There probably is some K2CO3, but if the KNO3 was in excess, then it would have been partially reduced to
KNO2:
2 KNO3 + H2CO (relative empirical formula for sucrose) → CO2 + H2O + 2
KNO2
Vs. the correct stoichiometry of the oxidation reaction:
2 KNO3 + 2 H2CO (relative empirical formula for sucrose) → 2 CO2 + 2
H2O + K2O
And finally, the CO2 reacts partially with the K2O to produce K2CO3:
CO2 + K2O ↔ K2CO3
. At STP, the reaction goes to completion, but
at the high temperatures of the reaction, it's at equilibrium.
[Edited on 21-7-2014 by Zyklon-A]