Question

Problema 27 .
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Answer 1

CxH2x + [O] + H2O ⇒ CxH2x(OH)2
M = 14x                      M = (14x + 34)
Δm = 34 =1,21428·M     M = 28g/mol     14x = 28          x = 2      CH2=CH2
m etena = 28n    m diol = 28n + 4,25
m/28 = (m+ 4,25)/62   62m = 28m +263,5     34m = 263,5    m = 7,75 g
PV = nRT   V = 7,75·0,082·400/(28·2) = 4,539 l
CH2=CH2 + 1/2 O2 ⇒ (CH2)2O
nO2 necesar = n etena/2 = 7,75/56 moli
nO2 folosit = 150/100 ·7,75/56 = 0,2 moli   V = 0,2x22,4 = 4,48 l