Question

You add 0.58 mL of n-propanol (0.803 g/mL, 60.1 g/mol) to an excess of acetic acid to create the pear ester propyl acetate (102.1 g/mol). If you obtain 0.53 grams of the product, what was your percent yield?

Answer 1

Ansdes Balaneed reaction: oH OH + +H20 As acetic aeid remains present in excess therefore n-Propanok is the imiting heagent mags Density Volume 0.58 mL of n-psopanol- (o.803 x 0.58)g of n-popanol mass Number of moles molas masSS :0-803x0.58) a of n-propanol O 00370-58 fo alaue n-Propanol 60.1 A ccosaing to belancea reaction, 1 mol of n- propanol produces 1 mol off Propyl acetate 0-803x0.5S moles moles of Propy a eatate 66.1 60-1 t 0xE08.0 O.803x0-58 .: Theosetical yield of Psopyl acetcte X 102:1 60-1 aetual yield theoxetical y iela . Pareent yialA 100 7 O.803x6-58 102.1 9 60.1