The mass of the ester obtained by reacting 6 g of acetic acid with 4 g of methanol is …

Given:
m (CH3COOH) = 6 g
m (CH3OH) = 4 g
To find:
m (ester) -?
Decision:
1) CH3COOH + CH3OH => CH3COOCH3 + H2O;
2) n (CH3COOH) = m (CH3COOH) / M (CH3COOH) = 6/60 = 0.1 mol;
3) n (CH3OH) = m (CH3OH) / M (CH3OH) = 4/32 = 0.125 mol;
4) n (CH3COOCH3) = n (CH3COOH) = 0.1 mol;
5) m (CH3COOCH3) = n (CH3COOCH3) * M (CH3COOCH3) = 0.1 * 74 = 7.4 g.
Answer: The mass of CH3COOCH3 is 7.4 g.