# Question c7e82

Apr 7, 2017

Here's what I got.

#### Explanation:

You need a balanced chemical equation to work with. Formic acid will react with ethanol to produce ethyl formate and water

${\text{HCOOH"_ ((aq)) + "CH"_ 3"CH"_ 2"OH"_ ((aq)) -> "C"_ 3"H"_ 6"O"_ (2(aq)) + "H"_ 2"O}}_{\left(l\right)}$

Notice that it takes $1$ mole of formic acid to react with $1$ mole of ethanol in order to produce $1$ mole of ethyl formate.

Your goal here is to figure out which one of the two reactants, if any, acts as a limiting reagent, i.e. is completely consumed before all the moles of the other reactants get the chance to take part in the reaction.

Convert the masses of the two reactants to moles by using the molar masses of formic acid and of ethanol, respectively

12.2 color(red)(cancel(color(black)("g"))) * "1 mole HCOOH"/(46.025 color(red)(cancel(color(black)("g")))) = "0.2651 moles HCOOH"

8.16 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"CH"_2"OH")/(46.068 color(red)(cancel(color(black)("g")))) = "0.1771 moles CH"_3"CH"_2"OH"#

As you can see, you don't have enough moles of ethanol to ensure that all the moles of formic acid will get the chance to react $\to$ ethanol will act as the limiting reagent.

To find the theoretical yield of the reaction, calculate the number of moles of ethyl formate produced by the reaction.

You know that $0.1771$ moles of ethanol will be completely consumed. The reaction will also consume $0.1771$ moles of formic acid and produce $0.1771$ moles of ethyl formate $\to$ this is because to the $1 : 1$ mole ratios that we pointed out earlier.

To convert the number of moles of ethyl formate to grams, use the compound's molar mass

$0.1771 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"_3"H"_6"O"_2))) * "74.08 g"/(1color(red)(cancel(color(black)("mole C"_3"H"_6"O"_2)))) = color(darkgreen)(ul(color(black)("13.1 g}}}}$

The answer is rounded to three sig figs.