Chemistry Forum

I have found the answers to balancing this equation, but am still unable to figure out how they get the numbers.
KNO₃+C₁₂H₂₂O₁₁=N₂+CO₂+H₂O+K₂CO₃

reactants finish off as 48KNO₃+5C₁₂H₂₂O₁₁ and i cannot figure out how we get the 5

Please Help

Redox pair NO₃- and N₂

From the sugar you can eliminate water only the carbon burns

C /CO₂

2 NO₃ - + 12 H+ + 10 e- => N₂ + 6 H₂O

C + 2 H₂O => CO₂ + 4 H+ + 4 e-

12 C + 24 H₂O => 12 CO₂ + 48 H+ + 48 e-

KGV = 24 for electron

48 NO₃ - + 288 H+ + 240 e- => 24 N₂ + 144 H₂O

60 C + 120 H₂O => 60 CO₂ + 240 H+ + 240 e-


Addition

48 NO₃ - + 60 C + 48 H+ => 60 CO₂ + 24 N₂ + 24 H₂O

add water for sugar

48 NO₃- + 5 C₁₂ H₂₂O₁₁ + 48 H+ => 60 CO₂ +24 N₂ + 79 H₂O

add K+ substitute water

48 KNO₃ + 5 C₁₂ H₂₂O₁₁ + => 60 CO₂ +24 N₂ + 55 H₂O + 24 K₂O

48 KNO₃ + 5 C₁₂ H₂₂O₁₁ + => 36 CO₂ + 24 N₂ + 55 H₂O + 24 K₂CO₃

finished.