JEE MAIN - Chemistry (2024 - 5th April Evening Shift - No. 29)
$$\mathrm{X} \mathrm{~g}$$ of ethanamine was subjected to reaction with $$\mathrm{NaNO}_2 / \mathrm{HCl}$$ followed by hydrolysis to liberate $$\mathrm{N}_2$$ and $$\mathrm{HCl}$$. The $$\mathrm{HCl}$$ generated was completely neutralised by 0.2 moles of $$\mathrm{NaOH} . \mathrm{X}$$ is _________ g.
Answer
9
Explanation
$$\begin{aligned} & \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{mol} \text { of } \mathrm{HCl}=0.2 \mathrm{~mol} \end{aligned}$$
$$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2 \xrightarrow[+\mathrm{HCl}]{\mathrm{NaNO}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{N}_2^{+} \mathrm{Cl}^{-} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{N}_2+\mathrm{HCl}$$
$$0.2 \mathrm{~mol} \mathrm{~HCl}$$ would be generated by $$0.2 \mathrm{~mol}$$ $$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$$
$$\mathrm{x}=0.2 \times 45=9 \mathrm{~g}$$
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