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(C) The reaction of ethanamine $(CH_3CH_2NH_2)$ with $NaNO_2 / HCl$ produces ethyl diazonium chloride,which upon hydrolysis yields ethanol,$N_2$,and $

Vedclass · Accepted Answer

$9$

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(C) The reaction of ethanamine $(CH_3CH_2NH_2)$ with $NaNO_2 / HCl$ produces ethyl diazonium chloride,which upon hydrolysis yields ethanol,$N_2$,and $HCl$.The overall reaction is: $CH_3CH_2NH_2 + NaNO_2 + HCl ightarrow CH_3CH_2OH + N_2 + NaCl + H_2O$.Wait,the question states $HCl$ is liberated and neutralized by $0.2 \ mol$ of $NaOH$. Based on the stoichiometry,$1 \ mol$ of ethanamine produces $1 \ mol$ of $HCl$.Given that $0.2 \ mol$ of $NaOH$ is required to neutralize the $HCl$,it implies $0.2 \ mol$ of $HCl$ was produced.Therefore,$0.2 \ mol$ of ethanamine was used.The molar mass of ethanamine $(CH_3CH_2NH_2)$ is $12 	imes 2 + 1 	imes 7 + 14 = 45 \ g/mol$.Mass $X = 	ext{moles} 	imes 	ext{molar mass} = 0.2 \ mol 	imes 45 \ g/mol = 9 \ g$.Thus,$X = 9$.

$X \ g$ of ethanamine was subjected to reaction with $NaNO_2 / HCl$ followed by hydrolysis to liberate $N_2$ and $HCl$. The $HCl$ generated was completely neutralized by $0.2 \ mol$ of $NaOH$. $X$ is . . . .

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