simple gravimetric
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#1
August 18th, 2006, 02:13
 redXI Junior Member Join Date: Aug 2005 Posts: 14
simple gravimetric

How many grams of NaCl are required to rpecipitate practically all the Ag ions from 2,5 x 10^2 mL of 0,0113 M AgNO3 Solution?? How about the net ionic equation for the reaction.

How do I it? I know the concepts fine but when the question is modified like this, I can't do it. Thanks for helping me out...I really need to study for my test.
:roll:
#2
August 18th, 2006, 07:02
 sdekivit Member Join Date: Jul 2005 Location: Holland Posts: 27

NaCl(s) + Ag(+) --> Na(+) + AgCl (s)

now you have the stoichiometry and you can calculate the amount of mol NaCl needed. Then you only need to convert that to gram.
#3
August 18th, 2006, 19:46
 redXI Junior Member Join Date: Aug 2005 Posts: 14

Quote:
 Originally Posted by sdekivit NaCl(s) + Ag(+) --> Na(+) + AgCl (s) now you have the stoichiometry and you can calculate the amount of mol NaCl needed. Then you only need to convert that to gram.
Okay. So I don't get it. It is AgNO3 that has volume and molarity then how is it possible for me calculate it? Where has that NO3 gone?? Sorry for asking too much
#4
August 19th, 2006, 00:38
 adrian Member Join Date: Apr 2006 Location: Bucharest, Romania Posts: 34
simple for others

You observed in the equation that you have mole/mole stoechiometry. Then your real problem is how many moles of AgNO3 you have in the solution.
The answer is: 2500 mL=2.5Lx0.0113=.. calculate yourself.
The result is exact No of NaCl moles you must use.
The last step is calculating the quantity of NaCl/ grams.
Concerning the "loss of NO3 ions": pls do not complicate the problem, is irrelevant.
#5
August 19th, 2006, 06:52
 sdekivit Member Join Date: Jul 2005 Location: Holland Posts: 27

Quote:
 Originally Posted by redXI Okay. Where has that NO3 gone??
You add solid NaCl to a solution containing Ag(+)-ions and NO3(-)-ions. since NaNO3(-) is soluble in water, it isn't necessary to write NO3(-) in the reaction equation, since NaNO3(aq) = Na(+)(aq) + NO3(-)(aq)

So we would get:

NaCl(s) + Ag(+) + NO3(-) --> Na(+) + NO3(-) + AgCl(s)

This is equal to:

NaCl(s) + Ag(+) --> Na(+) + AgCl(s)
#6
August 20th, 2006, 00:42
 redXI Junior Member Join Date: Aug 2005 Posts: 14
Re: simple for others

Quote:
 Originally Posted by adrian You observed in the equation that you have mole/mole stoechiometry. Then your real problem is how many moles of AgNO3 you have in the solution. The answer is: 2500 mL=2.5Lx0.0113=.. calculate yourself. The result is exact No of NaCl moles you must use. The last step is calculating the quantity of NaCl/ grams. Concerning the "loss of NO3 ions": pls do not complicate the problem, is irrelevant.
how do I calculate the quantity of NaCl/grams?
Divide the mass with molar mass of NaCl??

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