Hard to Balance equations
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Katleigh315
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Hard to Balance equations


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
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ChenBeier
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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.
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