Calculate the mass and give the name of the ester formed as a result of the reaction of 120 g

Calculate the mass and give the name of the ester formed as a result of the reaction of 120 g of acetic acid with 140 g of propyl alcohol.

According to the condition of the problem, we compose the equation:
1. СН3СООН + С3Н7ОН = СН3СОО – С3Н7 + Н2О – esterification, acetic-propionic ether was obtained;
2. Calculations:
M (CH3COOH) = 60 g / mol;
M (C3H7OH) = 60 g / mol;
M (word ether) = 102 g / mol.
3. Determine the amount of starting substances:
Y (CH3COOH) = m / M = 120/60 = 2 mol (deficient substance);
Y (C3H7OH) = m / M = 140/60 = 2.3 mol (substance in excess);
The calculation is based on the substance in deficiency.
Y (ester) = 2 mol since the amount of these substances according to the equation is 1 mol.