Răspuns:
volumul de CH4 ce intra in reactii: 8l
numarul de moli ce intra in reactii: niu= 8 l/22,4l/mol=0,3571molCH4
4a 4a 4a 4a
R1: CH4 + Cl2---> CH4Cl + HCl
3a 6a 3a 6a
R2 CH4 + 2Cl2---> CH2Cl2+2HCl
1a 3a 1a 3a
R3 CH4 + 3Cl2--->CHCl3 +3HCl
total moli Ch4= (4a+3a+1a)= 0,3571---> a=0,0446 mol
moli Cl2 in R2: niu=6X0,0446mol=0,2676mol
V Cl2= niuxVm=0,2676molx22,4 l/mol= 6l
Explicație: