Question

In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-butanol (C5H120) were mixed in the round-bottomed flask and allowed to cool for 2 -3 minutes. Then while swirling the flask, 10 mL sulfuric acid (H2SO4) was slowly added to the flask and the mixture was refluxed and later distilled. The mass of the pure product obtained was 6.48 g. Sodium hydrogen carbonate (NaHCO3) was used to remove impurities. (Density of 3-methyl-1-butanol is 0.810 g/mL). The percentage yield will be?​

Answer 1

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004