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Topic: Unknown Hgx(NO3)2 salt  (Read 4962 times)

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Offline Rutherford

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Unknown Hgx(NO3)2 salt
« on: August 04, 2012, 04:27:40 AM »
A piece of Al (m=3.55g) was thrown in 100ml of a Hgx(NO3)2 solution (c=0.5M). After the reaction ends the mass of the non-dissolved species was 17.65g. What was the unknown salt?

Wrote the equation first:
Al+3/2Hgx(NO3)2 :rarrow: Al(NO3)3+3x/2Hg
n(Al)=0.131mol
n(Hgx(NO3)2)=0.05mol
Al is in excess, n=0.131-0.05/(3/2)=0.098mol, m=0.098*27=2.65g
The non-dissolved species should be Al and Hg, mass of Hg is m=17.65-2.65=15g, then n=15/200.5=0.075mol.
From the reaction I see that from 0.05mol of Hgx(NO3)2, 0.05xmol of Hg is made, so 0.05x=0.075, x=1.5  ??? ?
Where I've mistaken?
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Offline Hunter2

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Re: Unknown Hgx(NO3)2 salt
« Reply #1 on: August 05, 2012, 07:38:11 AM »
Two errors

n=0.131-0.05/(3/2)=0.098mol is wrong

n=15/200.5=0.075mol. You missed the factor 1.5
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Offline Rutherford

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Re: Unknown Hgx(NO3)2 salt
« Reply #2 on: August 05, 2012, 08:03:22 AM »
Why is n=0.131-0.05/(3/2)=0.098mol wrong?
n=15/200.5=0.075mol I didn't miss any factor, it is the mass of Hg divided by the molar mass of Hg. Nothing to do with the reaction coefficients.
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Offline Hunter2

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Re: Unknown Hgx(NO3)2 salt
« Reply #3 on: August 05, 2012, 08:07:19 AM »
Maybe you have different calculator ;D.

My gives n = 0.131 -0.05(3/2) = 0.131 - 0.075 = 0,056 you got 0.098

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Offline Rutherford

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Re: Unknown Hgx(NO3)2 salt
« Reply #4 on: August 05, 2012, 08:28:04 AM »
It is 0.05/(3/2)=0.05*2/3=0.1/3=0.03333...
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Offline Hunter2

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Re: Unknown Hgx(NO3)2 salt
« Reply #5 on: August 05, 2012, 08:44:42 AM »
Yes correct
Probably its a mixture of Hg-I and Hg-II- nitrate.
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Offline Rutherford

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Re: Unknown Hgx(NO3)2 salt
« Reply #6 on: August 05, 2012, 08:53:28 AM »
Well, it isn't so that's why I am confused. It is Hg(NO3)2.
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Offline Hunter2

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Re: Unknown Hgx(NO3)2 salt
« Reply #7 on: August 05, 2012, 09:02:45 AM »
Probably the optained weights to inaccurate, also the use of atomic mass should use the real number. Or it is the truncation.  1.1-1.4 ~ 1, 1.5 - 1.9 ~ 2
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Offline Rutherford

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Re: Unknown Hgx(NO3)2 salt
« Reply #8 on: August 05, 2012, 09:21:48 AM »
I don't think that it is because of that.
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Offline AWK

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Re: Unknown Hgx(NO3)2 salt
« Reply #9 on: August 05, 2012, 09:51:32 AM »
This example has rather unpredictable results. At the excess of Al, Al in the presence of metallic Hg undergoes mercuric corrosion since Al/Hg amalgam reacts quite fast with water forming Al(OH)3. After a few hours you can expect a complete conversion of an excess of Al into Al(OH)3. Hence the solid residue may content Al, Al(OH)3 and Hg. Moreover H2 produced during decomposition of al amalgam may reduce nitrates up to ammonia that afterwards can react with Al nitrate forming more  more Al(OH)3.
This is reality!
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Offline Rutherford

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Re: Unknown Hgx(NO3)2 salt
« Reply #10 on: August 05, 2012, 10:05:15 AM »
So, there is no easy way to predict all that for solving the problem. Thanks for the explanation, I am giving up from this one.
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Offline AWK

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Re: Unknown Hgx(NO3)2 salt
« Reply #11 on: August 05, 2012, 10:53:18 AM »
Use copper salt instead of mercury salt and all be predictable.
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Offline Rutherford

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Re: Unknown Hgx(NO3)2 salt
« Reply #12 on: August 05, 2012, 10:58:34 AM »
That was the original problem, I can't change anything.
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Offline Borek

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Re: Unknown Hgx(NO3)2 salt
« Reply #13 on: August 06, 2012, 02:06:42 PM »
As far as I can tell none of the obvious reactions gives the correct answer.
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