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April 16, 2014

Homework Help: Chemistry 101

Posted by Gweedo8 on Monday, June 8, 2009 at 12:18pm.

I had a snag from a question I asked yesterday, here is the thread. (My snag being the final post)

I am trying to find the theoretical yield for this lab I am working on. The steps to finidng it would be great.

In my lab I took 40mls of H2O + 4g of AgNO3+ 8g of NaCL then I stirred it and filtered in a buchner funnel to be left with 6.7496g of silver chloride (AgCL)

The question I am asked is to find the percent yield which I know how to do, and since I have the actual yield I just cannot recall how to get the theoretical yield, do I do something with moles???

Responses

* Chemistry - DrBob222, Sunday, June 7, 2009 at 5:30pm

I assume your actual yield is 6.7496 g AgCl.
Theoretical yield:
AgNO3 + NaCl ==> AgCl + NaNO3.
Convert AgNO3 to moles. moles = g/molar mass.
Convert NaCl to moles.
Now you must determine which reagent (AgNO3 or NaCl0 is in excess. Since the reaction shows a 1:1 ratio for moles AgNO3 vs moles NaCl, the one with the fewer moles will be the limiting reagent. My guess is that will be AgNO3.
Then convert moles AgNO3 to moles AgCl and go from there to grams AgCl. That will be the theoretical yield. I did a quick calculation (but you need to check it out) and came out with about 3.4 g AgCl as theoretical yield. If I didn't goof, then you have something extra in your product for you have over 100% yeild and that's a no, no. My guess would be you didn't wash all of the NaNO3 and excess NaCl out of the product.


* Chemistry - Gweedo8, Monday, June 8, 2009 at 12:03pm

In my math I took 8g of AgNo3/1.69.8731 g/mol=.0471 mol of AgNo3

Then since there is only one AgNO3 and one AgCL, then AgCl's has a mol of .0471, multiply that by its molecular weight of 143.321 and I recieve 6.75 grams. not 3.4, where did I did I do something wrong?

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