answer gave 44.0g.

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He showed to do it 2(27)/(2*27)+3(16))= .79
.79(.86)x=20g x=29.4g was the correct answer for the test.

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Now it showed up on review for the final but his answer is not on the multiple choice. Is the solution I did incorrect?
 
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Start with the target 20 grams of Aluminum metal. Work toward the unknown mass of bauxite. You arrange a chain of ratios:

Aluminum metal to alumina;
alumina to bauxite.
 
Thank you for taking the time to respond. Just found the exact same problem with different numbers on another page.

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 x 10^13 g of aluminum produced each year in the United States? (Assume 100% conversion.)

Using the same method I used in the first example
1/.5=2 => 2(102)=204
54/204=.26457
.26457x=5.0x10^13
x=1.9x10^14g (which website said was correct)

So I think I will go to him & say I think he was incorrect & go with my original assumption that the correct answer is D.
 
44 g (D) is the correct answer to the original question. Your approach gives correct result, although I must admit the reasoning is unusual and hard to follow.