# For each pound (454g) of ammonium nitrate that exploded (products being nitrogen gas, oxygen gas, and water vapor) how many kJ of thermal energy were released and how many L of gas (measured at...

For each pound (454g) of ammonium nitrate that exploded (products being nitrogen gas, oxygen gas, and water vapor) how many kJ of thermal energy were released and how many L of gas (measured at 1atm and 25degreeC) were formed? We know deltaHf for ammonium nitrate is -87.37 kcal/mol

jerichorayel | Certified Educator

The chemical equation for this reaction can be written as:

`4 NH_3NO_3 _(s) -> 4 N_2 _(g) + 3 O_2 _(g) + 6 H_2O _(g)`

Heat of ammonium nitrate

`454 grams NH_4NO_3 *(1 mol e NH_4NO_3)/(80.052 grams NH_4NO_3)`

`= 5.67 mol es NH_4NO_3* (-87.37 kcal)/(mol e NH_4NO_3)`

`= -496 kcal`

Moles of the gases produced:

`454 grams NH_4NO_3 *(1 mol e NH_4NO_3)/(80.052 grams NH_4NO_3) * (13 mol es total gases)/(4 mol esNH_4NO_3)`

`= 18.432 mol es gases`

Volume of the gases occupied:

`PV = nRT`

• T = 25 + 273.15 = 298.15K
• R = 0.08206 atm-L/mol-K
• P = 1 atm
• n = total number of moles of gases

`V = (nRT)/(P)`

`V = ((18.432)*(0.08206) *(298.15))/(1)`

V = 450.96 L = 451 L