Using the data in the table, calculate both the mass and volume of n‑butyl alcohol required to completely react with 220.0 L of acetic acid.

Question

As part of a team of chemists at a food flavoring company, you need to produce n‑butyl acetate (C6H12O2), a sweet-smelling food additive that occurs naturally in apples, pears, and bananas. Your team plans to prepare this compound from the reaction of acetic acid (HC2H3O2) and n‑butyl alcohol (C4H10O):

HC2H3O2(l)+C4H10O(l)⟶C6H12O2(l)+H2O(l)

Name Formula Boiling Point Molar Mass Density

Acetic Acid HC2H3O2 119 degrees C 60.06g/mol 1.049g/mL

n-Butyl alcohol C4H10O 118 degrees C 74.14g/mol 0.810g/mL

n-Butyl acetate C6H12O2 126 degrees C 116.16 g/mol 0.883 g/mL

water H2O 100 degrees C 18.02 g/mol 1.00g/mL

1. mass of n‑butyl alcohol: ?kg
2. volume of n‑butyl alcohol: ?L

Chemists often use a large excess of n‑butyl alcohol for this reaction. If you begin with 750.0 L of this reagent, what volume will be left over after the reaction is complete?

1. volume of n‑butyl alochol left over: ? L

What is the theoretical yield of n‑butyl acetate in this reaction? Report your answer in grams, kilograms, and liters.

1. theoretical yield of n‑butyl acetate: ?g
2. theoretical yield of n‑butyl acetate: ?kg
3. theoretical yield of n‑butyl acetate: ?L

After completing this process and purifying the products, your team isolates 483.3 L of n‑butyl acetate. Calculate your percent yield for this reaction.

1. percent yield: ? %

Answer 1

HC₂H₃O₂(l) + C₄H₁₀O(l) ==> C₆H₁₂O₂(l) + H₂O(l) .. balanced equation

acetic acid.....butyl alcohol....butyl acetate....water...........

We must work in moles, to begin with, so we will first convert 220.0 L acetic acid to grams and then moles:

220.0 L x 1000 ml / L x 1.049 g / ml x 1 mol / 60.06 g = 3842.5 mols acetic acid

Next, from the stoichiometry of the balanced equation, we will calculate moles of n-butyl alcohol needed, and then convert that to kg of n-butyl alcohol:

1.) 3842.5 mol acetic acid x 1 mol n-butyl / mol acetic acid = 3842.5 mols n-butyl alcohol needed

3842.5 mols n-butyl x 74.14 g / mol x 1 kg / 1000 g = 284.9 kg of n-butyl alcohol needed

2). Volume of n-butyl needed = 284.9 kg x 1 L/ 0.810 kg = 351.7 Liters of n-butyl alcohol needed

750.0 L n-butyl alcohol x 0.810 kg / L x 1000 g / kg x 1 mol / 74.14 g = 8194 mols n-butyl alcohol used

moles n-butyl alcohol left over = 8194 mol - 3843 mol = 4351 mols n-butyl left over

1.) Volume n-butyl left over = 4351 mols x 74.14 g / mol x 1 ml / 0.810 g x 1 L/1000 ml = 398.3 L left over

Since n-butyl alcohol is present is EXCESS, that means acetic acid is the limiting reactant, and thus moles of acetic acid will determine the theoretical yield of butyl acetate.

1.) Theoretical yield butyl acetate = 3843 mols acetic acid x 1 mol butyl acetate/ mol acetic acid = 3843 mols of butyl acetate.

3843 mols butyl acetate x 116.16 g / mol = 446403 g butyl acetate

2.) 44640 g butyl acetate x 1 kg / 1000 g = 446.4 kg

3.) 446.4 kg butyl acetate x 1 L / 0.883 kg = 505.5 L of butyl acetate

Percent yield = actual yield / theoretical yield (x100%)

actual yield = 484.3 L x 0.883 kg / L = 427.6 kg

theoretical yield = 446.4 kg

% yield = 427.5 kg / 446.4 kg (x100%) = 95.80% yield

(BE SURE TO CHECK ALL CALCULATIONS)