10 moli a moli b moli c moli
2C6H3N3O7 + 8O2 —> 12CO2 + 3H2O(g) + 3N2
2 12 3 3
=> a = 60 moli CO2
=> b = 15 moli H2O
=> c = 15 moli N2
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total gaze = 90 moli obtinute in c.n.
PV = nRT
=> V = nRT/P
= 90x0,082x(273+500)/1
= 5704 L = 5,704 m3 gaze