Answer: A mass of 124457.96 g ammonia is produced by reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.
Explanation:
Given: Volume = 450 L
Temperature = 450 K
Pressure = 300 atm
Using ideal gas equation, moles of nitrogen are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = tempertaure
Substitute values into the above formula as follows.
![PV = nRT\\300 atm \times 450 L = n \times 0.0821 L atm/mol K \times 450 K\\n = \frac{135000}{36.945}\\= 3654.08 mol](https://tex.z-dn.net/?f=PV%20%3D%20nRT%5C%5C300%20atm%20%5Ctimes%20450%20L%20%3D%20n%20%5Ctimes%200.0821%20L%20atm%2Fmol%20K%20%5Ctimes%20450%20K%5C%5Cn%20%3D%20%5Cfrac%7B135000%7D%7B36.945%7D%5C%5C%3D%203654.08%20mol)
According to the given equation, 1 mole of nitrogen forms 2 moles of ammonia. So, moles of ammonia formed by 3654.08 moles of nitrogen is as follows.
![2 \times 3654.08 mol\\= 7308.16 mol](https://tex.z-dn.net/?f=2%20%5Ctimes%203654.08%20mol%5C%5C%3D%207308.16%20mol)
As moles is the mass of substance divided by its molar mass. So, mass of ammonia (molar mass = 17.03 g/mol) is as follows.
![Moles = \frac{mass}{molar mass}\\7308.16 = \frac{mass}{17.03 g/mol}\\mass = 124457.96 g](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C7308.16%20%3D%20%5Cfrac%7Bmass%7D%7B17.03%20g%2Fmol%7D%5C%5Cmass%20%3D%20124457.96%20g)
Thus, we can conclude that a mass of 124457.96 g ammonia is produced by reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.