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What product will be obtained get when a sucrose and potassium nitrate mix, which are the ingredients for smoke bomb, are burned in a combustion container?

$$\ce{KNO3 + C12H22O11 -> ?}$$

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closed as off-topic by ashu, Ben Norris, Philipp, bobthechemist, user467 Oct 11 '13 at 0:21

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You will get potassium carbonate ($\ce{K2CO3}$), carbon dioxide ($\ce{CO2}$), water ($\ce{H2O}$), and nitrogen gas ($\ce{N2}$): $$\ce{48KNO3 + 5C12H22O11 -> 24K2CO3 + 36CO2 + 55H2O + 24N2}$$

See this yahoo question and this science forums thread.

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The reaction equation is as follows:

$$\ce{9.6 KNO3 + C12H22O11 -> 4.8 K2CO3 + 7.2 CO2 + 11 H2O + 4.8 N2}.$$

Therefore $\pu{970 g}$ $\ce{KNO3}$ and $\pu{342 g}$ sucrose, which is $74\%$ $\ce{KNO3}$ and $26\%$ sucrose. This produces $\pu{662 g}$ of potassium carbonate.

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