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