How many grams of ethyl alcohol is needed to obtain 800 grams of ethyl acetate?

To solve the problem, let’s write down the process:

C2H5OH + CH3COOH = CH3COO – C2H5 + H2O – esterification, ethyl acetate was formed;
Calculations by formulas:
M (C2H5OH) = 46 g / mol.

M (word ether) = 88 g / mol.

Let’s determine the amount of the product:
Y (sl. ether) = m / M = 800/88 = 9.09 mol.

Y (C2H5OH) = 9.09 mol since their number is equal to 1 mol according to the equation.

We find the mass of the original substance:
m (C2H5OH) = Y * M 9.09 * 46 = 418 g.

Answer: you need ethanol weighing 418 g.