Inha C. answered • 02/27/21

BA in Chemistry with 4 classroom teaching experience as a TA

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 NH_{4}NO_{3 }has molar mass of ~80 g/mol. 86.6 kg would be 6.93*10^{6} mol.

The decomposition of NH_{4}NO_{3} to N_{2}, O_{2}, and H_{2}O would have a chemical equation of

NH_{4}NO_{3} => N_{2} + (1/2)O_{2} +2H_{2}O. This means each mole of NH_{4}NO_{3} 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*10^{6})*0.082*580 = 1.15*10^{9} L. That's a lot of gas.

Let's see if that's reasonable. We have 6.93*3.5*10^{6} = 2.43*10^{7} 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*10^{8} 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*10^{9} L is a bit more than double of 4.5*10^{8} L.