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Topic: yield %  (Read 635 times)

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Offline magnus

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yield %
« on: February 09, 2019, 03:09:49 AM »
Hi,
from the reaction of 3 moles of fluorine (F₂) with 5 moles of hydrogen (H₂), 5.82 moles of HF
H₂ + F₂ -> 2HF

the stechimetric ratio is 1:1, and the limiting element is F₂.
The yield should be 58%  ???

Offline AWK

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Re: yield %
« Reply #1 on: February 09, 2019, 03:36:10 AM »
How many moles of HF you can obtain from 3 moles of F2?
AWK

Offline magnus

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Re: yield %
« Reply #2 on: February 09, 2019, 03:39:10 AM »
5.82 moles

Offline AWK

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Re: yield %
« Reply #3 on: February 09, 2019, 03:40:02 AM »
This is the real yield. I asked for the theoretical one. So the yield is ...?
AWK

Offline AWK

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Re: yield %
« Reply #4 on: February 09, 2019, 04:00:08 AM »
Sometimes we ask for the yield in reference to the specified reagent. But this should be clearly stated.
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Offline magnus

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Re: yield %
« Reply #5 on: February 09, 2019, 04:07:13 AM »
This is the real yield. I asked for the theoretical one. So the yield is ...?

14.55

Offline AWK

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Re: yield %
« Reply #6 on: February 09, 2019, 04:11:14 AM »
?
5H₂ + 3F₂ -> ?HF
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Offline magnus

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Re: yield %
« Reply #7 on: February 09, 2019, 04:18:09 AM »
3 moles of H₂ react with 3 moles of F₂ -> 6 moles HF according to the stoichiometric ratio

H₂ + F₂ -> 2HF

Offline AWK

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Re: yield %
« Reply #8 on: February 09, 2019, 04:20:37 AM »
This is a theoretical yield.
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Offline magnus

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Re: yield %
« Reply #9 on: February 09, 2019, 04:30:03 AM »
5.82/6 * 100= 97%

Offline AWK

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Re: yield %
« Reply #10 on: February 09, 2019, 04:31:10 AM »
Solved.
AWK

Offline magnus

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Re: yield %
« Reply #11 on: February 09, 2019, 04:48:07 AM »
Thanks

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