Wondering where I went wrong in this problem...Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the tragic explosion...

Question

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 86.6 kg of ammonium nitrate to nitrogen, oxygen, and water vapor?

I must've rounded it wrong or too much. I got 20000 L gas.

Answer 1

How many liters of gas at 307°C and 1.00 atm are formed by the explosive decomposition of 86.6 kg of ammonium nitrate to nitrogen, oxygen, and water vapor?

Ammonium nitrate NH₄NO₃ has molar mass of ~80 g/mol. 86.6 kg would be 6.93*10⁶ mol.

The decomposition of NH₄NO₃ to N₂, O₂, and H₂O would have a chemical equation of

NH₄NO₃ => N₂ + (1/2)O₂ +2H₂O. This means each mole of NH₄NO₃ generates 3.5 mol gas.

307°C = 580 K.

Gas constant R = 0.082 atm*L/(mol*K).

Using PV = nRT, I get V = (3.5*6.93*10⁶)*0.082*580 = 1.15*10⁹ L. That's a lot of gas.

Let's see if that's reasonable. We have 6.93*3.5*10⁶ = 2.43*10⁷ moles of gas. The ideal gas has 22.4 L volume at STP, which is 0 °C and 1 atm. At STP, 2.43*107 moles of gas should have ~4.5*10⁸ liters of volume. 307 °C is more than double of 0 °C (in terms of kelvin temperature), so the gas volume should be also more than double. That checks out! 1.15*10⁹ L is a bit more than double of 4.5*10⁸ L.