Question

1-butanol yields 1-bromobutane in the presence of concentrated sulfuric acid and an excess of sodium bromide. CH3CH2CH2CH2OH (l) → CH3 CH2CH2CH2Br (l) If 15.89 mL of 1-butanol produced 12.23 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.)

Answer 1

Answer:

51.34 %

Explanation:

You first need to calculate the theoretical yield of Bromobutane taking into account the amount of initial Butanol doing some stoichiometric calculations:

$15.89 mL Butanol*\frac{0.81 g Butanol}{1 mL}*\frac{1 mol Butanol}{74 g} *\frac{1 mol Bromobutane Produced}{1 mol Butanol Consumed} *\frac{137g }{1 mol Bromobutane} =23.83 g Bromobutane Produced$

Then you calculate the yield giving the real amount you produced:

$Yield=\frac{12.23 g}{23.82 g} *100=51.34%$