Question

Cum rezolv problemele 2 si 3?

Answer 1

2.
acid :   CxH2xO2         M  = 14x + 32
acid + C2H6O ⇒ ester + H2O
ester : C(x+2)H(2x+4)O2   Me = M + 46 - 18 =  M + 28
M + 28 = 146,66/100 · M     M + 28 = 1,4666M     0,4666M = 28    M = 60 g/mol
14x + 32 = 60     x= 2         acid = CH3COOH
3. acid maleic; acid fumaric  : C4H6O4
C4H4O4 + H2 ⇒ C4H6O4
n acizi = 5,8/ 116 = 0,05 kmoli = 50 moli = nH2
PV = nRT   V = 50·0,082·273·760/800 = 1064 l