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

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.


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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