I need help with calculating the net change of enthalpy heat. The question is as follows,  Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia:N2+3H2 --> 2NH   specific heat= -92.kJ In the second step, ammonia and oxygen react to form nitric acid and water:NH3 + 2O2 --> HNO3 + H20   specific heat= -330.kJ Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.Round your answer to the nearest kJ,

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I need help with calculating the net change of enthalpy heat.

The question is as follows, 

Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia:
N2+3H2 --> 2NH   specific heat= -92.kJ

In the second step, ammonia and oxygen react to form nitric acid and water:
NH3 + 2O2 --> HNO3 + H20   specific heat= -330.kJ

Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions.
Round your answer to the nearest kJ,

 

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