Ammonia burns in the presence of a copper catalyst to form nitrogen gas. 4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g) ΔΗ = -1267 kJ What is the e

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Ammonia burns in the presence of a copper catalyst to form nitrogen gas. 4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g) ΔΗ = -1267 kJ What is the enthalpy change to burn 38.4 g of ammonia?​

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Hải Đăng 4 years 2021-07-21T20:36:00+00:00 1 Answers 81 views 0

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  1. nguyetanh
    0
    2021-07-21T20:37:12+00:00
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    Answer:

    -713 kJ

    Explanation:

    Step 1: Write the balaned thermochemical equation

    4 NH₃(g) + 3 O₂(g) → 2 N₂(g) + 6 H₂O(g)  ΔΗ = -1267 kJ

    Step 2: Calculate the moles corresponding to 38.4 g of NH₃

    The molar mass of NH₃ is 17.03 g/mol.

    38.4 g × 1 mol/17.03 g = 2.25 mol

    Step 3: Calculate the  enthalpy change to burn 2.25 mol of ammonia

    According to the thermochemical equation, 1267 kJ are released per 4 moles of ammonia that react.

    2.25 mol × (-1267 kJ/4 mol) = -713 kJ

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