Question

Prin clorurarea fotochimica a metanului se obtine un amestec de clorura de metil, diclorometan, cloroform si metan netransformat in raport molar 4:3:2:1. Pentru neutralizarea HCL rezultat din reactie se folosesc 240g solutie NaOH de concentratie 30%.
a) Calculati masa de metan si volumul de clor consumate in reactie.
b) Calculati conversia utila si conversia totala.

Answer 1

x...........x..........x............x
CH₄ +  Cl₂ ⇒ CH₃Cl + HCl
y..........2y............y..........2y
CH₄ + 2Cl₂⇒ CH₂Cl₂ + 2HCl
z..........3z..........z............3z
CH₄ + 3Cl₂ ⇒ CHCl₃ + 3HCl
w...........w
CH₄ ⇒ CH₄ (netransformat)

md NaOH = 30×240/100 = 72g

n NaOH = 72/40 = 1,8 moli

NaOH + HCl ⇒ NaCl + H₂O

n NaOH = n HCl = 1,8 moli

x/y = 4/3 ⇒ y = 3x/4 = 0,54 moli
x/z = 4/2 ⇒ z = 2x/4 = 0,36 moli
x/w = 4/1 ⇒ w = x/4 = 0,18 moli 

x+y+z+w = 1,8

x+3x/4+2x/4+x/4 = 1,8 ⇔ (4x+3x+2x+x)/4 = 1,8 ⇔ 10x = 7,2 ⇒ x = 0,72 moli

a) n CH₄ = x+y+z = 0,72+0,54+0,36 = 1,62 moli

m CH₄ = 1,62×16 = 25,92g

n Cl₂ = x+2y+3z = 0,72+2×0,54+3×0,36 = 2,88 moli
V Cl₂ = 2,88×22,4 = 64,512 L

b) Cu = n util/n total×100 = x/x+y+z+w = 0,72/1,8×100 = 40%

Ct = n u /n t ×100 = x+y+z/x+y+z+w = 1,62/1,8×100 = 90%

nu sunt sigur daca asa e bine :))