Missing 4 $\ce{H_2O}$ (per glucose) in Cellular Respiration... Where can they be?
First thing to make clear is that net $6 \, \ce{H_2O}$ go out of the reaction.($12 \,\ce{H_2O}-6\,\ce{H_2O}$)

Let me tell you my calculation, you should then be able to figure out what went wrong.

For the Left hand Side, $6\,\ce{H_2O}$ are accounted here :


$2\,\ce{H_2O}$ go in conversion of 2-Phosphoglycertae to phosphoenolpyruvate.
$2\,\ce{H_2O}$ in TCA from conversion of Oxoloacetate to Citrat by combining with Acetyl CoA.(One for each Acetyl CoA molecule)
The aconitase -cis-aconitase-isocitrate water molecules cancel out. Equal no go in and out.
$2\,\ce{H_2O}$ in TCA from conversion of fumarate to Malate.


For the right hnd side, if we calculate the no. of NADH formed = 10 and assume that each participates in respiratory chain, we have $10\,\ce{H_2O}$ from there. The other $2$ will be from the $2$ FADH produced.

So we have 6 on the LHS and 12 on the RHS. Hope this makes sense !