n = m/M = 68/17 = 4 kmoli NH3
N2 + 3H2 => 2NH3
4 kmoli amestec (N2+3H2)......W kmoli NH3
X kmoli amestec...........4 kmoli NH3
X = 4*4/2 = 8 kmoli amestec
T = 273+t°C = 390K
pV = nRT => V = nRT/p = 8*0,082*390/4 = 63,96 m3 amestec = 63960L amestec
pV =nRT => V = nRT/p = 4*0,082*390/4 = 31,98 m3 NH3