9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is _____ $$\times 10^{-1}$$ g.
(Given molar mass in g mol$$^{-1}$$: N = 14, O = 16, C = 12, H = 1)

The reaction of aniline with acetic anhydride:

$$C_6H_5NH_2 + (CH_3CO)_2O \to C_6H_5NHCOCH_3 + CH_3COOH$$

Molar mass of aniline = $$6(12) + 7(1) + 14 = 93$$ g/mol.

Molar mass of acetanilide = $$8(12) + 9(1) + 14 + 16 = 135$$ g/mol.

Moles of aniline = $$\frac{9.3}{93} = 0.1$$ mol.

Moles of acetanilide = 0.1 mol (1:1 ratio).

Mass of acetanilide = $$0.1 \times 135 = 13.5$$ g = $$135 \times 10^{-1}$$ g.

The answer is $$\boxed{135}$$.