1. etena + [O] ----KMnO4+H2O----> etandiol
C=98% = md *100/2480 => md = 2430,4 g etandiol
randament = 60% = 2430,4 * 100/Ct => Ct = 4050,66 g etandiol
n = m/M => n = 4050,66/62 => n= 65,3 moli
pV = nRT => 2 * V = 65,3 * 0,082 * 500=> V = 2678,53/2 => V = 1339,265 L
2. (oxidare energica)
a) C--COOH + C--C--C--C--COOH <---------C--C=C--C--C--C--C (2 heptena)
b) C--C--CO--C + C--COOH <---------- C--C--(CH3)C=C--C (3 metil 2 pentena)