Can someone help me with this percentage yield question
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# Can someone help me with this percentage yield question

[From: ] [author: ] [Date: 11-05-18] [Hit: ]
I cant seem to get the right answer to this question so can someone please explain how you do this one?molar mass butan-1-ol = 74.14.8g = 14.8/74. = 0.......
The question is:
14.8g of butan-1-ol were refluxed with NaBr and sulphuric acid, after which 17.8g of 1-bromobutane were collected. What's the percentage yield of this conversion?

I can't seem to get the right answer to this question so can someone please explain how you do this one?

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Equation
CH3CH2CH2CH2OH + NaBr → CH3CH2CH2CH2Br + NaOH
1mol butan-1-ol produces 1 mol bromobutane

molar mass butan-1-ol = 74.g/mol
14.8g = 14.8/74. = 0.20mol

Molar mass 1 bromobutane = 137 g/mol
0.2 mol = 137*0.2 = 27.4g = theoretical yield

Actual yield = 17.8g

% yield = 17.8/27.4*100 = 64.96%

% yield = 65.0%

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you lost me at yield :(
1
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