percentage yield PLS HELP
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(d) In an experiment starting with 5.05 g of phenylamine, 4.82 g of purified product wereobtained in step 1.
Calculate the percentage yield in this reaction.
Give your answer to the appropriate number of significant figures.
the mark scheme is
(d) Mr product = 135.0
1
Expected mass = 5.05 × = 7.33 g
1
Percentage yield = × 100 = 65.75 = 65.8(%)
i tried getting mr of phenyl amine which is 72 from the c6h5
but that is not right i also calculated mr of expected product which is 17+16+12+4 from the NCOCH3 i added this +the c6h5 =126 and this is still wrong
Calculate the percentage yield in this reaction.
Give your answer to the appropriate number of significant figures.
the mark scheme is
(d) Mr product = 135.0
1
Expected mass = 5.05 × = 7.33 g
1
Percentage yield = × 100 = 65.75 = 65.8(%)
i tried getting mr of phenyl amine which is 72 from the c6h5
but that is not right i also calculated mr of expected product which is 17+16+12+4 from the NCOCH3 i added this +the c6h5 =126 and this is still wrong
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Rossg123
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#2
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#2
RFM of phenylamine is 93. Moles of phenylamine= 5.05/93=0.0543 moles. 1:1 ratio therefore moles of product also=0.0543. RFM of product=135 therefore mass of product= 135*0.0543=7.3305g which is the theoretical yield. %yield= 4.82/7.3305*100= 65.8%
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#3
(Original post by Rossg123)
RFM of phenylamine is 93. Moles of phenylamine= 5.05/93=0.0543 moles. 1:1 ratio therefore moles of product also=0.0543. RFM of product=135 therefore mass of product= 135*0.0543=7.3305g which is the theoretical yield. %yield= 4.82/7.3305*100= 65.8%
RFM of phenylamine is 93. Moles of phenylamine= 5.05/93=0.0543 moles. 1:1 ratio therefore moles of product also=0.0543. RFM of product=135 therefore mass of product= 135*0.0543=7.3305g which is the theoretical yield. %yield= 4.82/7.3305*100= 65.8%
and to calculate Ram of product its mass/moles i did 4.82 ÷ 0.0543 =88.7 so im confused
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#4
Is it a past paper question and if it is, what year and exam board
(Original post by Hi freinds)
the part im stuck at is the RAM of product because the moles of phenylamine=0.0543 moles of product
and to calculate Ram of product its mass/moles i did 4.82 ÷ 0.0543 =88.7 so im confused
the part im stuck at is the RAM of product because the moles of phenylamine=0.0543 moles of product
and to calculate Ram of product its mass/moles i did 4.82 ÷ 0.0543 =88.7 so im confused
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#5
what is this question i dont know i just am answering a bunch of physical questions
i honestly dont understand how you got thatt mr of the product could you pleasse explain your working out
i honestly dont understand how you got thatt mr of the product could you pleasse explain your working out
(Original post by Rossg123)
Is it a past paper question and if it is, what year and exam board
Is it a past paper question and if it is, what year and exam board
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Rossg123
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#6
Is it definitely a 1:1 ratio?
(Original post by Hi freinds)
what is this question i dont know i just am answering a bunch of physical questions
i honestly dont understand how you got thatt mr of the product could you pleasse explain your working out
what is this question i dont know i just am answering a bunch of physical questions
i honestly dont understand how you got thatt mr of the product could you pleasse explain your working out
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#7
(Original post by Rossg123)
Is it definitely a 1:1 ratio?
Is it definitely a 1:1 ratio?
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#8
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#8
It works out at 135 if it's a 2:3 ratio
(Original post by Hi freinds)
i dont know i went by how you said it was
i dont know i went by how you said it was
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#9
but how can you tell its a 2:3 ratio
(Original post by Rossg123)
It works out at 135 if it's a 2:3 ratio
It works out at 135 if it's a 2:3 ratio
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