Question

Ce masa de clorura de metil se obtine in reactia dintre metal si 44,8L de clor de puritate 75%? ​

Answer 1

Răspuns:

CH4+Cl2---->CH3Cl+HCl

V=44,8L(Vi)

P=Vp/Vi × 100=>Vp/44,8=75/100=>Vp=33,6L Cl

M CH4=12+4=16g/mol

m CH4=16×22,4/33,6=10,66g CH4