Question

Buna, am si eu o problema care imi da batai de cap
Se supun clorurarii 179,2 L metan. In masa de reactie se gasesc monoclorometan, diclorometan si metan nereactionat in raport molar de 3:3:2. Volumul de clor consumat (c.n.) este de?

Answer 1

avem doar mono si diclorurare
raport de 3:3:2 => 3x moli:3x moli:2x moli

n=V/Vm=179,2/22,4=8moli metan
1mol..1mol...1mol
CH4 + Cl2 -> CH3-Cl + HCl

3x.....3x.......3x

1mol...2moli.....1mol

CH4 + 2Cl2 -> CH2Cl2 + 2HCl

3x......6x.........3x

1mol....1mol

CH4 -> CH4

2x........2x

obtinem ca 8x=8 => x=1mol

nCl2=9moli => VCl2=9*22,4=201,6l Cl2 (c.n)