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…

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