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Author Topic: yield %  (Read 374 times)

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magnus

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yield %
« on: February 08, 2019, 09:09:49 PM »

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%  ???
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AWK

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Re: yield %
« Reply #1 on: February 08, 2019, 09:36:10 PM »

How many moles of HF you can obtain from 3 moles of F2?
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magnus

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Re: yield %
« Reply #2 on: February 08, 2019, 09:39:10 PM »

5.82 moles
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Re: yield %
« Reply #3 on: February 08, 2019, 09:40:02 PM »

This is the real yield. I asked for the theoretical one. So the yield is ...?
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Re: yield %
« Reply #4 on: February 08, 2019, 10:00:08 PM »

Sometimes we ask for the yield in reference to the specified reagent. But this should be clearly stated.
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magnus

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Re: yield %
« Reply #5 on: February 08, 2019, 10:07:13 PM »

This is the real yield. I asked for the theoretical one. So the yield is ...?

14.55
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Re: yield %
« Reply #6 on: February 08, 2019, 10:11:14 PM »

?
5H₂ + 3F₂ -> ?HF
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magnus

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Re: yield %
« Reply #7 on: February 08, 2019, 10:18:09 PM »

3 moles of H₂ react with 3 moles of F₂ -> 6 moles HF according to the stoichiometric ratio

H₂ + F₂ -> 2HF
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Re: yield %
« Reply #8 on: February 08, 2019, 10:20:37 PM »

This is a theoretical yield.
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magnus

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Re: yield %
« Reply #9 on: February 08, 2019, 10:30:03 PM »

5.82/6 * 100= 97%
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Re: yield %
« Reply #10 on: February 08, 2019, 10:31:10 PM »

Solved.
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magnus

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Re: yield %
« Reply #11 on: February 08, 2019, 10:48:07 PM »

Thanks
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