X of enthanamine was subjected to reaction with \(NaNO_2/HCl\) followed by hydrolysis to liberate \(N_2\) and HCl. The HCl generated was completely neutralised by 0.2 moles of NaOH. X is ____ g.
Correct Answer: 9
Solution and Explanation
The reaction sequence is as follows:
Step 1: Reaction of ethanamine with \(\text{NaNO}_2\) and \(\text{HCl}\)
Ethanamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)) reacts with \(\text{NaNO}_2\) and \(\text{HCl}\) to form an unstable diazonium salt (\(\text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^-\)):
\[\text{CH}_3\text{CH}_2\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^- + \text{NaCl} + \text{H}_2\text{O}.\]
Step 2: Hydrolysis of the diazonium salt
The diazonium salt decomposes upon hydrolysis, liberating \(\text{N}_2\), \(\text{HCl}\), and ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)):
\[\text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 + \text{HCl}.\]
Step 3: Calculation of the mass of ethanamine
The reaction generates \(0.2\) moles of \(\text{HCl}\), which neutralizes \(0.2\) moles of \(\text{NaOH}\). From the stoichiometry of the reaction, 1 mole of ethanamine produces 1 mole of \(\text{HCl}\). Therefore, the moles of ethanamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)) required are \(0.2\) moles.
The molar mass of ethanamine is:
\[\text{Molar mass of } \text{CH}_3\text{CH}_2\text{NH}_2 = 12 + 3 + 3 + 12 + 2 + 14 + 1 = 44 \, \text{g/mol}.\]
The mass of ethanamine is:
\[\text{Mass} = \text{Moles} \times \text{Molar mass} = 0.2 \times 44 = 8.8 \, \text{g} \, (\text{Nearest Integer to } 9).\]
Final Answer: \(X = 9\).
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