Question

A reaction of $0.1$ $\mathrm{mole}$ of Benzyl amine 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) [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}$]

Answer 1

Step 1: Write the balanced the chemical equation

So $3$ moles of bromoethane is required to form $1$ mole of benzyl trimethyl ammonium bromide.

Step 2: Number of moles of benzyl trimethyl ammonium bromide present in $23\mathrm{g}$.

Molecular weight of benzyl trimethyl ammonium bromide $\left({\mathrm{C}}_{10}{\mathrm{NBrH}}_{16}\right)$

$$\begin{gathered} {\mathrm{C}}_{10}{\mathrm{NBrH}}_{16}=12\times 10+1\times 14+1\times 80+16\times 1 \\ =230\mathrm{u} \end{gathered}$$

Number of moles in $23\mathrm{g}$ is

$\frac{23}{230}=0.1\mathrm{mole}$

Step 3: Moles of bromoethane consumed.

For $1$mole product $3$ mole bromoethane is required.

So for $0.1$ mole product $0.3$ mole bromoethane will be required.

Hence the answer is $\mathbf{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{mole}$.