Given 7.45 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? express your answer in grams to three significant figures.

The equation for the reaction is:

C₄H₈O₂ + C₂H₅OH = C₆H₁₂O₂ + H₂O

Now you see that the number of the moles of butanoic acid and etyl butyrate is equal in

the reaction. That means;

number of moles of C₄H₈O₂ = number of moles of C₆H₁₂O₂

mass of C₄H₈O₂ / Molar mass of C₄H₈O₂ = mass of C₆H₁₂O₂ / molar mass of C₆H₁₂O₂

mass of C₆H₁₂O₂ = molar mass of C₆H₁₂O₂ x mass of C₄H₈O₂ / Molar mass of C₄H₈O₂

Now, assuming 100% yield, the mass of ethyl butyrate produced is:

= 7.45/88.11 x 116.16

=9.82g

Thus, the theoretical yield of ethyl butyrate is 9.82g.