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Newbie

Joined: 21 Feb 2018
Posts: 1

# 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

Sr. Staff Member

Joined: 27 Sep 2017
Posts: 229
Location: Berlin, Germany
 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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