**please show work for all three parts. Thank You. Ammonium nitrate, NH4N03 and
ID: 822930 • Letter: #
Question
**please show work for all three parts. Thank You.
Ammonium nitrate, NH4N03 and trinitroglycerine, C3H5N3O9 are explosive materials. The balanced equations for their decomposition reactions are given below: 2 NH4N03(s) - 2 N2 + 4 H20(g) + 02(g) 4 C3H5N309(1) right arrow 6 N2(g) + 12 C02(g) + 10 H20(g) + 02(g) delta rxn(2) Use the enthalpies of formation given, to calculate the enthalpy of reaction for each reaction. How much heat per gram of explosive is released? Answer rxn(l):_kJ/g of explosive Answer rxn(2):_kJ/g of explosive How many moles of gases are released per gram of explosive? Answer rxn(l):_moles gas/g of explosive Answer rxn(2):_moles gas/g of explosiveExplanation / Answer
(a) 2 NH4NO3 => 2 N2 + 4 H2O + O2
DH(1) = DHf(products) - DHf(reactants)
= 2 x DHf(N2) + 4 x DHf(H2O) + DHf(O2) - 2 x DHf(NH4NO3)
= 2 x 0 + 4 x (-241.82) + 0 - 2 x (-365.6)
= -236.08 kJ
4 C3H5N3O9 => 6 N2 + 12 CO2 + 10 H2O + O2
DH(2) = DHf(products) - DHf(reactants)
= 6 x DHf(N2) + 12 x DHf(CO2) + 10 x DHf(H2O) + DHf(O2) - 2 x DHf(C3H5N3O9)
= 6 x 0 + 12 x (-393.5) + 10 x (-241.82) + 0 - 4 x (-353.6)
= -5725.8 kJ
(c) Heat released per mole for rxn(1) = DH(1)/2
= 236.08/2 = 118.04 kJ
Heat relesed per g for rxn(1) = heat release per mole/mole mass of NH4NO3
= 118.04/80.04
= 1.475 kJ/g of explosive
Heat released per mole for rxn(2) = DH(2)/4
= 5725.8/4 = 1431.45 kJ
Heat relesed per g for rxn(2) = heat release per mole/mole mass of C3H5N3O9
= 1431.45/227.05
= 6.305 kJ/g of explosive
(c) Moles of NH4NO3 = mass/molar mass of NH4NO3
= 1/80.04 = 0.012494 mol
Moles of gas for rxn(1) = (2 + 4 + 1)/2 x moles of NH4NO3
= 7/2 x 0.012494
= 0.04373 moles gas/g of explosive
Moles of C3H5N3O9 = mass/molar mass of C3H5N3O9
= 1/227.05 = 0.0044043 mol
Moles of gas for rxn(2) = (6 + 12 + 10 + 1)/4 x moles of C3H5N3O9
= 29/4 x 0.0044043
= 0.03193 moles gas/g of explosive
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