Use the following reaction: C4H9OH NaBr H2SO4 C4H9Br NaHSO4 H2O If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction

Percent yield = 61.58%

Explanation:

C4H9OH + NaBr + H2SO4 --> C4H9Br + NaHSO4 + H2O

From the reaction;

1  mol of C4H9OH reacts with 1 mol of NaBr and 1 mol of H2SO4 to form 1 mol of  C4H9Br

We have to obtain the limiting reagent. We do this by calculatig the mols of each reactant in the reaction;

Number of moles = Mass / Molar mass

C4H9OH; 15 / 74g/mol = 0.2027

NaBr; 22.4 / 102.89 = 0.2177

H2SO4; 32.7 / 98 = 0.3336

In this case, the limiting reagent is C4H9OH, it determines the amount of product formed.

From the stoichometry of the reaction; one mole of C4H9OH forms one mole of C4H9Br. This means 0.2027 mol of C4H9Br was formed.

Mass = Number of moles * Molar mass

Mass of C4H9Br formed = 0.2027 * 137

Mass = 27.77g

Percent yield =  (Practical yield /  theoretical yeield ) * 100%

Percent Yield = 17.1 / 27.77 * 100

Percent yield = 61.58%