The decomposition of 74.2g of ammonium nitrate yields how many liters of water at 0.954 atm and 387 k

Question

Am I correct in thinking that you would use PV=nRT to get the moles, then use stoichiometry to get liters, or just  pay no attention to the atm and kelvin and just convert the grams into litems with stiochiometry?

Answer 1

This is the thought process that I used to solve this problem:

N₂H₄O₃ (Ammonium Nitrate)--->N₂O + 2H₂O (g)

74.2 g 

                                                ?L

P= 0.954 atm

T= 387K

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:

 

PV=nRT

 

0.954 (V)= 1.854 (0.0821)(387)

 

V= 61.74 L H₂O (g)

 

I hope this was helpful.