Reaction: 4 NH_{3}(l) + 3 O_{2}(g) → 2 N_{2}(g) + 6 H_{2}O(g)

$\overline{){\mathbf{\u2206}}{{\mathbf{H}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\u2206}}{{\mathbf{H}}}_{{\mathbf{f}}{\mathbf{,}}{\mathbf{}}{\mathbf{products}}}{\mathbf{-}}{\mathbf{\u2206}}{{\mathbf{H}}}_{{\mathbf{f}}{\mathbf{,}}{\mathbf{}}{\mathbf{reactants}}}}$

Ammonia (NH_{3}) boils at -33^{o}C; at this temperature it has a density of 0.81 g/cm^{3}. The enthalpy of formation of NH_{3}(g) is -46.2 kJ/mol, and the enthalpy of vaporization of NH_{3}(l) is 23.2 kJ/mol.

Calculate the enthalpy change when 4 L of liquid NH_{3} is burned in air to give N_{2}(g) and H_{2}O(g).

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