Question

\( X \, \text{g} \) of ethylamine is subjected to reaction with \( \text{NaNO}_2/\text{HCl} \) followed by water; evolved dinitrogen gas which occupied \( 2.24 \, \text{L} \) volume at STP. \( X \) is ______ \( \times 10^{-1} \, \text{g} \).

Answer 1

Correct Answer: 45

The reaction of ethylamine (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)) with \( \text{NaNO}_2/\text{HCl} \) followed by hydrolysis produces ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)) and dinitrogen gas (\( \text{N}_2 \)) as follows:

\[ \text{CH}_3\text{CH}_2\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 \]

Volume of \( \text{N}_2 \) Produced:

Given that \( \text{N}_2 \) evolved occupies 2.24 L at STP.

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 2.24 L corresponds to:

\[ \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mole} \]

Calculating the Mass of Ethylamine (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)):

Molar mass of \( \text{CH}_3\text{CH}_2\text{NH}_2 = 45 \, \text{g/mol}. \)

Since 0.1 mole of \( \text{N}_2 \) is produced, this means 0.1 mole of ethylamine was used.

Mass of ethylamine (\( X \)):

\[ X = 0.1 \times 45 = 4.5 \, \text{g}. \]

Expressing \( X \) in Terms of \( 10^{-1} \):

\[ X = 4.5 \, \text{g} = 45 \times 10^{-1} \, \text{g}. \]

Conclusion:

The value of \( X \) is \( 45 \times 10^{-1} \, \text{g}. \)