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

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 $45 \times 10^{-1} \, \text{g}.$