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.