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
Grams of ethyl butyrate produced : 11.27 g
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) :
[tex]\tt \dfrac{8.55}{88.11}=0.097[/tex]
[tex]\tt \%yield=\dfrac{actual}{theoretical}\times 100\%[/tex]
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)
[tex]\tt 0.097\times 116,16 g/mol=11.27~g[/tex]
