Se consuma acelasi volum de K2Cr2O7 in solutie acida pentru oxidarea a:
2 moli de 2,3 dimetil-2-butena;
1 mol de 2-butena;
1 mol de 2-metil-1-butena;
1 mol de 3-metil-1-butena.

3CH3-C(CH3)=C(CH3)-CH3 + 2K2Cr2O7 + 8H2SO4 ---> 3CH3-C(=O)-CH3 + 3CH3-C(=O)-CH3 + 2K2SO4 + 2Cr2(SO4)3 + 8H2O

C⁰ -2e⁻--> C⁺² se oxideaza (ag.red) ||6||3

C⁰ -2e⁻--> C⁺² se oxideaza (ag.red) ||6||3

Cr2⁺⁶ +6e⁻--> Cr2⁺³ se reduce (ag.ox)||4||2

3moli C6H12..............................2moli K2Cr2O7
2moli...........................................x=1,333 moli*22,4=29,859L

nu este bine..

3CH3-CH=CH-CH3 + 4K2Cr2O7 + 16H2SO4 ---> 3CH3-COOH + 3CH3-COOH + 4K2SO4 + 4Cr2(SO4)3 + 16H2O

C⁻¹ -4e⁻--> C⁺³ se oxideaza (ag.red) ||6||3

C⁻¹ -4e⁻--> C⁺³ se oxideaza (ag.red) ||6||3

Cr2⁺⁶ +6e⁻--> Cr2⁺³ se reduce (ag.ox) ||8||4

3 moli C4H8..................................4 moli K2Cr2O7
1mol..............................................x=1,333 moli

3CH2=C(CH3)-CH2-CH3 + 4K2Cr2O7 + 16H2SO4 ---> 3CO2 + 16H2O + 3CH3-C(=O)-CH2-CH3 + 4K2SO4 + 4Cr2(SO4)3

C⁻² -6e⁻--> C⁺⁴ se oxideaza (ag.red) ||6||3

C⁰ -2e⁻--> C⁺² se oxideaza.. ||6||3

Cr2⁺⁶ +6e⁻--> Cr2⁺³ se reduce.. ||8||4

nr de moli este la fel.. 1,333 moli K2Cr2O7.. => V=n*Vm=1,333*22,4=29,859L K2Cr2O7

3CH2=CH-CH(CH3)-CH3 + 5K2Cr2O7 + 20H2SO4 ---> 3CO2 + 20H2O + 3HOOC-CH2-CH2-CH3 + 5K2SO4 + 5Cr2(SO4)3

C⁻² -6e⁻--> C⁺⁴ se oxideaza.. ||6||3

C⁻¹ -4e⁻--> C⁺³ se oxideaza.. ||6||3

Cr2⁺⁶ +6e⁻--> Cr2⁺³ se reduce.. ||10||5

3 moli C5H10..........................5 moli K2Cr2O7
1mol........................................x=1,666 moli

Se vede ca in cele 3 cazuri volumele de K2Cr2O7 (bicromat de potasiu) este la fel, iar in ultimul caz difera..