To solve this problem, we can use the ideal gas law to find the number of moles of N2 gas produced, and then use stoichiometry to relate the moles of N2 gas to the moles of alanine.
First, we need to convert the given volume of N2 gas to liters and the pressure to atm to match the units in the ideal gas law.
Given:
- Volume of N2 gas = 12.4 mL
- Pressure = 755 mmHg
- Temperature = 25°C
We convert the volume to liters and the pressure to atm:
- Volume = 12.4 mL = 0.0124 L
- Pressure = 755 mmHg * (1 atm / 760 mmHg) = 0.9934 atm
Now, we can use the ideal gas law to find the number of moles of N2 gas:
\[ PV = nRT \]
Where:
- \( P \) is the pressure in atm
- \( V \) is the volume in liters
- \( n \) is the number of moles
- \( R \) is the ideal gas constant (\( 0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K} \))
- \( T \) is the temperature in Kelvin
\[ n = \frac{PV}{RT} \]
\[ n = \frac{(0.9934 \, \text{atm})(0.0124 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K})(298 \, \text{K})} \]
\[ n \approx 5.75 \times 10^{-4} \, \text{moles} \]
Now, let's consider the stoichiometry of the reaction:
\[ 1 \, \text{mole of alanine} + 1 \, \text{mole of HNO}_2 \rightarrow 1 \, \text{mole of N}_2 \]
From the balanced equation, we see that 1 mole of alanine reacts to produce 1 mole of N2.
So, the number of moles of alanine is also \( 5.75 \times 10^{-4} \, \text{moles} \).
The molar mass of alanine (\( \text{NH}_2\text{CH(CH}_3\text{)COOH} \)) is:
\[ \text{Molar mass of alanine} = 1(14.01) + 2(1.01) + 1(12.01) + 2(1.01) + 12.01 + 16.00 + 16.00 + 1.01 + 16.00 = 89.09 \, \text{g/mol} \]
Finally, we can use the number of moles of alanine and its molar mass to find the mass of alanine:
\[ \text{Mass of alanine} = \text{number of moles} \times \text{molar mass} \]
\[ \text{Mass of alanine} = (5.75 \times 10^{-4} \, \text{moles})(89.09 \, \text{g/mol}) \]
\[ \text{Mass of alanine} \approx 0.0516 \, \text{grams} \]
So, approximately 0.0516 grams of alanine was originally present in the sample.