Question

196,8 L metan (27 grade C si 3 atm) reactioneaza cu 10% exces de clor gazos la lumina, rezultand un amestec organic care contine in raport molar CH3Cl:CH2Cl:CHCl3:CCl4=1:2:3:2. Sa se calculeze:
a) masa de cloroform care se obtine
b)Volumul de clor (c.n)

Nu imi dau ca la sfarsit, va rog frumos..

Answer 1

n = PV/(RT) = 196,8·3/(0,082·300) = 24 moli CH4
n1 moli CH4 + Cl2 ⇒ CH3Cl + HCl
n2 moli CH4 + 2Cl2 ⇒ CH2Cl2 + 2HCl
n3 moli CH4 + 3Cl2 ⇒ CHCl3 + 3HCl
n4 moli CH4 + 4Cl2 ⇒ CCl4 + 4HCl
n1/1 = n2/2 = n3/3 = n4/2 = n/8      n1 = n/8 =3moli     n2 = n/4 = 6moli      n3 =3n/8 = 9 moli    n4 = n/4 = 6 moli
a) n CHCl3 = n3 = 9 moli    m = 9·119,5 = 1075,5g
nCl2 reactionat = n1 + 2n2 + 3n3 + 4n4 = 3+ 12 + 27 + 24 = 66moli
nCl2 folosit = 110/100 ·66 = 72,6 moli     V = 72,6·22,4 = 1626,24 l