Question

Volumul de solutie de K2Cr2O7 1/3 M care se consuma la oxidarea a 35,2 g amestec echimolecular de alcooli tertiari si secundari cu formula moleculara C5H12O este egal cu:.....

Answer 1

Doar alcooli secundari se oxideaza, cei tertiari sunt rezistenti la agentii oxidantii dicromat si permanganat
avem alcooli:
CH3-CH2-CH2-CH(OH)-CH3
CH3-CH2-CH(OH)-CH2-CH3
CH3-CH(CH3)-CH(OH)-CH3
CH3-C(CH3)(OH)-CH2-CH3
fiind izomeri => f.m: C5H12O
amestec echimolecular => x:x:x:x
avem 4x moli amestec
n=m/M => m=n*M => m=4x*88=352x
352x=35,2 => x=0,1moli
doar 0,3moli reactioneaza cu solutia de dicromat.
1mol.........1mol
C5H12O + [O] -> C5H10O + H2O
0,3moli....0,3moli
1mol.............................3moli
K2Cr2O7 + 4H2SO4 -> 3[O] + K2SO4 + Cr2(SO4)3 + 4H2O
0,1moli.........................0,3moli
Cm=n/Vs => Vs=n/Cm=0,1/0,33333=0,3l K2Cr2O7=300ml K2Cr2O7