Calculate the amount of methyl alcohol and acetic acid required to obtain 148 g of methyl acetic acid if the yield is 80%

Let’s execute the solution, compose the equation:
CH3COOH + CH3OH = CH3COO – CH3 + H2O – esterification reaction, methyl acetate was obtained;
Let’s make calculations using the formulas of substances:
M (CH3COOH) = 60 g / mol;
M (CH3OH) = 32 g / mol;
M (ether) = 74 g / mol.
Let’s determine the amount of ether moles:
Y (ether) = m / M = 148/74 = 2 mol (practical).
We find the theoretical mass of ether by the formula:
W = m (practical) / m (theoretical) * 100;
m (theoretical) = 148 * 0.80 = 118.4 g;
Y (theoretical) = m / M = 118.4 / 74 = 1.6 mol.
According to the reaction equation, the number of moles of acetic acid, methanol and ether are equal to 1 mol, which means that Y (CH3COOH) = 1.6 mol, Y (CH3OH) = 1.6 mol.
Answer: the number of moles of acetic acid and methanol is 1.6 moles.