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c.n. → P =1atm; T=273.15Kmoli C 3 H 8 = (1×2,74)/(0.082×273.15)=0.122 molimoli O 2 = (0.954×20.5)/(0.082×300.15)=0.794 moli /5 = 0.159 (eccesso)moli CO2 = 0.122×3= 0.366 moliV CO2 = (0.366×0.082×273.15)/1= 8.2LEs. 2E’ data una miscela di 100 ml di N 2 e 400 ml di H 2 a c.n. Si ha laformazione di NH 3 fino a esaurimento del reagente in difetto. Qual è lapressione esercitata in un recipiente di 700 ml a 225°C?3H 2 + N 2 → 2NH 3moli H 2 = (1×0.4)/(0.082×273.15)=0.0178 molimoli N 2 = (1×0.1)/(0.082×273.15)=0.00446 moli3H 2+ N 2→ 2NH 30.0178 0.004460-0.0133 -0.00446 +0.008920.0044200.00892n tot = 0.00442+0.00892 = 0.0133Ptot = (0.0133×0.08206×498.15)/0.7 = 0.777atm
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