Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on September 03, 2012, 12:48:45 PM
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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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Sounds like a ********* alloy
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Sounds like a ********* alloy
Please either post a full answer, or none - don't spoil the fun for others.
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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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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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Still, an informative link. Thanks.
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Still no correct answer though.
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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.
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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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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.
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Still no correct answer.
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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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Or is there another product (maybe a gas?) ?
None that I know about.
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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
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80:20 Al:Ga by weight.
And these nice, round numbers were what we were looking for :)