Given 7.20 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

The reaction between butanoic acid and ethanol is:

CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH3CH2 + H2O

Based on the reaction stoichiometry:

1 mole of butanoic acid forms 1 mole of ethyl butyrate

Now,

Molar mass of Butanoic acid = 88.0 g/mol

Given mass of butanoic acid = 7.20 g

Therefore, # moles of butanoic acid reacted = 7.20/88.0 = 0.0818 moles

# moles of ethyl butyrate formed = 0.0818 moles

Molar mass of ethyl butyrate = 116 g/mol

Mass of ethyl butyrate synthesized = 0.0818 * 116 = 9.49 g