Question #d2807

1 Answer
Sep 9, 2016

Answer:

#"% yield" = 70.3%#

Explanation:

The idea here is that you need to convert the volumes to moles by using the densities and the molar masses of the three chemical species of interest.

You will have

  • acetic acid

#4.40 color(red)(cancel(color(black)("mL"))) * "1.05 g"/(1color(red)(cancel(color(black)("mL")))) = "4.62 g"#

#4.62 color(red)(cancel(color(black)("g"))) * "1 mole acetic acid"/(60.0color(red)(cancel(color(black)("g")))) = "0.0770 moles"#

  • isopentyl alcohol

#3.35color(red)(cancel(color(black)("mL"))) * "0.81 g"/(1color(red)(cancel(color(black)("mL")))) = "2.7135 g"#

#2.7135 color(red)(cancel(color(black)("g"))) * "1 mole isopentyl alcohol"/(88.2color(red)(cancel(color(black)("g")))) = "0.03077 moles"#

  • isopentyl acetate

#3.25 color(red)(cancel(color(black)("mL"))) * "0.867 g"/(1color(red)(cancel(color(black)("mL")))) = "2.818 g"#

#2.818color(red)(cancel(color(black)("g"))) * "1 mole isopentyl acetate"/(130.2 color(red)(cancel(color(black)("g")))) = "0.02164 moles"#

Now, acetic acid and isopentyl alcohol react in a #1:1# mole ratio, which basically means that the reactant that has the fewest moles available will act as a limiting reagent, i.e. it will be completely consumed before all the moles of the other reactant get the chance to take part in the reaction.

In this case, isopentyl alcohol will be the limiting reagent, since

#"0.03077 moles " < " 0.0770 moles"#

Therefore, the reaction will consume #0.03077# moles of acetic acid and #0.03077# moles of isopentyl alcohol.

Consequently, the reaction should theoretically produce #0.03077# moles of isopentyl acetate, since both reactants have #1:1# mole ratios with this product.

However, you calculated that the reaction produced only #0.02164# moles of isopentyl acetate.

The percent yield for this reaction, which basically tells you how much product you should expect to see for every #100# moles (or grams, it works the same way) that can be theoretically produced

#color(purple)(bar(ul(|color(white)(a/a)color(black)("% yield" = "what you actually get"/"what you should theoretically get" xx 100)color(white)(a/a)|)))#

will be equal to

#"% yield" = (0.02164 color(red)(cancel(color(black)("moles"))))/(0.03077color(red)(cancel(color(black)("moles")))) xx 100 = color(green)(bar(ul(|color(white)(a/a)color(black)(70.3%)color(white)(a/a)|)))#

The answer is rounded to three sig figs.