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Topic: Another Problem Regarding moles.  (Read 3031 times)

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Offline rannekk

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Another Problem Regarding moles.
« on: June 27, 2007, 10:17:18 AM »
Hello, me again. I have come across another question which I really can't seem to be able to get my head around. If you could help, I would be extremely happy. (not kidding, since im very frustrated at the moment).

Q:
A student was asked to find the relative atmic mass of an element X. Crystals of the choride of X were known to have the formula XCl2.6H2O.
The student dissolved 2.03g of the crystals in water, and then added an axcess of silver nitrate solution to the solution formed. A white precipitate of silver chloride was formed, which was filtered, dried and weighed. 2.87g were formed.

Ag+(aq) + Cl-(aq) -> AgCl (s)

   Calculate:
(a) the number of moles of silver chloride formed
(b) the number of moles of X chloride in the solution (you may assume that all the chlorine in X chloride is in solution as Cl-(aq) ions
(c) the mass of 1 mole of XCl2.6H2O
(d) the relative atmic mass of X.



I have already calculated (a), which was quite easy:
2.87/143.5 = 0.02mol
Yet i have not a clue how to start on (b)  ???

Thanks

Offline enahs

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Re: Another Problem Regarding moles.
« Reply #1 on: June 27, 2007, 10:54:34 AM »
You know for every 2 mole of AgCl formed there was one mole of XCL2.6H2O
You know that because there are two Cl's in the unknown and one in the AgCl, correct (and none from anywhere else) ?

Now that you know the number of moles, in a specific mass (what you started with) you then known the Molecular Weight (c). And knowing the molecular weight it is just addition/subtraction to find part (d).

Offline rannekk

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Re: Another Problem Regarding moles.
« Reply #2 on: June 27, 2007, 11:06:28 AM »
Ack, you are right. I didn't realize that for ever 2 mol 1 mole of crystal formed. This is the kind of bad thinking that gets me each time.  >:( will have to be alot more analytical in the future  :-\.
Thank you very much for the *delete me* It is very much appreciated

Offline rannekk

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Re: Another Problem Regarding moles.
« Reply #3 on: June 27, 2007, 12:12:46 PM »
aaah hang on . im a bit confused as the answer to B in the book is 0.01mol




edit:

Heh really have no idea about this question. Just cant seem to make ANY sence of it.

Offline enahs

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Re: Another Problem Regarding moles.
« Reply #4 on: June 27, 2007, 03:23:21 PM »
aaah hang on . im a bit confused as the answer to B in the book is 0.01mol




edit:

Heh really have no idea about this question. Just cant seem to make ANY sence of it.

0.01 mol is correct. That is half of what you said was the amount for (A).

If 1 mol of the unknown turns into 2 moles of the known, then 2 moles of the known turns into 1 mol of the unknown, other wise known as half in this case.


It is important you look at that concept and try and understand it from many viewpoints, working forward and backwards, as it is the essence of stoichiometry, something you will be doing a lot of.
http://www.shodor.org/unchem/basic/stoic/index.html

*note* You might see it better if you write out a complete balanced equation for this reaction.
And as always when learning, even if it is not needed but you have the information and ability, go ahead and write on the balanced equation, practice is good.

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