Question

The experiment to get Isoamyl Bromide from Isoamyl Alcohol was done. chemicals used were: 9 grams (.087 moles) of sodium bromide, 8.2 mL (6.6 grams; .075 moles) of isoamyl alcohol, 8mL (14.7 grams;.150 moles) of concentrated sulfuric acid, and 8mL of 10% aqueous sodium carbonate.

What are the limiting reagents?

What is the theoretical yield?

What is the % yield?

Answer 1

The reaction invoved is as follows

Isoamyl Alcohol + NaBr    Isoamyl Bromide

1 mole of isoamyl alcohol will react with 1 mole NaBr to produce isoamyl alcohol.

Now, 8.2 mL (6.6 grams; .075 moles) of isoamyl alcohol & 9 grams (.087 moles) of sodium bromide is used in this reaction .

Amount of concentrated H2SO4 and aq Na2CO3 is not directly related to the yield of the reaction.

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this limiting reagent.

So in this above reaction isoamyl alcohol is the limiting reagent. (Answer 1)

So, 0.075 moles of isoamyl alcohol will react with 0.075 mole of NaBr to produce 0.075 moles of isoamyl bromide

MW of  Isoamyl Bromide is 151.04

So, 0.075 moles of isoamyl bromide = 11.328 g of isoamyl bromide

= 9.44 mL of  isoamyl bromide (because density of  isoamyl bromide is 1.2 g /mL)

Therefore theoritical yield of product isoamyl amyl is 11.33 g or 9.44 mL (Answer 2)

Now, % yield = ( Actual yield / Theoritical yield ) x 100

Since in this problem actual yield is not given, % yield cannot be calculated.