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HNO3 (g) + 4H2 (g) ---> NH3 (g) + 3H2O (g) deltaH = -637 kJ
Calculate the enthalpy change when one mole of hydrogen reacts; and What is the enthalpy change when 10.00 g of ammonia is made to react with excess of steam to form nitric acid and hydrogen gases?

  • chem -

    637 kJ/4 mols H2. So 637/4 = ?? for 1 mol H2.

    For #2,
    637 kJ/mol NH3 x [10 g NH3 x (1 mol NH3/17 g NH3)] = xx

  • chem -

    thank Dr.Bob222

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