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..
