what mass of ethanoic acid and what mass of ethanol would be needed to produce 100 gram of ethyl ethanoate assuming the reaction went to completion
The balanced equation for the reaction between ethanoic acid and ethanol to form ethyl ethanoate is:
CH3COOH + C2H5OH -> CH3COOC2H5 + H2O
From the balanced equation, we can see that the molar ratio between ethanoic acid and ethyl ethanoate is 1:1. This means that for every 1 mole of ethanoic acid, we will produce 1 mole of ethyl ethanoate.
Since we want to produce 100 grams of ethyl ethanoate, we first need to determine the molar mass of ethyl ethanoate. The molar mass of ethyl ethanoate is 88.11 g/mol.
Next, we need to convert the mass of ethyl ethanoate to moles. We can use the formula:
moles = mass / molar mass
moles of ethyl ethanoate = 100 g / 88.11 g/mol = 1.135 mol
Since the molar ratio between ethanoic acid and ethyl ethanoate is 1:1, we need 1.135 moles of ethanoic acid to produce 1.135 moles of ethyl ethanoate.
The molar mass of ethanoic acid is 60.05 g/mol.
Finally, we can calculate the mass of ethanoic acid needed:
mass of ethanoic acid = moles of ethanoic acid * molar mass of ethanoic acid
mass of ethanoic acid = 1.135 mol * 60.05 g/mol = 68.11 grams
Therefore, you would need 68.11 grams of ethanoic acid and 1.135 grams of ethanol to produce 100 grams of ethyl ethanoate.