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
Hard to Balance equations
Moderators: Xen, expert, ChenBeier
- ChenBeier
- Distinguished Member
- Posts: 1563
- Joined: Wed Sep 27, 2017 7:25 am
- 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.
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.