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
Approach Solution - 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: \(X = 9\).
Approach Solution -2
Ethylamine (C₂H₅NH₂) reacts with NaNO₂ and HCl to form an unstable diazonium salt, which decomposes to liberate N₂ gas and form ethanol along with HCl:
\[ C_2H_5NH_2 + HNO_2 + HCl \rightarrow C_2H_5OH + N_2 + H_2O + HCl \]
Step 2: Analyze the reaction stoichiometry.
From the equation, 1 mole of ethylamine produces 1 mole of HCl.
Given that the amount of HCl formed is completely neutralized by 0.2 moles of NaOH:
\[ NaOH + HCl \rightarrow NaCl + H_2O \]
Therefore, moles of HCl = moles of NaOH = 0.2 mol.
Step 3: Determine moles of ethylamine (X).
Since 1 mole of ethylamine produces 1 mole of HCl,
moles of ethylamine = 0.2 mol.
Step 4: Calculate mass of ethylamine.
Molar mass of ethylamine (C₂H₇N) = (2×12) + (7×1) + (14) = 24 + 7 + 14 = 45 g/mol
Mass = moles × molar mass
\[ m = 0.2 × 45 = 9 \, \text{g} \]
Step 5: Final Answer.
The mass of ethylamine (X) is 9 g.
Final Answer: 9 g
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