An aqueous solution of ammonium nitrite decomposes when heated to give off nitrogen gas and water. This reaction can be used to prepare pure nitrogen gas. How many grams of ammonium nitrite must have reacted if 3.75 liters of nitrogen gas was collected over water at 25˚C and 97.8 kPa? I solved it and got 11.8g but my teacher said its wrong because of the over water thing. Can you show me steps to get the correct answer. Thanks
To solve this problem, you need to use the Ideal Gas Law equation to account for the presence of water vapor in the collected nitrogen gas. Here are the steps to get the correct answer:
Step 1: Convert the given temperature from Celsius to Kelvin:
Temperature (K) = Temperature (˚C) + 273.15
Temperature (K) = 25 + 273.15 = 298.15 K
Step 2: Convert the given pressure from kilopascals (kPa) to atmospheres (atm):
Pressure (atm) = Pressure (kPa) / 101.325
Pressure (atm) = 97.8 / 101.325 = 0.965 atm
Now, let's calculate the number of moles of nitrogen gas (N2) using the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (atm)
V = volume (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature (K)
Step 3: Calculate the volume of dry nitrogen gas:
Volume of dry nitrogen gas (V) = Volume collected – Volume of water vapor
Since the water vapor pressure at 25˚C is 23.76 mmHg (or 0.0313 atm), we need to subtract its contribution from the total pressure to get the partial pressure of nitrogen gas:
Partial pressure of nitrogen gas (P) = Total pressure – Vapor pressure of water
Partial pressure of nitrogen gas (P) = 0.965 atm – 0.0313 atm = 0.9337 atm
Step 4: Substitute the values into the Ideal Gas Law equation:
(0.9337 atm) × (V) = (n) × (0.0821 L·atm/(K·mol)) × (298.15 K)
Step 5: Rearrange the equation to solve for the number of moles (n):
n = (0.9337 atm) × (V) / (0.0821 L·atm/(K·mol)) × (298.15 K)
Step 6: Convert the given volume of nitrogen gas from liters to moles using the calculated number of moles:
n = (0.9337 atm) × (3.75 L) / (0.0821 L·atm/(K·mol)) × (298.15 K)
Step 7: Calculate the number of moles:
n = 0.145 mol
Finally, let's convert the moles of ammonium nitrite (NH4NO2) to grams using its molar mass:
Molar mass of NH4NO2:
NH4NO2 = 14.01 g/mol (N) + 4(1.01 g/mol) (H)
+ 14.01 g/mol (N) + 2(16.00 g/mol) (O)
= 80.06 g/mol
Step 8: Calculate the mass of ammonium nitrite:
Mass = Number of moles × Molar mass
Mass = 0.145 mol × 80.06 g/mol
The correct mass of ammonium nitrite is 11.6 g (rounded to two decimal places). Hence, your teacher may have rounded differently, resulting in a slightly different answer.