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

Consider the above reaction. The percentage yield of amide product is (Round off to the Nearest Integer). (Given : Atomic mass : $\mathrm{C}:12.0\mathrm{u},\mathrm{H}:1.0\mathrm{u}$ $\mathrm{N}:14.0\mathrm{u},\mathrm{O}:16.0\mathrm{u},\mathrm{Cl}:35.5\mathrm{u})$

In the above reaction, $3.9 \mathrm{g}$ of benzene on nitration gives $4.92 \mathrm{g}$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _________ $\%$. (Round off to the Nearest Integer).

(Given atomic mass : $\mathrm{C} :12.0 \mathrm{u}, \mathrm{H} :1.0 \mathrm{u}$$\mathrm{O} :16.0 \mathrm{u},\mathrm{N} :14.0 \mathrm{u}$)