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  • 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 in this reaction are w

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 in this reaction are n \times 10^{-1}, when n = _________. (Round off to the Nearest Integer). [Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]  
 

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best_answer

Given,

A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide.

Product of the mole = 0.1.

The mole of product and reactant are the same which is 0.1 mole.

So, moles of CH3Br = 3 x 0.1 = 3 x 10-1

Ans = 3

Posted by

Kuldeep Maurya

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