The decomposition of 128.4g ammonium nitrate yields how many liters of water at 1.89atm and 314K?

NH₄NO₃ (s) →N₂O (g) + 2H₂O (g)

First, we calculate the moles of ammonium nitrate:

Moles = mass / molecular mass

Moles = 128.4 / 80

Moles = 1.60

From the equation, it is visible that the molar ratio between ammonium nitrate and water is 1 : 2. The moles of water formed will be:

1.6 * 2 = 3.2 moles

Next, we may apply the ideal gas law equation to find the volume:

PV = nRT

V = nRT/P

V = (3.2 * 0.082 * 314) / 1.89

V = 43.6 liters

43.6 liters of water will be produced