A reaction of $$0.1$$ mole of benzyl amine with bromomethane gave $$23g$$ 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: $$12.0{\text{ }}u$$ , H: $$1.0{\text{ }}u$$, N: $$14.0{\text{ }}u$$, Br: $$80.0{\text{ }}u$$ ]

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Vedantu Content Team · Accepted Answer

Hint: The structure of benzyl amine is as follows:Image: Structure of benzylamineThe structure of benzyl trimethyl ammonium bromide is as follows:Image: structure of benzyl trimethyl ammonium bromideComplete Step by Step Solution:There is a reaction between benzylamine and bromoethane produces benzyl trimethyl ammonium bromide. The balanced chemical reaction is as follows:Image: reaction between benzylamine and bromomethaneFrom the reaction stoichiometry, one mole of benzylamine reacts with three moles of bromomethane and produces one mole of benzyl trimethyl ammonium bromide.Here, $$0.1$$ mole of benzyl amine reacts with bromomethane and gave $$23g$$ of benzyl trimethyl ammonium bromide. First of all, we have to calculate the molar mass of benzyl trimethyl ammonium bromide.From the structure, the chemical formula of benzyl trimethyl ammonium bromide is $${C_{10}}{H_{16}}NBr$$ .Calculate the molar mass of $${C_{10}}{H_{16}}NBr$$ from the given atomic masses as follows:$$  {\text{molar mass of }}{C_{10}}{H_{16}}NBr = 10\left( {12.0} \right) + 16\left( {1.0} \right) + 14.0 + 80.0   \\   \Rightarrow {\text{molar mass of }}{C_{10}}{H_{16}}NBr = 230   \\  $$Therefore, the molar mass of $${C_{10}}{H_{16}}NBr$$ is $$230$$.We have to calculate the moles of $${C_{10}}{H_{16}}NBr$$ from given mass ( $$23g$$ ).We know that,$$n = \dfrac{{weight}}{{mol.mass}}$$Substituting the values,$$  {n_{{C_{10}}{H_{16}}NBr}} = \dfrac{{23}}{{230}}   \\   \Rightarrow {n_{{C_{10}}{H_{16}}NBr}} = 0.1   \\  $$From the reaction stoichiometry,  $$1$$ mole of $${C_{10}}{H_{16}}NBr$$ formed from $$3$$ moles of bromomethane.Calculate the number of moles of bromomethane requires for $$0.1$$ mole of $${C_{10}}{H_{16}}NBr$$ formation as follows:$$  {n_{C{H_3}Br}} = 0.1{\text{ }} \times \dfrac{3}{1}   \\   \Rightarrow {n_{C{H_3}Br}} = 0.3   \\  $$The number of moles consumed in the reaction are $$n \times {10^{ - 1}}$$ .So, we can say that,$$  n \times {10^{ - 1}} = {n_{C{H_3}Br}}   \\   \Rightarrow n \times {10^{ - 1}} = 0.3   \\   \Rightarrow n = 3   \\  $$Therefore, the value of $$n$$ is $$3$$ .Note: Mole, or mol, is a common scientific unit in chemistry that is used to measure vast amounts of tiny objects such as molecules, atoms, or other particles. The number of moles of a solute in a litre of solution is known as molarity. A mole is a unit of measurement that helps us match the particles of a substance to its mass. The molecular weight, also known as molar mass, is the sum of the masses of each atom in grammes that make up a mole of a molecule.