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Topic: Is it correct my caculation?  (Read 10785 times)

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Offline avanwz

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Is it correct my caculation?
« on: March 02, 2008, 05:05:19 PM »
The question is
The Hall process for the production of aluminum involves the
reaction of aluminum oxide with elemental carbon to give aluminum metal and carbon
monoxide. If the yield of this reaction is 75%, what mass of aluminum metal can be
produced from the reaction of 1.65 ´ 106 of aluminum oxide with 1.50 ´ 106 g of
carbon?

And my caculation for this,

Because Al2O3 is a limited agent(by comparing mol of Al2O3 with 3C), the product 2Al also have to have the same mole as Al2O3.

Mole(Al2O3) = 1.62 X 10^4 = Mole(2Al)

Let suppose Mw(2Al) = 27 X 2 = 54 = ? gram / 1.62 X 10^4 mole

So, Grams of produced aluminum = 54 X 1.62 X 10^4 = 8.75 X 10^5 -> 75% is 6.56 X 10^5 gram

Is is correct?

Offline Borek

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Re: Is it correct my caculation?
« Reply #1 on: March 02, 2008, 05:17:34 PM »
6.56 X 10^5 gram

No idea what you did, but number seems OK - I got 6.55x105 g.
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