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Topic: Problem of the week - 03/09/2012  (Read 10805 times)

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Offline Borek

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Problem of the week - 03/09/2012
« on: September 03, 2012, 12:48:45 PM »
Several years ago researchers at one of the US universities produced an alloy, that can revolutionize our ways of producing and storing energy. Compared with alkali or alkaline earth metals alloy is quite safe to store and use. When 5.00 g pellets of the alloy were put into water, they reacted vigorously creating hydrogen and heat (but the mixture didn't get hot enough to ignite the hydrogen). Solid remains were dissolved in a strong acid, producing water and even more hydrogen. Total volume of the produced hydrogen was 5.940 L (measured at 20°C and 0.9876 atm). What was the exact composition of the alloy?
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Offline Sophia7X

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Re: Problem of the week - 03/09/2012
« Reply #1 on: September 03, 2012, 01:53:18 PM »
Sounds like a ********* alloy
« Last Edit: September 03, 2012, 02:00:43 PM by Borek »
Entropy happens.

Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #2 on: September 03, 2012, 02:01:32 PM »
Sounds like a ********* alloy

Please either post a full answer, or none - don't spoil the fun for others.
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Offline Sophia7X

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Re: Problem of the week - 03/09/2012
« Reply #3 on: September 03, 2012, 02:50:15 PM »
Sorry! I was a bit tentative to post a full answer because I wasn't sure of the method I used to arrive at the answer.

Anyway, I got 0.244 mol H2 with PV=nRT, multiply by (2 mol Group 3 Metal/3 mol H2) because I thought it could be a group 3 metal which are capable of producing H2 in water (2 M + 6 H2O --> 2 M(OH)3 + 3H2) , if not group 1 or 2.  5.00 g / 0.163 mol H2 = 30.7 g/mol, which is close to the molar mass of aluminum. Since even more H2 is produced in acid, gallium is probably also present (~8.79%) since it doesn't react in water to produce hydrogen, but it can react in something like sulfuric acid. Also, aluminum doesn't react with water by itself because of its oxide coating so gallium could slow down the oxide formation.
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Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #4 on: September 03, 2012, 03:20:45 PM »
Your logic is sound, your result is off.

See http://www.greencarcongress.com/2007/08/purdue-research.html (it would be better to find the original paper, but I occupied with something else at the moment).
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Offline Sophia7X

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Re: Problem of the week - 03/09/2012
« Reply #5 on: September 03, 2012, 09:07:30 PM »
Still, an informative link. Thanks.
Entropy happens.

Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #6 on: September 05, 2012, 07:20:57 AM »
Still no correct answer though.
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Re: Problem of the week - 03/09/2012
« Reply #7 on: September 05, 2012, 10:05:03 AM »
I though the problem is finished...

so there is 0.240328mol of H2 liberate by PV=nRT

(Taking its Ga and Al-anyway it seems to be okay because it's group III: so reaction is

M + 3H20---> 1.5H2 + M(OH)3 or M + 3HCl---> 1.5H2 + MCl3 (just take acid as HCl then its easier to calc...)

If its totally Al then there will be 4.16g of the alloy, if its Ga then 11.05g of alloy. So it makes cents as 5 is between. (though it doesn't make dollars haha!)

Take there is X mol of Al, Ymol of Ga.

Then 26.98X+69.72Y=5.00,
1.5(X +Y) = 0.240328

solving stimutaneous equations X=0.14437, Y=0.015847

So there's 0.14437mol of Al (3.8951g) and 0.015847mol of Ga (1.1049g)

I hope this is correct... Whee this is so fun! Will definitely look out for the next one!  ;D

EDIT: Which seems like it corresponds quite nicely to the 30-70 ratio stated in the paper.

Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #8 on: September 05, 2012, 10:12:27 AM »
I got 0.244 mol H2 with PV=nRT

so there is 0.240328mol of H2 liberate by PV=nRT

One of you must be wrong.
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Offline 123456789

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Re: Problem of the week - 03/09/2012
« Reply #9 on: September 05, 2012, 10:29:00 AM »
One of you must be wrong.

I think it has something to do with the significant figures. This time I bundled everything together and got another value.

PV/T1=PV/T2

0.9876 x 5.940/293=V/298
V=5.966452259L (okay everything from calculator-over precise...)
Then V/24 as it is at rtp so no. of mol = 0.24860mol -yet another answer. The first time I calculated I took everything to 5sf. I think that's the difference.

Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #10 on: September 10, 2012, 06:36:58 AM »
Still no correct answer.
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Offline Schrödinger

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Re: Problem of the week - 03/09/2012
« Reply #11 on: September 10, 2012, 08:48:00 AM »
Upon reaction with water, I assume the only products are those 'solid remains' and hydrogen. Or is there another product (maybe a gas?) ?
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Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #12 on: September 10, 2012, 09:14:45 AM »
Or is there another product (maybe a gas?) ?

None that I know about.
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Offline DrCMS

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Re: Problem of the week - 03/09/2012
« Reply #13 on: September 10, 2012, 12:14:04 PM »
80:20 Al:Ga by weight.

5.940L at 20°C and 0.9876atm = 0.243871 moles Hydrogen.  So there were 0.16258 moles of group III metal, 5g in total works out to 4.000g Al and 1.000g Ga

Offline Borek

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Re: Problem of the week - 03/09/2012
« Reply #14 on: September 17, 2012, 12:56:52 PM »
80:20 Al:Ga by weight.

And these nice, round numbers were what we were looking for :)
« Last Edit: September 18, 2012, 04:10:28 PM by Borek »
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