<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol
<u>Explanation:</u>
To calculate the mass of water, we use the equation:

Density of water = 1 g/mL
Volume of water = 25.0 mL
Putting values in above equation, we get:

To calculate the heat released by the reaction, we use the equation:

where,
q = heat released
m = Total mass = [1.25 + 25] = 26.25 g
c = heat capacity of water = 4.18 J/g°C
= change in temperature = 
Putting values in above equation, we get:

To calculate the number of moles, we use the equation:

Given mass of ammonium nitrate = 1.25 g
Molar mass of ammonium nitrate = 80 g/mol
Putting values in above equation, we get:

To calculate the enthalpy change of the reaction, we use the equation:

where,
q = amount of heat released = -0.428 kJ
n = number of moles = 0.0156 moles
= enthalpy change of the reaction
Putting values in above equation, we get:

Hence, the enthalpy change of the reaction is -27. kJ/mol