You can work this as a ratio/proportion because you know 2 mols NH3 (2*17=34 g NH3) will require +92.6 kJ (I just changed the sign because the question you asked is the reverse of the equation).
(92.6 kJ/34g) = (? kJ/30.5 g) and solve for ? kJ.
I do these this way.
92.6 kJ x (30.5/2*17) = ? kJ.
Given the following equation:
N2 (g) + 3H2 (g) → 2NH3 (g)
ΔH = -92.6 kJ
Calculate ΔH (in kJ) for the decomposition of 30.5 g NH3 (g) into N2 (g) and H2 (g)
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