Question

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.

Answer 1

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: $X = 9$.