Answer
$\Delta$H= - 4.2 kJ, or +4.2 kJ of energy released
Work Step by Step
0.2 L $\times$$\frac{0.4 mol HNO3}{1 L}$ = 0.08 mol HNO3
0.15 L $\times$$\frac{0.5mol KOH}{1 L}$ = 0.075 mol KOH --> LR because neutralization reactions all have 1:1 mole ratios
0.075 mol H2O $\times$$\frac{-56 kJ}{1 mol H2O}$ = -4.2 kJ
