Answer:
c) 22
Explanation:
Let's consider the following balanced equation.
N₂(g) + 3 H₂(g) ----> 2 NH₃(l)
According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:
[tex]170gNH_{3}.\frac{1molN_{2}}{34.0gNH_{3}} =5.00molN_{2}[/tex]
According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:
[tex]170gNH_{3}.\frac{3molH_{2}}{34.0gNH_{3}} =15.0molH_{2}[/tex]
The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.
We can calculate the pressure (P) using the ideal gas equation.
P.V = n.R.T
where
V is the volume (50.0 L)
n is the number of moles (20.0 mol)
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (400.0 + 273.15 = 673.2K)
[tex]P=\frac{n.R.T}{V} =\frac{20.0mol\times (0.08206atm.L/mol.K)\times 673.2K ) }{50.0L} =22.0atm[/tex]