Question

Prin reactia de clorurare a metanului se obtine ch2cl2 si chcl3 in raport molar de 1:3.
ce cantitati de produsi (in kg si kmoli) se obtin 4m cubi CH4 la 1,6 atm si 127°c?

Answer 1

pV=nRT=>n=pV/RT=1,6*4/0,082*400 = 0,195121 kmoli CH4

m CH4= 0,195121*16=3,12195kg CH4

x/y=1/3 => y=3x

85x+119,5y=3,12195 => 443,5x=3,12195 -> x=0,007 kmoli CH2Cl2

y=0,021118 kmoli CHCl3

mCH2Cl2=0,5983444 kg 

mCHCl3=2,5236kg