The balanced redox equation considering the ferrous ammonium sulfate hexahydrate is as follows:
$10\left[\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}\right]+2 \mathrm{KMnO}_4+8 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow$
$5 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{MnSO}_4+10\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+\mathrm{K}_2 \mathrm{SO}_4+68 \mathrm{H}_2 \mathrm{O}$
Now, we are asked to find the number of water molecules produced for 2 molecules of KMnO₄.
From the balanced equation, we can observe that 2 moles of KMnO₄ produce 68 moles of H₂O. Therefore, for 2 molecules of KMnO₄, the number of water molecules produced will be:
2 molecules $\times$ 34 water molecules per molecule of KMnO₄ = 68 water molecules
So, 68 water molecules are produced for 2 molecules of KMnO₄ in this titration.