A reaction of \( 0.1 \) mole of Benzylamine with bromomethane gave \( 23 \, \text{g} \) 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: \( \text{C} : 12.0 \, \text{u} \), \( \text{H} : 1.0 \, \text{u} \), \( \text{N} : 14.0 \, \text{u} \), \( \text{Br} : 80.0 \, \text{u} \)]