Final answer:
To find the mass of ammonium nitrite that decomposed, the partial pressure of nitrogen gas was calculated by subtracting the vapor pressure of water from the total pressure. The ideal gas law was then used to determine the number of moles of nitrogen gas, and by stoichiometry, the same number of moles of ammonium nitrite. Multiplying the moles by the molar mass gave an estimated decomposed mass of 1.23 grams (option C).
Step-by-step explanation:
To determine the mass of ammonium nitrite that decomposed, we must first calculate the pressure of the nitrogen gas. The total pressure is given as 745 torr, and the partial pressure of water vapor at 26°C is 25.21 torr. We subtract the vapor pressure from the total pressure to get the pressure of the nitrogen gas.
PN2 = Ptotal - PH2O = 745 torr - 25.21 torr = 719.79 torr
Converting this to atmospheres (since the ideal gas law uses pressure in atmospheres), we get:
PN2 in atm = 719.79 torr / 760 torr/atm ≈ 0.9471 atm
Now we can use the ideal gas law, PV = nRT, to find the number of moles (n) of N2. We are given the volume (V) as 511 mL, which we convert to liters (0.511 L), R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin (26°C + 273 = 299 K).
n = PV / RT = (0.9471 atm × 0.511 L) / (0.0821 L·atm/K·mol × 299 K) ≈ 0.0196 mol
The stoichiometry of the decomposition reaction is 1:1 for NH4NO2 to N2, so the moles of NH4NO2 is the same as the moles of N2. The molar mass of NH4NO2 is 64.06 g/mol. We calculate the mass of ammonium nitrite that decomposed:
Mass = moles × molar mass = 0.0196 mol × 64.06 g/mol = 1.26 g
The answer that is closest to our calculated mass is:
C) Mass of NH4NO2 = 1.23 g