Question

HEEELP
La obtinerea CH3Cl din metan se supun clorurarii 168l de metab, clorura consumandu-se integral. Din reactie rezulta un amestec de: CH3Cl,CH2Cl2,CHCl3,CCl4 si CH4 nereactionat in raport molar: 5:3:2:1:5. Nr de moli de CH4 nereactionat? Conversia utila ade CH4?
Multumesc anticipat

Answer 1

n1 moli    CH4 + Cl2 ⇒ CH3Cl + HCl
n2 moli     CH4 + 2Cl2⇒ CH2Cl2 + 2HCl
n3 moli    CH4 + 3Cl2 ⇒ CHCl3 + 3HCl
n4moli     CH4 + 4Cl2 ⇒ CCl4 + 4HCl
n5 moli CH4 nereactionat
n1/5 = n2/3 = n3/2 = n4/1 = n5/5 = (168/22,4)/16 = 7,5/16
n1 = 37,5/16 moli   n2 = 22,5/16 ;  n3 = 15/16;  n4 = 7,5/16 ; n5 = 37,5/16
n5 = 37,5/16 = 2,34375 moli
Cu = n1·100/n = 234,375/7,5 = 31,25%