what mass of ethanol would be needed to produce 100 gram of ethyl ethanoate assuming the reaction went to completion when ethanoic mass is equal to 68.11 gram
To determine the mass of ethanol needed to produce 100 grams of ethyl ethanoate, we need to consider the balanced chemical equation and the stoichiometry of the reaction involved.
The chemical reaction for the formation of ethyl ethanoate (also known as esterification) is as follows:
Ethanol + Ethanoic acid -> Ethyl ethanoate + Water
The molar mass of ethanoic acid (CH3COOH) is 60.05 g/mol, and the molar mass of ethanol (CH3CH2OH) is 46.07 g/mol.
From the equation, we can see that 1 mole of ethanol reacts with 1 mole of ethanoic acid to produce 1 mole of ethyl ethanoate. Therefore, the molar mass of ethyl ethanoate (CH3COOC2H5) is also 88.11 g/mol.
The given information states that the mass of ethanoic acid is 68.11 grams. To find the mass of ethanol needed, we need to use stoichiometry.
The molar ratio between the reactant (ethanoic acid) and the desired product (ethyl ethanoate) is 1:1. Therefore, the number of moles of ethanoic acid is equal to the number of moles of ethyl ethanoate.
Number of moles of ethanoic acid = mass / molar mass
Number of moles of ethanoic acid = 68.11 g / 60.05 g/mol = 1.135 mol
Since the stoichiometry of the reaction is 1:1, the number of moles of ethanol needed is also 1.135 mol.
The mass of ethanol needed can be calculated as follows:
Mass of ethanol = number of moles x molar mass
Mass of ethanol = 1.135 mol x 46.07 g/mol = 52.34 grams
Therefore, approximately 52.34 grams of ethanol would be needed to produce 100 grams of ethyl ethanoate when the reaction goes to completion.