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

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