Answer
20.8 kJ
Work Step by Step
When 2 moles of sodium reacts with excess water, 368 kJ heat is liberated.
$\implies$ 1 mol of sodium reacts with excess water to liberate $\frac{368\,kJ}{2}=184\,kJ$ heat.
$\implies$ Heat liberated when 0.113 mol of sodium reacts with excess water=
$0.113\,mol\,Na\times\frac{184\,kJ}{1\,mol\,Na}=20.8\,kJ$
(The negative sign is neglected as it only indicates that the heat is liberated.)
