N2 + 3 H2 ---> 2 NH3
1 mol N2.......3 moli H2........2 * 17 g NH3
x kmoli N2.......y kmoli H2.....68 kg NH3
x = 2 kmoli N2 si y = 6 kmoli H2
x + y = 2 + 6 = 8 kmoli amestec
pV=nr moli * R * T
V = nr moli * R * T / p = 8 * 0,082 * (117+273) / 4 = 63,96 m cubi
V cond normale = 22,4
V = nr moli * V cn = 8 * 22,4 = 179,2 m cubi
