One mole of Ethylamine when reacts with nitrous acid will produce dinitrogen gas (at `0^@C` and 1 atmsopheric pressure) equal to

Options

A. 22.4 L
B. 1 L
C. 11. 2 L
D. 24.8 L

Answer

To solve the problem of how much dinitrogen gas (N₂) is produced when one mole of ethylamine (C₂H₅NH₂) reacts with nitrous acid (HNO₂), we can follow these steps:

### Step 1: Write the reaction
Ethylamine reacts with nitrous acid to form a diazonium salt, which is unstable and decomposes to produce nitrogen gas, alcohol, and ethene. The overall reaction can be summarized as follows:

\[ C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5N_2^+ + H_2O \]
\[ C_2H_5N_2^+ \rightarrow C_2H_5OH + CH_2=CH_2 + N_2 \]

### Step 2: Identify the moles of nitrogen gas produced
From the reaction, we can see that one mole of ethylamine produces one mole of nitrogen gas (N₂). Therefore, when one mole of ethylamine reacts, it will yield one mole of nitrogen gas.

### Step 3: Calculate the volume of nitrogen gas at STP
At standard temperature and pressure (STP), which is defined as 0°C and 1 atmosphere, one mole of any ideal gas occupies a volume of 22.4 liters. Since we have one mole of nitrogen gas produced:

\[ \text{Volume of } N_2 = 1 \text{ mole} \times 22.4 \text{ L/mole} = 22.4 \text{ L} \]

### Conclusion
Thus, the volume of dinitrogen gas produced when one mole of ethylamine reacts with nitrous acid at STP is **22.4 liters**.

### Final Answer
The correct option is **A: 22.4 liters**.

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