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

Answer 1

2 NH4NO3 —> N2 + 2 O2 + 2 H2O

Molar mass of NH4NO3,

MM = 2*MM(N) + 4*MM(H) + 3*MM(O)

= 2*14.01 + 4*1.008 + 3*16.0

= 80.052 g/mol

mass(NH4NO3)= 30.8 Kg

= 30800 g

use:

number of mol of NH4NO3,

n = mass of NH4NO3/molar mass of NH4NO3

=(3.08*10^4 g)/(80.05 g/mol)

= 3.847*10^2 mol

From balanced equation:

2 mol of NH4NO3 forms 5 mol of gas

so,

mol of gas = (5/2)*mol of NH4NO3

= (5/2)*(3.847*10^2 mol)

= 961.75 mol

Given:

P = 1.0 atm

n = 961.75 mol

T = 307.0 oC

= (307.0+273) K

= 580 K

use:

P * V = n*R*T

1 atm * V = 961.75 mol* 0.08206 atm.L/mol.K * 580 K

V = 4.58*10^4 L

Answer: 4.58*10^4 L