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

True, the standard unit for volume is liters which can be represented by either a lower case "l" or a capital "l"