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  1. $6\ce{CO_2} + 6\ce{H_2O}\xrightarrow{some\;catalyst}9\ce O_2 + \ce{C_6H_{12}}$
  2. $\ce{C_6H_{12}} + 6\ce O_2\to 6\ce{CO_2} + 6\ce H_2$
  3. $6\ce H_2 + 3\ce O_2\to 6\ce{H_2O}$

At one point, I had made a balanced set of equations while starting college. Those equations are listed above. I had wanted to get into chemistry to see if there was any possible way those equations could be used as a means for a new kind of energy source; one that could catalyze its fuel, hexene, that it could later use, but then also take the byproducts of the combustion of hexene and produce the same chemicals it needs to produce hexene, and repeat the process. I don't know if it's possible, and I don't know what, if any catalyst would be needed to make it possible. However, as time went on, my major changed to software development and I gave up on becoming a chemist. So this is both a question and a challenge: Can anyone figure out if the system described above is possible, and what would it require to make it available as an energy source?

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    $\begingroup$ To make this scheme available as an energy source, it should have an energy source. Now it has none. A catalyst, no matter how sophisticated, is not an energy source. $\endgroup$
    – Ivan Neretin
    Sep 12 '15 at 20:51
  • $\begingroup$ The energy source is in the second formula, in the same method as a combustion engine. A spark ignites the hexene fuel. $\endgroup$
    – Lee S
    Sep 12 '15 at 20:54
  • $\begingroup$ That, or I'm confused. The catalyst is definitely something relating to photosynthesis, but that's as far as I got with it. $\endgroup$
    – Lee S
    Sep 12 '15 at 20:56
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    $\begingroup$ Well, then 2 and 3 put together will bring you the exact amount of energy needed to pull off 1, minus the inevitable losses. $\endgroup$
    – Ivan Neretin
    Sep 12 '15 at 20:57
  • $\begingroup$ No, I meant hexene, since hexene is a combustible fuel. The formulas can't work with glucose. $\endgroup$
    – Lee S
    Sep 12 '15 at 21:03

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