Verified Solution • 2025–26 Session moderate

QJEEM24Ch7f77: JEE Main 2024 Chemistry - Class 12 JEEM24Ch7f77 Solution: of ethanamine was subjected to reaction with followed by…

X g\mathrm{X} \mathrm{~g} of ethanamine was subjected to reaction with NaNO2/HCl\mathrm{NaNO}_2 / \mathrm{HCl} followed by hydrolysis to liberate N2\mathrm{N}_2 and HCl\mathrm{HCl} . The HCl\mathrm{HCl} generated was completely neutralised by 0.2 moles of NaOH.X\mathrm{NaOH} . \mathrm{X} is _________ g.

Step-by-Step Verified Solution

amp;HCl+NaOH→NaCl+H2Oamp;mol of HCl=0.2 mol\begin{aligned}& \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\& \mathrm{mol} \text { of } \mathrm{HCl}=0.2 \mathrm{~mol}\end{aligned}

CH3CH2NH2→+HClNaNO2CH3CH2N2+Cl−→H2OCH3CH2OH+N2+HCl\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2 \xrightarrow[+\mathrm{HCl}]{\mathrm{NaNO}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{N}_2^{+} \mathrm{Cl}^{-} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{N}_2+\mathrm{HCl}

0.2 mol HCl0.2 \mathrm{~mol} \mathrm{~HCl} would be generated by 0.2 mol0.2 \mathrm{~mol} C2H5NH2\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2

x=0.2×45=9 g\mathrm{x}=0.2 \times 45=9 \mathrm{~g}

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