Given 8.50 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Amount of ethyl butyrate formed = 11.3 g

Explanation:

The reaction between butanoic acid (C3H7COOH) and excess ethanol (C2H5OH) is:

Since ethanol is the excess reagent, the formation of the product i.e. ethyl butyrate is influenced by the amount of butanoic acid present

Based on the reaction stoichiometry:

1 mole of butanoic acid produces 1 mole of ethyl butyrate

Mass of butanoic acid = 8.50 g

Molar mass of butanoic acid = 88 g/mol

Moles of ethyl butyrate = 0.097

Molar mass of ethyl butyrate= 116 g/mol

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