CH4 + NH3 + 3/2 O₂ ⇒ HCN + 3H₂O
0,025 kmoli CH4 = 25 moli CH4
p=n pur/n imp*100 ⇒ np = 60*25/100 = 15 moli CH4 pur
a) MHCN=27g/mol
1 mol CH4...........................27g HCN
15 moli CH4.........................x=405g HCN (mt)
n(eta) = mp/mt*100 => mp=n*mt/100 = 80*405/100 = 324g HCN
b) 1 mol CH4..................22,4 L NH3
15 moli CH4.....................x = 336 L NH3 (V teoretic)
n(eta) = Vp/Vt*100 => Vp = n*Vt/100 = 80*336/100 = 268,8 L NH3 pur
p=V pur/Vimp*100 => Vimp=Vpur*100/p = 268,8*100/90 = 298,666 L NH3 impur
c) 1 mol CH4................1,5*22,4 L O2
15 moli CH4...................x=504 L O2 (V teoretic)
n(eta) = Vp/Vt*100 => Vp=n*Vt/100 = 80*504/100 = 403,2 L O2
Vaer(20%O2) = 5*VO2 = 5*403,2 = 2016 L aer