5x 5x 5x
CH4 + Cl2 --> CH3Cl + HCl
1 1 1
3x 3x 6x
CH4 + 2Cl2 --> CH2Cl2 + 2HCl
1 1 2
x x 3x
CH4 + 3Cl2 --> CHCl3 + 3HCl
1 1 3
ms.HCl = 1022 g, c% = 25%
stim ca c% = mdx100/ms
=> md = msxc%/100
= 1022x25/100 = 255,5 g HCl
avem rezultati 5x+6x+3x = 14x moli HCl
=> 14x36,5 = 255,5 => x = 0,5
=> metan intrat = 5x+3x+x = 9x moli
=> m.CH4 = 9x16 = 72 g