For the reaction below determine the limiting reagent, the theoretical yield and

Question

For the reaction below determine the limiting reagent, the theoretical yield and the percent yield given an actual yield of 4.5 grams. See the Technique Book Section 3.2 on page 35 for help with setting this up. Please show all set-ups and provide answers to the correct number of significant figures

The procedure specified 10.8 mL of 1-bromobutane and 10.2 grams of sodium ethoxide.

CH3CH2CH2CH2Br + CH3CH2O-Na+ CH3CH2CH=CH2 + NaBr 1-bromobutane sodium ethoxide 1-butene MM 137 g/mol MM 68.1 g/mol MM 56 g/mol Denisty 1.27 g/mL

Explanation / Answer

The mass of 1-bromobutane = 1.27 x 10.8 = 13.716 g

Number of moles of 1-bromobutane = 13.716 / 137 = 0.1 mol

Number of moles of sodium ethoxide = 10.2 / 68.1 = 0.15 mol

Since equimolar amounts of both the reagents are used in a well balanced reaction , so the limiting reagent is 1-bromobutane.

So the number of moles of 1-buten formed will be 0.1 mol

Theoretical yield = 0.1 x 56 = 5.6 grams

Percent yield = (actual yield / theoretical yield) x 100

= (4.5 / 5.6) x 100

= 80. %

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