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Question 6b

http://pmt.physicsandmathstutor.com/...t-B/Arenes.pdf
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.
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Lilith2400
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Calculate the moles of phenol : 100/ (Mr of phenol)
Calculate the theoretical moles of the product using the 1:1 ratio of the equation
Calculate the theoretical mass of the product by doing moles x Mr of product
Multiply this mass by 0.27 to find the mass produced if the yield is 27%

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(Original post by Lilith2400)
Calculate the moles of phenol : 100/ (Mr of phenol)
Calculate the theoretical moles of the product using the 1:1 ratio of the equation
Calculate the theoretical mass of the product by doing moles x Mr of product
Multiply this mass by 0.27 to find the mass produced if the yield is 27%

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http://pmt.physicsandmathstutor.com/...0Compounds.pdf

Question 5b

1g=1000mg
250x3x7= 5.25 g

moles chloral hydrate= 5.25/165.5 = 0.0317 mol
mass trichloro= mols x Mr = 0.0317 x 147.5 = 4.6789 g

percentage yield= actual/theoretical = 60%
Actual = theoretical x 0.60 = 4.6789 x 0.60 = 2.81 g

Answer is 7.80....
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Lilith2400
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(Original post by chemquestion)
http://pmt.physicsandmathstutor.com/...0Compounds.pdf

Question 5b

1g=1000mg
250x3x7= 5.25 g

moles chloral hydrate= 5.25/165.5 = 0.0317 mol
mass trichloro= mols x Mr = 0.0317 x 147.5 = 4.6789 g

percentage yield= actual/theoretical = 60%
Actual = theoretical x 0.60 = 4.6789 x 0.60 = 2.81 g

Answer is 7.80....
5.25 g is what you get for 60% yield.
(5.25 / 60 ) ×100
Then divide this new mass by the Mr
Use 1:1 ratio and multiply by reactant Mr which gives you 7.8

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username3835832
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(Original post by Lilith2400)
5.25 g is what you get for 60% yield.
(5.25 / 60 ) ×100
Then divide this new mass by the Mr
Use 1:1 ratio and multiply by reactant Mr which gives you 7.8

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Thank you
As I can see this question requires a slightly different method than the one above...
what's the difference between this question and the one I asked before?
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Lilith2400
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(Original post by chemquestion)
Thank you
As I can see this question requires a slightly different method than the one above...
what's the difference between this question and the one I asked before?
The first one is finding the mass of a product formed from a certain mass of reactant given and the second is finding the mass of a reactant required to form a certain amount of product given.
The second is the same method but reversed - you're just working backwards.
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(Original post by Lilith2400)
The first one is finding the mass of a product formed from a certain mass of reactant given and the second is finding the mass of a reactant required to form a certain amount of product given.
The second is the same method but reversed - you're just working backwards.
I meant in terms of the yield calculation part
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