Question

Given 8.55 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer 1

Answer: 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

brainless, please

Answer 2

Grams of ethyl butyrate produced : 11.27 g

Further explanation

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

The reaction between Butanoic acid-C₄H₈O₂ and an excess Ethanol-C₂H₅OH to produce ethyl butyrate-C₆H₁₂O₂ :

C₄H₈O₂+C₂H₅OH⇒C₆H₁₂O₂+H₂O

mass Butanoic acid- C₄H₈O₂= 8.55 g, so mol C₄H₈O₂ (MW=88,11 g/mol) :

`\tt (8.55)/(88.11)=0.097`

`\tt \%yield=(actual)/(theoretical)* 100\%`

For 100% yield ⇒ actual=theoretical

Because Ethanol as an excess, so Butanoic acid as a limiting reactant and mol Ethyl butyrate based on mol Butanoic acid

From equation, mol ratio of mol C₆H₁₂O₂ : mol C₄H₈O₂ = 1 : 1, so : mol C₆H₁₂O₂=mol C₄H₈O₂= 0.097

mass C₆H₁₂O₂(MW=116.16 g/mol)

`\tt 0.097* 116,16 g/mol=11.27~g`