Answer to Question #169152 in General Chemistry for Dawson

Question

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the tragic explosion at the Oklahoma City federal building in 1995. How many liters of gas at 307°C and 1.00 atm are formed by the explosive decomposition of 80.3 kg of ammonium nitrate to nitrogen, oxygen, and water vapor? Do not answer in scientific notation.

Answer 1

2NH₄NO₃ = 2N₂ + O₂ + 4H₂O

80.3kg = 80300g

2 mole ammonium nitrate = 2 moles nitrogen gas

2 × 80g NH₄NO₃ = 2mole N₂

1g NH₄NO₃ = 2/160 mole nitrogen gas

80300g 2NH₄NO₃ = 2 × 80300/160 mole N2

= 1003.75mol

2 mole NH₄NO₃ = 1 mole oxygen gas

2×80g NH₄NO₃ = 1 mole oxygen gas

1g NH₄NO₃ = 1/160 mole oxygen gas

80300g NH₄NO₃ = 80300/160 mole oxygen gas

= 501.875 mole oxygen gas

2 mole NH₄NO₃ = 4 mole H20

2×80 g NH₄NO₃ = 4mole H20

1g NH₄NO₃ = 4/160 mole H2O

80300g NH₄NO₃ = 4×80300/160 mole H2O

= 2007.5 mole of H20

PV=nRT . T= 307°C = 34K

V = nRT / P

For nitrogen V= 1003.75 × 0.082 × 34 / 1 = 2798.46L

For oxygen V = 501.875 × 0.082 × 34 = 1399.23L

For water vapour V = 2007.5 × 0.082 × 34 = 5596.91L

Total volume = 2798.46L + 1399.23L + 5596.91L

= 9794.6L