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Topic: Unknown chloride  (Read 3316 times)

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Offline Rutherford

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Unknown chloride
« on: May 15, 2012, 12:01:37 PM »
In a solution of a chloride whose mass is 7.95g, Ag2SO4 is added. 23.55g of precipitate is made which is partially dissolved in ammonia. Write the formula of the chloride.
This seems though to me. I wrote the equation:
MCl+Ag2SO4 :rarrow: MSO4+AgCl
Can't balance it because I don't know if it is a mono-, di-... valent metal.
The precipitate is AgCl and MSO4 together I think. Thought that M is Ba but didn't fit when I checked the equation and the masses. What to do?

Offline Borek

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Re: Unknown chloride
« Reply #1 on: May 15, 2012, 02:23:59 PM »
Do you know any other insoluble sulfates?

Information given is enough to calculate equivalent mass of the element M, so there is no need for guessing.
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Offline Rutherford

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Re: Unknown chloride
« Reply #2 on: May 15, 2012, 02:31:01 PM »
How to use them if I can't balance the equation?

Offline Borek

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Re: Unknown chloride
« Reply #3 on: May 15, 2012, 02:34:55 PM »
Start assuming it is M+ - balance the reaction equation and calculate molar mass of M. If no element fits, check if assuming it is M2+ doesn't help - and so on.

Note that you don't have to solve the problem separately each time, if you calculate molar mass assuming it is M+, molar mass of M2+ is just twice larger, that's how equivalents work.
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Offline Rutherford

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Re: Unknown chloride
« Reply #4 on: May 15, 2012, 03:10:58 PM »
Done how you said and I got SrCl2. How can I assume that the molar mass is 2 times larger?

Offline Borek

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Re: Unknown chloride
« Reply #5 on: May 15, 2012, 03:41:02 PM »
Do you know what a chemical equivalent is?
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Offline Rutherford

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Re: Unknown chloride
« Reply #6 on: May 16, 2012, 03:12:59 AM »
Found out what it is, but how to use it here? I found some simple problems but don't know how to use it here.
« Last Edit: May 16, 2012, 03:29:49 AM by Raderford »

Offline Rutherford

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Re: Unknown chloride
« Reply #7 on: May 16, 2012, 03:36:00 AM »
I think I understood now, correct me if I am wrong. First I assume that it is a monovalent metal (it can react with 1 mole of H so I will calculate its equivalent mass). Then I balance the equation and the mass I get is the equivalent mass. Got 44g/mol but there isn't any metal with that mass. If I assume that it is a divalent metal (it can react with 2 moles of H), then the equivalent mass would be 2 times higher. Is it so?

Offline Borek

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Re: Unknown chloride
« Reply #8 on: May 16, 2012, 04:14:36 AM »
Yes.
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Offline Rutherford

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Re: Unknown chloride
« Reply #9 on: May 16, 2012, 04:17:39 AM »
Ok, thanks.

Offline AWK

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Re: Unknown chloride
« Reply #10 on: May 16, 2012, 05:24:43 AM »
In this case both sulfates of (Pb, Ba, Ca or Sr) and AgCl (which is soluble in amonia) are insoluble!
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