Answer:
To solve this problem, we can use the ideal gas law and stoichiometry.
First, we need to convert the given temperature to Kelvin:
208.00 degrees Celsius + 273.15 = 481.15 K
Next, we can use the ideal gas law to find the volume of NH3 produced:
PV = nRT
P = 4.50 torr = 0.00592 atm (converting to atm)
V = unknown (what we are trying to find)
n = moles of NH3 produced = 26.00 moles N2 (from stoichiometry)
R = 0.0821 L·atm/K·mol (gas constant)
T = 481.15 K
Solving for V:
V = nRT/P
V = (26.00 mol)(0.0821 L·atm/K·mol)(481.15 K) / (0.00592 atm)
V = 3671.46 L
However, this is the volume of NH3 produced at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 K) and 1 atm. We need to convert this to the volume at the given temperature and pressure using the combined gas law:
(P1V1/T1) = (P2V2/T2)
P1 = 1 atm (STP pressure)
V1 = 3671.46 L (volume at STP)
T1 = 273.15 K (STP temperature)
P2 = 0.00592 atm (given pressure)
V2 = unknown (what we are trying to find)
T2 = 481.15 K (given temperature)
Solving for V2:
V2 = (P1V1T2) / (P2T1)
V2 = (1 atm)(3671.46 L)(481.15 K) / (0.00592 atm)(273.15 K)
V2 = 315491.48 L or 315491 L (rounded to two decimal places)
Therefore, 315491 L of NH3 will be produced at a temperature of 208.00 degrees Celsius and 4.50 torr pressure to consume 26.00 moles of N2.