M,toluen=92g/mol
n=13,8g/92g/mol= 0,15mol toluen-teoretic;practic se oxideaza doar
n=60x0,15/100=0,09mol
C6H5 -CH3+3O--> C6H5-COOH+H2O
1mol toluen.......3molO
0,09mol..........0,27molO
2molKMnO4.....5molO*
n=........................o,27mol
n=0,108molKMnO4
c=n/V---> 0,6= 0,108/ V------> CALCULEAZA V !!!!!!
*2KMnO+3H2SO4--> 2MnSO4+K2SO4+3H2O+5O