First thing to make clear is that net $6$ $H_2O$ go out of the reaction.($12$$H_2O$ $-$ $6$$H_2O$)

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

For the Left hand Side, $6$ $H_2O$ are accounted here :

• $2$ $H_2O$ go in conversion of 2-Phosphoglycertae to phosphoenolpyruvate.
• $2$ $H_2O$ 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$$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$$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 !

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 !