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Author: Subject: Calculating Theoretical Yield
ech310n
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Calculating Theoretical Yield

I've been wanting to know how to calculate the theoretical yield of a reaction for sometime. I have recently had enough time to learn this by using whatever sources I can find. Please take note that this IS NOT a homework question. Now could someone please kindly tell me if this is correct?

H2SO4 + 2KOH > K2SO4 + 2H2O
0.267 moles of KOH react with 0.165 moles of H2SO4 to yield 0.165 moles of K2SO4 and 0.33 moles of H2O. The H2SO4 is the limiting reagent in this reaction. Theoretical yield should then be:

K2SO4 – 0.165 moles x 174g = 28.71g
H2O – 0.33 moles x 18g = 5.94g
bogus poster
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No.
1 mol H2SO4 reacts with 2 mole KOH to yield 1 mol K2SO4 and 2 mole H2O

Now you can insert the gram values of the compounds:

98,08 gram H2SO4 react with 112,21 gram KOH (2 x 56,11g) to yield 174,25 gram K2SO4 and 36,03 gram H2O (2 x 18,02g)

So if you start with 10 gram H2SO4 this equals 0,10 mole H2SO4 (10g/98,08g). You will get as shown before 0,10 mole K2SO4. So you will yield 17,43 gram K2SO4 (174,25g x 0,10).

(what looks like mistakes is caused by the rounding of the numbers )

Ok?

A mole calculator like the "Free Formula Wizard" is helpful here.
Geomancer
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Why do you assume that H2SO4 is limiting? What reagent is limiting is not a function of the reaction, it is a function of the amounts of the reagents you have. In particular, the limiting reagent is the one that is not in excess, that is, it is used up completely at the end of the reaction. In your example:
You have .165mol of H2SO4. Given the opportunity, this will react with twice as many moles of KOH.
.165 H2SO4+.33 KOH -> .165 K2SO4 + .33 H2O

But there are only .267mol of KOH present. This is not enough to use up all the H2SO4, so H2SO4 is not limiting.

Conversely, you have .267mol of KOH. Given the opportunity, this will react with half as much H2SO4.
.267.267 KOH + .134 H2SO4 -> .134 K2SO4 + .267 H2O

Clearly, there is more than enough ("excess" H2SO4 present, and the KOH will be used up completely. Therefore, the KOH is limiting in this situation.

Combining equations we get:
.267 KOH + .134 H2SO4 -> .134 K2SO4 + .267 H2O
.31 H2SO4 -> .31 H2SO4
.267 KOH + .165 H2SO4 -> .134 K2SO4 + .267 H2O + .31 H2SO4

Note that, since it was present in excess, there is some H2SO4 left over.

The following statements are equivalent for simple reactions:

• Reagent R is limiting
• Reagent R is used up completely
• Reagent R is not in excess
ech310n
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Thanks for your replies Geomancer and bogus poster, you both have cleared up a lot of things for me. The only sources of information I could find on calculating theoretical yield were either too vague or very hard to understand.

[Edited on 17-4-2004 by ech310n]

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