# Chem word problem

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.

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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?

Homework Helper
Gold Member
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.

Mentor
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.