Question

Exercițiul 3 roman a doua intrebare

Answer 1

MC6H6=78g/mol
MC6H5-CH3=92g/mol
MC6H5-COOH=122g/mol

5C6H5-CH3 + 6KMnO4 + 7H2SO4 ---> 5C6H5-COOH + K2SO4 + 6MnSO4 + 7H2O

C⁻³ -6e⁻---> C⁺³ se oxideaza (ag.red) ||5
Mn⁺⁷ +5e ---> Mn⁺² se reduce (ag.ox) ||6

5mol C6H5-CH3...................5 C6H5-COOH mol
x=0.2mol.............,..................0.2mol

n=m/M=24.4/122=0.2 mol C6H5-COOH

n=m/m==>m1=0.2*92=18.4g

m2=0.2*78=15.6g C6H6

m amestec=15.6+18.4=34g

5 moli C6H5-CH3.................6 moli KMNO4
0.2moli.................................x=0.24 moli

Cm=n/Vs==>Vs=n/Cm=0.24/0.1=2.4 L KMnO4