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

[ 128.4 / 80 ] X 2 <<< use a calculator ... gives moles of water

(1)(22.4)/ [ 1 X 273] = (1.89)(V) / [ answer from above X 314 ] <<< solve for V

V = [ answer from above X 314 ] (1)(22.4) / [ 1 X 273 x 1.89] <<< USE A CALCULATOR ... ANSWER IS LITERS OF GASEOUS WATER AT 1.89ATM AND 314 K.

At 1.89 atm and 314 K (41°C), water exists as liquid.

Molar mass of NH₄NO₃ = (14.01×2 + 1.008×4 + 16.00×3) = 80.05 g/mol
No. of moles of NH₄NO₃ reacted = (128.4 g) / (80.05 g/mol) = 1.604 mol

NH₄NO₃(s) → N₂O(g) + 2H₂O(l)
Mole ratio NH₄NO₃ : H₂O = 1 : 2
No. of moles of H₂O formed = (1.604 mol) × 2 = 3.208 mol

Molar mass of H₂O = (1.008×2 + 16.00) g/mol = 18.02 g/mol
Mass of H₂O formed = (18.02 g/mol) × (3.208 mol) = 57.81 g

Density of H₂O = 1000 g/L
Volume of H₂O formed = (57.81 g) / (1000 g/L) = 0.05781 L