Question

31-35. One of the steps in producing nitric acid using Oswald process is:

4NH3(g) + 5O2(g) → 4NO(g) + 6H20(I) ∆H = -904kJ

Determine the amount of heat in (kJ) given off per 65.3g of NH3 (17g/mol) used.

Answer 1

Moles of NH3 = $\frac{65.3}{17}= 3.841 moles$

Heat (KJ)= heat change/1000 ÷ moles

$∆H = \frac{-904}{1000×3.841} =-0.235kJ$