The decomposition of 74.2 g ammonium nitrate yields how many liters of water a
0.954 atm and 387 K?
NH, NO3(s) → NO g) + 2H, O(g)
Note: Keep extra digits in any intermediate
calculations! Round to the correct number of
significant figures at the very end.
For R, use 0.08206 L atm/(mol K).

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.