Answer
-29.475 kJ
Work Step by Step
Given Ag+ + Cl- --> AgCl $\Delta H = -65.5 kJ $, it means that for each mole of AgCl formed, 65.5 kJ of heat are released.
To find $\Delta H$ for the production of 0.450 mol, it would be 0.450 * -65.5 kJ = -29.475 kJ