Question

Prin clorurarea fotochimica a unei probe de metan se obtine un amestec format din CH3Cl,CH2Cl2 si CHCl3 in rap molar 5:4:1.Stiind ca HCl separat cantitativ din reactia de clorurare este neutralizat de 6,4kg de sol de NaOH,de concentratie 20%,determinati:
a) masa amestec de compusi halogenati
b) masa amestec de compusi organici rezultati prin hidroliza compusilor halogenati
c) volumul (c.n) de metan supus clorurarii daca 1 la suta din metanul introdus nu a reactionat

Answer 1

CH4 + Cl2 = CH3Cl + HCl
CH4 + 2 Cl2 = CH2Cl2 + HCl
CH4 + 3 Cl2 = CHCl3 + 3 HCl
NaOH + HCl = NaCl + H2O
c = md/ms * 100
md = 6,4*20/100 = 1,28 kg NaOH
40 g NaOH.....1 mol HCl
1,28 kg NaOH.............a
a = 0,032 kmoli HCl = 32 moli HCl
Notăm cu x,y,z - nr de moli de derivați obținuți
1 mol CH3Cl.........1 mol HCl
x..................................x'
x = x'
1 mol CH2Cl2.........2 moli HCl
y..................................y'
2y = y'
1 mol CHCl3..................3 moli HCl
z.......................................z'
z' = 3z
x'+y'+z' = 32
x/y = 5/4 => x = 5y/4 => x' = 5y/4 = x 
y/z = 4/1 => z = y/4 => z' = 3z = 3y/4 => z= y/4
5y/4+2y+3y/4 = 32
5y+8y+3y = 128
16y = 128
y = 8 moli CH2Cl2
x = 5*8/4 = 10 moli CH3Cl
z = 8/4 = 2 moli
m compuși halogenați = 10*50,5 + 8*85 + 2*119,5 = 1424 g 
b) CH3Cl + H2O = CH3OH + HCl
1 mol CH3Cl............32 g metanol
10 moli........................a
a = 320 g metanol
CH2Cl2 + H2O = CH2O + 2 HCl
1 mol CH3Cl............30 g metanal
8 moli........................b
b = 240 g metanal
CH3Cl + 2 H2O = HCOOH + 3 HCl
1 mol CHCl3............46 g acid metanoic
2 moli........................c
c = 92 g acid metanoic
m amestec = 320 + 240 + 92 = 652 g
n CH4 = 2+8+10 = 20 moli
V CH4 = n*22,4 = 20*22,4 = 448 l CH4
randament = V practic/V teoretic * 100
V teoretic = 448/99*100 = 452,52 l CH4