This is the thought process that I used to solve this problem:
N2H4O3 (Ammonium Nitrate)--->N2O + 2H2O (g)
P= 0.954 atm
R=0.0821 L-atm/ mole-K
Note: It is reasonable to assume that at 387K, all of the water that is formed will be in the form of water vapor (373K= 100°C [boiling]).
1) First find the number of moles of water that is yielded in the complete decomposition of ammonium nitrate via stoichiometry:
74.2 g N2H4O3 X (l mol N2H4O3 / 80.052 g N2H4O3) X (2 mol H2O / 1 mol N2H4O3)= 1.854 mol H2O (g)
2) Then use the number of moles of water vapor to calculate the volume that is produced:
0.954 (V)= 1.854 (0.0821)(387)
V= 61.74 L H2O (g)
I hope this was helpful.