Determine the mass of the ester obtained by the reaction of 12.0 acetic acid with 9.6 g of methanol.

To solve the problem, you need to write the equation:
CH3COOH + CH3OH = CH3COO – CH3 + H2O – esterification reaction, methyl acetate is released;
M (CH3COOH) = 60 g / mol;
M (ether) = 74 g / mol;
M (CH3OH) = 32 g / mol;
Determine the amount of mol of acetic acid, methanol:
Y (CH3COOH) = m / M = 12/60 = 0.2 mol (deficient substance)
Y (CH3OH) = m / M = 9.6 / 32 = 0.3 mol (substance in excess);
Further calculations are made for the substance in deficiency.
According to the equation, the number of moles of acetic acid and ether are equal to 1 mole, which means that Y (ether) = 0.2 mole.
Let’s calculate the mass of the ether:
m (ether) = Y * M = 0.2 * 74 = 14.8 g.
Answer: the mass of ether is 14.8 g.