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yield %
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Topic: yield % (Read 374 times)
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magnus
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yield %
«
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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 %
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Reply #1 on:
February 08, 2019, 09:36:10 PM »
How many moles of HF you can obtain from 3 moles of F
2
?
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magnus
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Re: yield %
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Reply #2 on:
February 08, 2019, 09:39:10 PM »
5.82 moles
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AWK
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Re: yield %
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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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AWK
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Re: yield %
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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 %
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Reply #5 on:
February 08, 2019, 10:07:13 PM »
Quote from: AWK on February 08, 2019, 09:40:02 PM
This is the real yield. I asked for the theoretical one. So the yield is ...?
14.55
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AWK
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Re: yield %
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Reply #6 on:
February 08, 2019, 10:11:14 PM »
?
5H₂ + 3F₂ -> ?HF
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magnus
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Re: yield %
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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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AWK
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Re: yield %
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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 %
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February 08, 2019, 10:30:03 PM »
5.82/6 * 100= 97%
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AWK
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Re: yield %
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Reply #10 on:
February 08, 2019, 10:31:10 PM »
Solved.
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magnus
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Re: yield %
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Reply #11 on:
February 08, 2019, 10:48:07 PM »
Thanks
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