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Question: Answered & Verified by Expert
A 90 g of ethyl amine on reaction with methyl chloride produced a tertiary amine as an exclusive product. The amount of methyl chloride required is
[Given mass in amu : H=1, C=12, N=14, Cl=35.5]
ChemistrySome Basic Concepts of ChemistryTS EAMCETTS EAMCET 2021 (05 Aug Shift 2)
Options:
  • A 50.5 g
  • B 101 g
  • C 202 g
  • D 303 g
Solution:
1888 Upvotes Verified Answer
The correct answer is: 202 g

The Standard Reaction is:

CH3CH2NH2 + 2CH3Cl  CH33N + HCl

Molar mass of ethyl amine = 45 g

The molar mass of methyl chloride = 50.5 g

Given Moles ethyl amine= 9045=2 

As per the Standard reaction,

1 mole of CH3CH2NH2 react with 2 mole of CH3Cl

So, 2 moles of CH3CH2NH2 react with 4 moles of CH3Cl

Hence, the amount methyl chloride obtained is 4×50.5 = 202 g

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