Final answer:
The mass of ammonia in a 6.64 L cylinder at 4.76 atm and 25.00°C is calculated to be approximately 19.6 g after using the ideal gas law and the molar mass of ammonia. However, this answer is not listed among the options provided to the student.
Step-by-step explanation:
To calculate the mass of ammonia in a 6.64 L cylinder at 4.76 atm and 25.00°C, we use the ideal gas law, which is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.08206 L atm/K mol), and T is temperature in Kelvin. First, we need to convert the temperature to Kelvin: T = 25.00 + 273.15 = 298.15 K. Using the ideal gas law:
PV = nRT
(4.76 atm)(6.64 L) = n(0.08206 L atm/K mol)(298.15 K)
Now solving for n (moles of NH₃):
n = (4.76 atm)(6.64 L) / (0.08206 L atm/K mol)(298.15 K) ≈ 1.150 moles
To find the mass of ammonia, we multiply the number of moles by the molar mass of ammonia (17.04 g/mol):
Mass = n × molar mass
Mass = 1.150 moles × 17.04 g/mol ≈ 19.596 g
The mass of ammonia closest to this calculation and offered in the options is 19.6 g rounded to 3 significant figures, therefore the correct answer is not listed among the options provided to the student.