# Calculating Yield--need some help finishing problem. Almost done, just need help one ONE step, please help!!?

Well we did a lab on limiting reagent/reactants.

The balanced equation is Sr(NO3)2 + 2KIO3 --> Sr(IO3)2 + 2KNO3

Mols of Sr(NO3)2 used was .005023 mol and
Mols of KIO3 used was .01179 mol.

If I did this correctly my limiting reactant is SR(NO3)2 because I only have .005023 mols and to fully reacnt I need .005895 mols.

How do I find out the yield then. I don't understand the next step. This would be our theoretical yield. Our actual yield was 2.151 grams.

So I need to finish figuring out the theoretical yield to get the percent yield.

I've tried for HOURS now and gone through two textbooks, but I can't get it to work out!

If you know the limiting reagent, use that to calculate how much Sr(IO3)2 should be produced. That is the theoretical yield for Sr(IO3)2. You said you obtained 2.151 grams. Grams of what? Did you weigh Sr(IO3)2? If so, then that is the actual yield.

I was actually having trouble figure out to to calculate the SR(IO3)2 with the mole ratio. I think I got it though... hopefully after 4 hours of working on it I do.

0.005023 mols Sr(NO3)2 x (1 mol Sr(IO3)2/1 mol Sr(NO3)2) = 0.005023 mols Sr(IO3)2.

grams Sr(IO3)2 = mols Sr(IO3)2 x molar mass Sr(IO3)2 = 0.005023 x 437.43 = 2.1972 g = theoretical yield.

percent yield = [2.1972/2.151]x100 =

Two notes of caution here. Sr(IO3)3 isn't all that insoluble. You might want to look up the Ksp and calculate the solubility using the KIO3 mols as a common ion. That is to say that some of the " lower yield" may have been due to solubility of the Sr(IO3)2 and not to any procedural problem. Of course, 97.9% yield isn't all that bad but Ksp work is usually better than that. The second note of caution is that you are using molarities without making a correction for ionic strength of the solution. Using molarities is the usual way of working these problems in school. I hope this added material is of help to you.

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