Question

Doua hidrocarburi aromatice mononucleare omoloage formeaza,prin oxidare energica,acelasi acid carboxilic.Amestecul echimolecular al celor doua arene formeaza,prin halogenare catalitica,un amestec de monocloroderivati cu un continut de 26.59% clor.Ce cantitate de acid carboxilic,de puritate 60%, se obtine daca se oxideaza 396 g de amestec,cu un randament al reactiei de oxidare de 80%?
R: 325.3g acid ftalic

Answer 1

CxH(2x-6) + Cl2 ⇒ CxH(2x-7)Cl + HCl       M;     35,5·n g Cl
CyH(2y -6) + CL2 ⇒ CyH(2y-7)Cl + HCl    y = x+1    M+14    35,5n g Cl
2·35,5·n·100/[n(M+34,5+M+48,5) = 26,59
267= 2M +83    ⇒  M = 92 g/mol    14x - 6 = 92  14x = 98    x = 7  y = 8
C6H5-CH3 +3 [O]    ⇒ C6H5-COOH + H2O
C6H5-CH2-CH3 +6[O] ⇒ C6H5-COOH + CO2 + 2H2O
acid benzoic  - nu acid ftalic
92n + 106n = 396       198n = 396      n = 2
n acid = 2+2 = 4 moli  acid pur - teoretic
n practic = 0,8·4 = 3,2 moli    m = 3,2·122 = 390,4g
m imp. = 390,4·100/60 = 650,66g