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redXI
Regular
Joined: 12 Aug 2005
Posts: 20
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 Posted: Fri Aug 18, 2006 7:13 am Post subject: simple gravimetric |
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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.
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sdekivit
Regular
Joined: 26 Jul 2005
Posts: 37
Location: Holland
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| Posted: Fri Aug 18, 2006 12:02 pm Post subject: |
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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.
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redXI
Regular
Joined: 12 Aug 2005
Posts: 20
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| Posted: Sat Aug 19, 2006 12:46 am Post subject: |
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| sdekivit wrote: |
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 
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adrian
Regular
Joined: 20 Apr 2006
Posts: 44
Location: Bucharest, Romania
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| Posted: Sat Aug 19, 2006 5:38 am Post subject: simple for others |
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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.
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sdekivit
Regular
Joined: 26 Jul 2005
Posts: 37
Location: Holland
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| Posted: Sat Aug 19, 2006 11:52 am Post subject: |
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| redXI wrote: |
| 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)
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redXI
Regular
Joined: 12 Aug 2005
Posts: 20
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| Posted: Sun Aug 20, 2006 5:42 am Post subject: Re: simple for others |
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| adrian wrote: |
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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