Redox equation - help!

[Expelliarmus9]

Is it correct the next equation: C₇H₁₄+H₂SO₄+K₂Cr₂O₇=C₃H₆O+C₄H₈O+K₂SO₄+Cr₂(SO₄)₃+H₂O?
Then how I can balance it?

[GrahamKemp]

Is it correct the next equation: C7H14+H2SO4+K2Cr2O7=C3H6O+C4H8O+K2SO4+Cr2(SO4)3+H2O?
Then how I can balance it?

Assuming C7H14 is 3-heptene, CH3CH2CH=CHCH2CH2CH3, then yes, it will produce 1-propanal and 1-butanal... to start with at least. In excess, chromic acid will continue to oxidize the alkanals further into carboxilic acids.

Firstly, let's just consider the active ingredients (remove the passive observers). With R as CH3CH2 and R' as CH3CH2CH2.

RCH=CHR' + Cr2O7^-2 + H⁺ => RCH=O + R'CH=O + Cr⁺³ + H2O

Balance the chromium:
RCH=CHR' + Cr2O7^-2 + H⁺ => RCH=O + R'CH=O + 2 Cr⁺³ + H2O

Balance the charges:
RCH=CHR' + Cr2O7^-2 + 8 H⁺ => RCH=O + R'CH=O + 2 Cr⁺³ + H2O

Balance the hydrogen:
RCH=CHR' + Cr2O7^-2 + 8 H⁺ => RCH=O + R'CH=O + 2 Cr⁺³ + 4 H2O

Balance the oxygen, and incidentally the organics:
1.5 RCH=CHR' + Cr2O7^-2 + 8 H⁺ => 1.5 RCH=O + 1.5 R'CH=O + 2 Cr⁺³ + 4 H2O

Adjust to the Lowest Integer Solution:
3 RCH=CHR' + 2 Cr2O7^-2 + 16 H⁺ => 3 RCH=O + 3 R'CH=O + 4 Cr⁺³ + 8 H2O

Add back the observer ions (as 4 potassium cations and 8 sulphate anions) to neutralise charges:
3 RCH=CHR' + 2 K2Cr2O7 + 8 H2SO4 => 3 RCH=O + 3 R'CH=O + 2 K2SO4 + 2 Cr2(SO4)3 + 8 H2O
Thus:
3 CH3CH2CH=CHCH2CH2CH3 + 2 K2Cr2O7 + 8 H2SO4 => 3 CH3CH2CH=O + 3 CH3CH2CH2CH=O + 2 K2SO4 + 2 Cr2(SO4)3 + 8 H2O
Or if you will:
3 C7H14 + 2 K2Cr2O7 + 8 H2SO4 => 3 C3H6O + 3 C4H8O + 2 K2SO4 + 2 Cr2(SO4)3 + 8 H2O