Calculate the mass of sodium acetate that will result from the interaction of glacial acid with excess sodium hydroxide.

Given:
m (CH3COOH) = 6 g

To find:
m (CH3COONa) -?

1) CH3COOH + NaOH => CH3COONa + H2O;
2) M (CH3COOH) = Mr (CH3COOH) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O ) + Ar (O) * N (O) + Ar (H) * N (H) = 12 * 1 + 1 * 3 + 12 * 1 + 16 * 1 + 16 * 1 + 1 * 1 = 60 g / mol ;
3) n (CH3COOH) = m (CH3COOH) / M (CH3COOH) = 6/60 = 0.1 mol;
4) n (CH3COONa) = n (CH3COOH) = 0.1 mol;
5) M (CH3COONa) = Mr (CH3COONa) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (C) * N (C) + Ar (O) * N (O ) + Ar (O) * N (O) + Ar (Na) * N (Na) = 12 * 1 + 1 * 3 + 12 * 1 + 16 * 1 + 16 * 1 + 23 * 1 = 82 g / mol ;
6) m (CH3COONa) = n (CH3COONa) * M (CH3COONa) = 0.1 * 82 = 8.2 g.

Answer: The mass of CH3COONa is 8.2 g.