Saponification of ethyl acetate gave 34.5 g of ethanol. Determine the mass of the reacted ester.

Let’s implement the solution:
1. According to the condition of the problem, we compose the equation:
CH3COO – C2H5 + NaOH = C2H5OH + CH3COONa – exchange reaction, obtained ethanol, sodium acetate;
2. Calculation of molar masses of organic substances:
M (ethyl acetate) = 88 g / mol;
M (C2H5OH) = 46 g / mol.
3. Calculate the number of moles of the reaction product:
Y (C2H5OH) = m / M = 34.5 / 46 = 0.75 mol;
Y (ethyl acetate) = 0.75 mol since the amount of these substances according to the equation is 1 mol.
4. Find the mass of the original substance:
m (ethyl acetate) = Y * M = 0.75 * 88 = 66 g.
Answer: for the saponification reaction, ethyl acetate weighing 66 g is required.