C
6H
12O
6 + C
12H
22O
11 + KNO
3 CO
2 + CO + H
2O + H
2 + N
2 + K
2CO
3 + KOH
This definitely isn't for the faint hearted.
What I did first was balance the K, there were 3 K on the left and one on the right, so I put a 2 in front of the KOH and a 4 in front of the KNO
3 and that's sorted. Then I gambled that there would only be one molecule of C
6H
12O
6 and C
12H
22O
11. So that left me with balancing the carbon. There were 18 C's. There was already one carbon on K
2CO
3 so that left me with 17 C's to split between CO
2 and CO. There were 29 O's and 5 were used in the potassium compounds so I was left with 24 O's to split between H
2O, CO
2 and CO. Since there was only 17 C's, there must be more CO than CO
2 and this would have to be an odd number since the number of O's added to an even number, 24 (odd + odd = even). I did this and I got 4CO
2 + 13CO + 3H
2O. Then the rest is pretty easy because all you have to do is balance out the N's and the H's. I'm sure you'll manage.
Howszat sound?