- #1
kuahji
- 394
- 2
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?
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?