Question

The reaction of 9.0 g of 1-butanol (MW = 76g/mol) with 13.0 g of sodium bromide...

The reaction of 9.0 g of 1-butanol (MW = 76g/mol) with 13.0 g of sodium bromide (MW = 103 g/mol) with excess acid produces 8.5 g of 1-bromobutane (MW 137 g/mol). a. What is the purpose of the acid? b. What is the % yield for the reaction? Show your calculations.

Homework Answers

Answer #1

C4H9OH + NaBr ------> C4H9Br

Moles of butanol present = mass/molar mass = 9/76 = 0.118

moles of NaBr present = mass/molar mass = 13/103 = 0.126

As per the balanced reaction , butanol and NaBr reacts in the molar ratio of 1:1

Thus, butanol is the limiting reagent

Hence theoretical moles of bromobutane produced = moles of butanol reacted = 0.118

Thus, theoretical mass of bromobutane produced = moles*molar mass = 16.22 g

Thus, % yield = (mass of bromobutane actually produced/theoretical mass of bromobutane)*100 = 52.4 %

Acid is used for the hydrolysis of the salt.

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