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

[kuahji]

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.

--------------

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.

--------------
Now it showed up on review for the final but his answer is not on the multiple choice. Is the solution I did incorrect?

 

[symbolipoint]

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.

 

[kuahji]

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.

 

[Borek]

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.