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Author Topic: Theoretical/Percent Yield of Tin(II) Iodide  (Read 3664 times)

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Theoretical/Percent Yield of Tin(II) Iodide
« on: February 10, 2010, 10:40:42 AM »

Does anyone know...
How do you calculate percent yield of SnI2 in a two-part reaction...

Sn(S) + 2HCl-(L) → SnCl2(S) + H2+(G)
0.5463g Sn and 15mL of Hcl were added to get SnCl2. Then...

SnCl2(S) + 2KI-(L) → SnI2(S) +2KCl(S)
to the SnCl2, 2mL of KI were added.

I think that you need to figure out the theoretical yield of SnCl2, but then I don't understand what to do with the second reaction? Do you do another theoretical yield with the KI?


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Re: Theoretical/Percent Yield of Tin(II) Iodide
« Reply #1 on: February 10, 2010, 10:44:05 PM »

Yes, do the theoretical twice.

Note that you can do it in one step - 1 mole of Sn gives 1 mole of SnI2, no matter what intermediate steps are.
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