DIR Return Create A Forum - Home
---------------------------------------------------------
Chemistry:The Central Science
HTML https://kolathechemist.createaforum.com
---------------------------------------------------------
*****************************************************
DIR Return to: General chemistry discussion
*****************************************************
#Post#: 1360--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Michel
Date: December 2, 2012, 1:46 pm
---------------------------------------------------------
--- Quote from: Chemist@ link ---
>
> I tried this 2 times, two days ago and few days after it was
posted. The math just wouldn't hold up :-\.
>
--- End Quote ---
Good to see you're attempting it:) I'd like to post a few
hints, but like i said earlier, I want to use your attempts to
give hints...
What compounds have you tried? Or specifically, what anions have
you considered.?
#Post#: 1361--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Michel
Date: December 2, 2012, 1:50 pm
---------------------------------------------------------
--- Quote from: Shiva link ---
>
> Heating in the absence of air means, the product or one among
the reactants might react with water or easily oxidized by the
air. Still thinking lol damn.
>
> Shiva
>
--- End Quote ---
by 'heating in the absence of air', I mean heating in a vacuum
#Post#: 1362--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Shiva
Date: December 2, 2012, 1:56 pm
---------------------------------------------------------
No, what I meant is, as they need to be heated in the absence of
air, it might be a compound that would be oxidized in the air
easily or does react with water. Can you plz give me clue
something like one among the mixture is solid, or both are
solids?
Shiva
#Post#: 1363--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Michel
Date: December 2, 2012, 2:02 pm
---------------------------------------------------------
They are all solid inorganic compounds
#Post#: 1364--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Shiva
Date: December 2, 2012, 2:06 pm
---------------------------------------------------------
Okay, thanks.
#Post#: 1365--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Chemist@
Date: December 2, 2012, 2:32 pm
---------------------------------------------------------
Okay, I shall write my attempt.
-First I subtracted the 1/(23+35.5)mol of Na from 3.42*0.02mol
of Na to conclude that in the 2 other compounds there is
0.0513mol of Na.
-Then I thought about the metal hydroxyde. It should be Al(OH)3
so there is 0.0203mol of Al in the mixture.
-The gas X that is made from heating and addition of a strong
acid should be CO2 meaning that there is 0.0116mol of CO32- in
the mixture.
-The gas Y could be Cl2 or ClO2. The reaction in the first case
would be ClO3-+5Cl-+6H+→3Cl2+3H2O and in the second case
2ClO3-+4H++2Cl-→2ClO2+Cl2+2H2O (the hydrogen ions could be
possibly made from the hydrolysis of Al3+). When taking into
account the Cl- ions from NaCl, the first case gives a rational
result→0.0047mol of Cl-.
I tried to combine the results I wrote here to make some
reasonable compounds that could be in the mixture, but the math
wouldn't work. One of the problems is the very big quantity of
sodium.
#Post#: 1367--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Michel
Date: December 2, 2012, 3:54 pm
---------------------------------------------------------
--- Quote from: Chemist@ link ---
>
> Okay, I shall write my attempt.
> -First I subtracted the 1/(23+35.5)mol of Na from 3.42*0.02mol
of Na to conclude that in the 2 other compounds there is
0.0513mol of Na
>
--- End Quote ---
correct.
--- Quote from: Chemist@ link ---
>
> -Then I thought about the metal hydroxyde. It should be
Al(OH)3 so there is 0.0203mol of Al in the mixture.
>
--- End Quote ---
Not necessarily true. There are other possible metal hydroxides
that fit just as well.
#Post#: 1368--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Michel
Date: December 2, 2012, 4:12 pm
---------------------------------------------------------
--- Quote from: Chemist@ link ---
>
> -The gas Y could be Cl2 or ClO2. The reaction in the first
case would be ClO3-+5Cl-+6H+→3Cl2+3H2O and in the second
case 2ClO3-+4H++2Cl-→2ClO2+Cl2+2H2O (the hydrogen ions
could be possibly made from the hydrolysis of Al3+). When taking
into account the Cl- ions from NaCl, the first case gives a
rational result→0.0047mol of Cl-.
>
--- End Quote ---
Uhmm..Y is neither Cl2 nor ClO2
The mixture is a dry mixture, and I indicated when it was
dissolved in water(in the OP). It was mixed with dry
KClO3.(hence no H+). You've done well so far, in identifying the
carbonate ion, and the extra Na+ in the mix.
Hint: There are three inorganic compounds, they are all common
compounds, you've identified Na+, Na+, [Al3+ isn't present],
Cl-, CO32-. You need to (guess?) one more Cation and one more
anion...then work up the math...sounds a bit easy now. The last
anion is KEY to solving the problem. Finding the last anion
gives a big clue to gas Y.
Good luck;)
#Post#: 1374--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Shiva
Date: December 3, 2012, 6:34 am
---------------------------------------------------------
A quick clarification:
You said, 5.00g of the mix was heated in stoichiometric amount
of KClO3, is it the same mix the one heated with 20% Nacl?
Again, When 5.00g of the mix is dissolved in 20 ml water, the
solution is alkaline, is it the same mix again?
I'm bit confused with this question.
Shiva
#Post#: 1375--------------------------------------------------
Re: Problem of the week-19/11/12
DIR By: Michel
Date: December 3, 2012, 7:33 am
---------------------------------------------------------
Sorry for ambiguity. Imagine it's a mixture, and 5.00g taken out
of the mixture for each experiment.
*****************************************************
Page 2 of 4
DIR Previous Page
DIR Next Page