By balancing redox reaction:

$10[{\mathrm{FeSO}}_4.({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4.6{\mathrm{H}}_2\mathrm{O}]+2{\mathrm{KMNO}}_4+8{\mathrm{H}}_2{\mathrm{SO}}_4\to 5{\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3+2{\mathrm{MnSO}}_4+10({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4+{\mathrm{K}}_2{\mathrm{SO}}_4+68{\mathrm{H}}_2\mathrm{O}$

$8{\mathrm{H}}^{+}+{\mathrm{MnO}}_4^{-}+5{\mathrm{Fe}}^{+2}⟶5{\mathrm{Fe}}^{+3}+{\mathrm{Mn}}^{+2}+4{\mathrm{H}}_2\mathrm{O}$

For $2$ moles of ${\mathrm{MnO}}_{4^{-}}, 8$moles of ${\mathrm{H}}_2\mathrm{O}.$ are released $5$ moles of Mohr's salt contain $30$ moles of ${\mathrm{H}}_2\mathrm{O}.$ So, $10$ moles of Mohr's salt contain $60$ moles of ${\mathrm{H}}_2\mathrm{O}$ So, total moles of ${\mathrm{H}}_2\mathrm{O}$ produced$=68$