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Ammonia reacts with oxygen according to the equation 4N H 3 (g)+5 O 2 (g)→4NO(g)+6 H 2 O(g),Δ H rxn =−906 kJ Calculate the heat (in kJ ) associated with the complete reaction of 155 g of N H 3 .

Respuesta :

dsdrajlin

Answer:

-2.07 × 10³ kJ

Explanation:

Let's consider the following balanced equation.

4 NH₃(g) + 5O₂(g) → 4 NO(g) + 6H₂O(g)    ΔHrxn =−906 kJ

The enthalpy of reaction is -906 kJ, that is, 906 kJ are released upon the reaction of 4 moles of NH₃. Taking into account that the molar mass of NH₃ is 17.0 g/mol, the heat associated with the complete reaction of 155 g of NH₃ is:

[tex]155gNH_{3}.\frac{1molNH_{3}}{17.0gNH_{3}} .\frac{-906kJ}{4molNH_{3}} =-2.07 \times 10^{3} kJ[/tex]