1-butanol yields 1-bromobutane in the presence of concentrated sulfuric acid and an excess of sodium bromide. CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br (l) If 18.54 mL of 1-butanol produced 15.65 g of 1-bromobutane, the percentage yield of the product equals: (Assume the density of 1-butanol is 0.81 g/mL, the molar mass of 1-butanol is 74 g/mol, and the molar mass of 1-bromobutane is 137 g/mol.)

Question

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joseaaronlara joseaaronlara · Answer 1 · 2019-10-23T01:01:07+02:00

Answer:

The answer to your question is: yield = 56.27%

Explanation:

Data

CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br

18.54 ml 1-butanol 15.65 g of 1-bromobutane

% yield = ?

density = 0.81 g/ml

MM = 74 g 1- butanol

MM = 137 g 1-bromobutane

Process

Calculate mass of 1- butanol

density = mass/volume

mass = density x volume

mass = 0.81 x 18.54

mass = 15.02 g of 1-butanol

Theoretical yield

74 g of 1- butanol ----------------- 137 g of 1-bromobutane

15.02 g of 1- butanol ------------- x

x = (15.02 x 137) / 74

x = 27.81 g of 1-bromobutane

% yield = experimental yield / theoretical yield x 100

% yield = 15.65 / 27.81 x 100

% yield = 56.28