# Calculated mass of the ester formed by the action of 100 g of acetic acid with methanol.

We carry out the solution according to the plan:
1. Let’s write down the reaction equation:
CH3COOH + CH3OH = CH3COO – CH3 + H2O – esterification reaction, methyl acetate was obtained;
2. Let’s calculate the molecular weights of substances:
M (CH3COOH) = 12 * 2 + 4 + 16 * 2 = 60 g / mol;
M (CH3COO – CH3) = 12 * 2 + 6 + 16 * 2 = 62 g / mol;
3. Determine the number of moles of the ester:
M (CH3COOCH3) = m / M = 100/60 = 1.6 mol;
4. Let’s make the proportion:
1.6 mol (CH3COOH) – X mol (CH3COOCH3);
-1 mol -1 mol, hence, X mol (CH3COOCH3) = 1.6 * 1/1 = 1.6 mol;
5. Let’s calculate the mass of methyl acetate:
M (CH3COOCH3) = Y * M = 1.6 * 62 = 99.2 g.
Answer: in this reaction, methyl acetate is formed with a mass of 99.2 g.

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