Răspuns :
CH4 + O2 ⇒ C↓ + 2H2O
V CH4 pur = 6,72·0,96 = 6,4512 n = 0,288 kmoli
nC = nCH4 = 0,288kmoli m = 0,288·12 = 3,456kg
V CH4 pur = 6,72·0,96 = 6,4512 n = 0,288 kmoli
nC = nCH4 = 0,288kmoli m = 0,288·12 = 3,456kg
CH4 + O2 ---> C↓ + 2H2O
n=V/Vm=6,72/22,4=0.3 kmoli CH4
puritatea = npur/nimpur*100==>npur=p*nimp/100=96*0.3/100=0.288 kmoli CH4 pur
n=m/M==>m=n*M=0.288*12=3,456kg C pur
AC=12
n=V/Vm=6,72/22,4=0.3 kmoli CH4
puritatea = npur/nimpur*100==>npur=p*nimp/100=96*0.3/100=0.288 kmoli CH4 pur
n=m/M==>m=n*M=0.288*12=3,456kg C pur
AC=12