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We know from Wikipedia that the combustion of rocket candy follow this equation :

$\ce{48 KNO3 + 5 C12H22O11 → 55 H2O + 36 CO2 + 24 K2CO3 + 24 N2}$

We can add small quantity of iron (III) oxide ($\ce{Fe2O3}$) as a catalyst.

The new equation look like this : $\ce{KNO3 + C12H22O11 + Fe2O3 -> H2O + CO2 + K2CO3 + N2 + ?}$

What product does the combustion of iron (III) oxide produce ?

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I think you may be missing a key point to catalysis; that the catalyst is not net consumed or altered by the reaction. If you were to put the catalyst in the chemical equation as in your second reaction as a reactant, you would technically need to also put it on the right hand side as a product. However, this is not the way a catalyzed reaction is typically indicated. A common way to indicate that a catalyst is used would be as follows:

$$\ce{48KNO3 + 5C12H22O11 ->[Fe2O3] 55H2O + 36CO2 + 24K2CO3 + 24N2}$$

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