A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n $$\times 10^{-1}$$, when n = ________. (Round off to the Nearest Integer).
(Given: Atomic masses: C: 12.0 u, H: 1.0 u, N: 14.0 u, Br: 80.0 u)

Benzylamine is $$\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$$. When it reacts with excess bromomethane ($$\text{CH}_3\text{Br}$$), the nitrogen undergoes successive N-alkylation. The final product mentioned is benzyl trimethyl ammonium bromide, $$[\text{C}_6\text{H}_5\text{CH}_2\text{N(CH}_3)_3]^+\text{Br}^-$$.

Starting from benzylamine, three methyl groups must be added to the nitrogen to form the quaternary ammonium salt. Each methylation step consumes one mole of $$\text{CH}_3\text{Br}$$: the first replaces one N-H with N-CH$$_3$$ (releasing HBr), the second replaces the second N-H, and the third adds a methyl to the tertiary amine to form the quaternary salt. So 3 moles of $$\text{CH}_3\text{Br}$$ are needed per mole of benzylamine.

Now we verify the yield. The molar mass of benzyl trimethyl ammonium bromide is: $$\text{C}_{10}\text{H}_{16}\text{NBr} = 10(12) + 16(1) + 14 + 80 = 120 + 16 + 14 + 80 = 230$$ g/mol. From 0.1 mol of benzylamine, the theoretical mass of product would be $$0.1 \times 230 = 23$$ g, which matches the given 23 g (100% yield for this problem).

The moles of bromomethane consumed are $$3 \times 0.1 = 0.3 = 3 \times 10^{-1}$$ mol. Therefore $$n = 3$$.