The mass of the ester obtained by the interaction of 200 g of a 60% solution of acetic acid with ethyl alcohol.

1. Let’s compose the equation of interaction of acetic acid with ethyl alcohol:

CH3COOH + C2H5OH → CH3COOC2H5 + H2O;

2.Calculate the mass of acetic acid:

m (CH3COOH) = w (CH3COOH) * m (solution) = 0.6 * 200 = 120 g;

3. find the chemical amount of acetic acid:

n (CH3COOH) = m (CH3COOH): M (CH3COOH);

M (CH3COOH) = 12 + 3 + 12 + 33 = 60 g / mol;

n (CH3COOH) = 120: 60 = 2 mol;

4.Set the amount and calculate the mass of ethyl acetate:

n (CH3COOC2H5) = n (CH3COOH) = 2 mol;

m (CH3COOC2H5) = n (CH3COOC2H5) * M (CH3COOC2H5);

M (CH3COOC2H5) = 12 + 3 + 12 + 32 + 2 * 12 + 5 = 88 g / mol;

m (CH3COOC2H5) = 2 * 88 = 176 g.

Answer: 176 g.