# Question e72f9

Apr 8, 2016

Here's what I got.

#### Explanation:

I assume that the question is incomplete because the only way to solve this problem is to use the densities of the two reactants, which are listed as

${\rho}_{\text{acetic acid" = "1.049 g mL}}^{- 1}$

${\rho}_{\text{ethanol" = "0.789 g mL}}^{- 1}$

to determine the masses of acetic acid, $\text{CH"_3"COOH}$, which I'll label as $\text{HAc}$, and ethanol, $\text{CH"_3"CH"_2"OH}$, which I'll label as $\text{EtOH}$.

Knowing the masses of the two reactants will allow you to determine if you're dealing with a limiting reagent.

To find the percent yield of the reaction, use the density of ethyl acetate, ${\text{CH"_3"COOCH"_2"CH}}_{3}$, which I'll label as $\text{EtOAc}$

${\rho}_{\text{ethyl acetate" = "0.902 g mL}}^{- 1}$

So, the balanced chemical equation for this synthesis reaction (known as the Fisher esterification) looks like this

${\text{HAc"_ ((aq)) + "EtOH"_ ((aq)) -> "EtOAc"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

Notice that the two reactants take part in the reaction in a $1 : 1$ mole ratio. This means that every mole of acetic acid that takes part in the reaction needs one mole of ethanol.

Use the volumes and densities of the two reactants to find their masses

$\text{For HAc: " 20.2 color(red)(cancel(color(black)("mL"))) * "1.049 g"/(1color(red)(cancel(color(black)("mL")))) = "21.19 g}$

$\text{For EtOH: " 20.1 color(red)(cancel(color(black)("mL"))) * "0.789 g"/(1color(red)(cancel(color(black)("mL")))) = "15.86 g}$

Now use the molar masses of the two compounds to find how many moles of each you're mixing

21.19 color(red)(cancel(color(black)("g"))) * "1 mole HAc"/(60.05color(red)(cancel(color(black)("g")))) = "0.3529 moles HAc"

15.86color(red)(cancel(color(black)("g"))) * "1 mole EtOH"/(46.07color(red)(cancel(color(black)("g")))) = "0.3443 moles EtOH"

Notice that you have fewer moles of ethanol than of acetic acid. This tells you that the ethanol will be completely consumed before all the moles of acetic acid will get the change to react $\to$ ethanol is the limiting reagent.

Now, the ethanol is completely consumed by the reaction, which means that you can use the $1 : 1$ mole ratio that exists between ethanol and ethyl acetate to say that the reaction will produce $0.3443$ moles of $\text{EtOAc}$.

Use the molar mass of ethyl acetate to find how many grams would contain this many moles

0.3443 color(red)(cancel(color(black)("moles EtOAc"))) * "88.11 g"/(color(red)(cancel(color(black)("moles EtOAc")))) = "30.34 g"

This is the theoretical yield of the reaction, i.e. what you would expect to get if all the moles of reactants that took part in the reaction actually ended up producing ethyl acetate and water.

Use the volume and density of ethyl acetate to determine the actual yield of the reaction

27.5color(red)(cancel(color(black)("mL"))) * "0.902 g"/(1color(red)(cancel(color(black)("mL")))) = "24.81 g"

So, the reaction should have theoretically produced $\text{30.34 g}$ of ethyl acetate, but it ended up producing $\text{24.81 g}$ instead. The percent yield of the reaction can be calculated using

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "what you actually get"/"what you should theoretically get} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this case, you will have

"% yield" = (24.81 color(red)(cancel(color(black)("g"))))/(30.34color(red)(cancel(color(black)("mL")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"81.8%"color(white)(a/a)|)))#

The answer is rounded to three sig figs.