HNO3 (g) + 4H2 (g) ---> NH3 (g) + 3H2O (g) deltaH = -637 kJ

Question

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?

Answer 1

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

Answer 2

thank Dr.Bob222