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I can balance equations but only if I already know both the sides. I am trying to find the following products:

$$\begin{aligned} \ce{Na2S2O5 + O2 ->}\ ?\\ \ce{Na2S2O5 + KNO3 ->}\ ?\\ \ce{Na2S2O5 + KNO3 + Fe2O3 ->}\ ?\\ \ce{Na2S2O5 + KNO3 + C12H22O11 ->}\ ?\\ \ce{KNO3 + C12H22O11 ->}\ ?\\ \ce{KNO3 + C12H22O11 + Fe2O3 ->}\ ? \end{aligned}$$

I'm guessing that the sodium metabisulfate will produce sulfur dioxide and sodium oxide, but that would just be plain decomposition, so I'm not sure where the oxygen would fit. I know that $\ce{KNO3 + C12H22O11}$ produces carbon dioxide and water, but I'm not sure where the nitrogen and potassium go.

Of course once I know the products, balancing and calculating specific energy is simple, if a touch laborious.

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    $\begingroup$ Welcome to Chemistry.SE. For homework-type questions, we want to see what your efforts have been so that we can better help you. $\endgroup$ – Ben Norris Jul 22 '16 at 13:50
  • $\begingroup$ This is the first step of a rather lengthy project/problem. I'll need to balance the equations, calculate bond energies, solve for specific energy, and go on from there. I just need the reaction products as a first step. $\endgroup$ – sevenperforce Jul 22 '16 at 14:00
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    $\begingroup$ Then share what you can in your question. $\endgroup$ – Ben Norris Jul 22 '16 at 14:08
  • $\begingroup$ Most (all?) of these seem like redox reactions. $\endgroup$ – a-cyclohexane-molecule Jul 22 '16 at 17:48
  • $\begingroup$ Yes, they are all redox reactions. I'm just very, very bad at chemistry. $\endgroup$ – sevenperforce Jul 22 '16 at 18:48
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I will assume that all of these reactions are happening in solution. Since we've established that all these reactions are redox reactions, we know the general structure of each reaction--- $$\text{reducing agent} + \text{oxidizing agent} \to \text{oxidized thing} + \text{reduced thing}$$ ---and we generally won't have to worry about other things (like what you've mentioned in your edit) happening. I will work out one reaction, and leave you with the rest. Let's look at reaction 4.

We can ignore $\ce{Na+}$ and $\ce{K+}$, because we know from prior experience that these are relatively inert spectator ions that won't participate in most reactions. Then we have the three species $\ce{S2O5^2-}$, $\ce{NO3^-}$, and $\ce{C12H22O11}$.

  • $\ce{N}$ has its maximum oxidation state of $+5$ in $\ce{NO3^-}$, so we expect that $\ce{NO3^-}$ will be a good oxidizing agent. $\ce{NO3^-}$ is commonly reduced to $\ce{NO2^-}$ in solution.
  • On the other hand, $\ce{S}$ has an oxidation state of $+4$ in $\ce{S2O5^2-}$ and a maximum oxidation state of $+6$, so we might expect $\ce{S2O5^2-}$ to be a reducing agent. Fully oxidized $\ce{S}$ is usually present as $\ce{SO4^2-}$ in solution.
  • Finally, $\ce{C}$ also has its maximum oxidation state of $\ce{+4}$ in $\ce{C12H22O11}$, so $\ce{C12H22O11}$ is likely to also be a reducing agent. Fully oxidized $\ce{C}$ is commonly present as $\ce{CO2}$.

This gets us the three half-reactions $$\begin{aligned} \ce{NO3^- + H2O + 2e- &-> NO2^- + 2OH-}\\ \ce{S2O5^2- + 6OH- &-> 2SO4^2- + 3H2O + 4e^-}\\ \ce{C12H22O11 + 48OH- &-> 12CO2 + 35H2O + 48e-} \end{aligned}$$ where we've also assumed basic conditions and hence added $\ce{H2O}$ and $\ce{OH-}$ as required to balance out charges and atoms on both sides of each equation. Now we simply need to take linear combinations of these half-reactions to cancel out electrons on both sides, and we're done.


In response to your comment, everyone starts out bad at chemistry. You'll get better.

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the result really depends on reaction conditions. These products are believable:

Na2S2O5 + O2 -> Na2SO4 + SO2

Na2S2O5 + KNO3 -> Na2SO4 + KNO2 +NO2

Na2S2O5 + KNO3 + Fe2O3 -> Na2SO4 + KNO2 + [Fe(NO3)3 + Fe(NO2)3]

This one makes little sense. Two reducing agents and one oxydizer. Try making this: Na2S2O5 + KNO3 + C12H22O11 -> Na2SO4 + K2SO4 + CO2 + H2O

KNO3 + C12H22O11 -> K2CO3 + N2 + CO2 + H2O

KNO3 + C12H22O11 + Fe2O3-> K2CO3 + N2 + CO2 + H2O + FeO

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