Given the following reaction equation: $$\ce{KNO3 + C + S -> SO2 + CO2 + N2 + K2O}$$ Here is my solution: \begin{align} & 2\times & \ce{2 N^{+V} + 2 * 5e- &-> N2^0}\\ & 5\times & \ce{C^0 &-> C^{+IV} + 4e-}\\ \end{align} That is why $$\ce{KNO3 + 5 C + S -> SO2 + CO2 + N2 + K2O}$$ Then I think it should be $$\ce{12 KNO3 + 5 C + 10 S -> 10 SO2 + 5 CO_2 + 6 N2 + 6 K2O}$$ Am I right?
Answer
Your equation balances so it could be right. However, there is no unique solution to your given equation because the $\ce{S}$ and $\ce{C}$ in the reactants and the $\ce{SO2}$ and $\ce{CO2}$ in the products allow you to dispose of any excess oxygen from the nitrate. Another balanced equation with different stoichiometry is:
$$\ce{4KNO3 + 3C + 2S -> 2SO2 + 3CO2 + 2N2 + 2K2O}$$
Now the interesting part. The equation you are given is for the burning of some form of gunpowder, although one without any charcoal, which is usually described as $\ce{C7H4O}$. The products of burning gunpowder are complicated to predict and strongly dependent on the exact composition of the powder sample being used. As suggested in the comments by @IvanNeretin, the $\ce{K2O}$ is likely to react with the acidic oxides $\ce{CO2}$ and $\ce{SO2}$ to form a mixture of carbonates and sulfates. There are many possible equations showing this (one is shown below) but the exact stoichiometry will depend of the ratio in which the reactants are mixed in the powder:
$$\ce{10KNO3 + 3S + 8C -> 2K2CO3 + 3K2SO4 + 6CO2 + 5 N2}$$
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