Answer:
61.75 L of water
Explanation:
We'll begin by calculating the number of mole in 74.2 g of ammonium nitrate, NH₄NO₃. This can be obtained as follow:
Mass of NH₄NO₃ = 74.2 g
Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)
= 14 + 4 + 14 + 48
= 80 g/mol
Mole of NH₄NO₃ =?
Mole = mass / molar mass
Mole of NH₄NO₃ = 74.2 / 80
Mole of NH₄NO₃ = 0.9275 mole
Next, the balanced equation for the reaction. This is illustrated below:
NH₄NO₃ —> N₂O + 2H₂O
From the balanced equation above,
1 mole of NH₄NO₃ decomposed to produce 2 moles of H₂O.
Therefore, 0.9275 mole of NH₄NO₃ will decompose to produce = (0.9275 × 2) = 1.855 moles of H₂O.
Finally, we shall determine the volume of water obtained from the reaction. This can be obtained as follow:
Pressure (P) = 0.954 atm
Temperature (T) = 387 K
Number of mole (n) of H₂O = 1.855 moles.
Gas constant (R) = 0.08206 L.atm/Kmol
Volume (V) of H₂O =?
PV = nRT
0.954 × V = 1.855 × 0.08206 × 387
Divide both side by 0.954
V = (1.855 × 0.08206 × 387) / 0.954
V = 61.75 L
Therefore, 61.75 L of water were obtained from the reaction.
