4C₃H₅(ONO₂)₃ = 12CO₂ + 10H₂O + 6N₂ + O₂ + Q
Vgaze = 29/4 = 7,25 kmoli gaze
0,75*0,908=0,681t = 681 kg TNT pur
n=m/M=681/227=3 kmoli TNT
4 kmoli TNT......................................19 kmoli gaze
3kmoli.............................................x=14,25 kmoli gaze
n=V/Vm ⇒ V=22,4*14,25=319,2m³ gaze