5C4H8 + 8KMnO4 + 12H2SO4 => 10H3C-COOH + 4K2SO4 + 8MnSO4 + 12H2O
5 moli C4H8.........8 moli KMnO4
4 moli C4H8.........X = 6,4 moli
C=n/Vs => Vs =n/Cm = 6,4/0,1 = 64 L KMnO4
5 moli C4H8..........10 moli acid
4 moli. C4H8........X = 8 moli acid
md acid = 8*60 = 480g
ms = 480+600 = 1080g
Cp = md/ms*100 = 480/1080*100 = 44,44℅
3. C2H4 + H2O => C2H5-OH (1)
5C2H5-OH + 4KMnO4 + 6H2SO4 --> 5CH3-COOH + 2K2SO4 + 4MnSO4 + 6H2O
n=V/Vm = 67,2/22,4 = 3 moli C2H4
nC2H4 = nC2H5-OH = 3 moli
0,6*3 = 1,8 moli care participă la reacție (alcool)
5 moli alcool....... 4 moli KMnO4
1,8 moli alcool.....X= 1,44 moli KMnO4
C = n/Vs => Vs = n/Cm = 1,44/0,2 = 7,2 L KMnO4