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For each pound (454g) of ammonium nitrate that exploded (products being nitrogen gas,...

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juliek21 | eNotes Newbie

Posted April 28, 2013 at 5:50 AM via web

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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 

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted April 28, 2013 at 8:19 AM (Answer #1)

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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


 

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