The decomposition of 86.5 g ammonium nitrate yields how many liters of dinitrogen monoxide at 1.28 atm and 328k
nh4no3-> n2o+2h2o

First, we calculate the moles of ammonium nitrate:
Moles = mass / molecular mass
Moles = 86.5 / 80
Moles = 1.08

From the equation, it is visible that the molar ratio between ammonium nitrate and dinitrogen monoxide is 1 : 1. The moles of dinitrogen monoxide formed will be:
1.08 moles

Next, we may apply the ideal gas law equation to find the volume:
PV = nRT
V = nRT/P
V = (1.08 * 0.082 * 328)/1.28
V = 22.69 liters

22.7 liters of dinitrogen monoxide will be produced

Elias is almost correct. he just got it backwards.the correct answer is that it demonstrates elastic potential energy transforming into kinetic energy.