[Astrokel]

KOH + C4H4O4 ---> K2C4H2O4 + H2O

1) Balance K first as it is easier compare to O and H

2KOH + C4H4O4 ---> K2C4H2O4 + H2O

2) Now check for H and O, since K and C are already balanced. You need 2 more H and O on the right, logically, you will assign the 2 to H2O and not K2C4H2O4.

2KOH + C4H4O4 ---> K2C4H2O4 + 2H2O

[upsidedown]

2KOH + C4H4O4 ---> K2C4H2O4 + 2H2O

This means that on the left hand side there are 2Ks, 6Os, 4Cs, 6Hs
                 and on the right hand side there is 2Ks, 6Os, 4Cs, 8Hs

[Arkcon]

2KOH + C4H4O4 ---> K2C4H2O4 + 2H2O

This means that on the left hand side there are 2Ks, 6Os, 4Cs, 6Hs
                 and on the right hand side there is 2Ks, 6Os, 4Cs, 8Hs

Yet again, you are messing up the numbers ...

2KOH + C₄H₄O₄ ---> K₂C₄H₂O₄ + 2H₂O

2 + 2*2 = 6 not 8

[upsidedown]

Thanks for the help.
I really have no idea why I made so many mistakes, it all looks so simple now.