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chemistry question?
the energy change, ΔH, associated with the following reaction is +81.0 kJ.
NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g)
What is the expected energy change for the reaction of 0.642 moles of NBr3 with 0.938 moles of H2O?
1 Answer
- Roger the MoleLv 72 years agoFavorite Answer
0.938 mole of H2O would react completely with 0.938 x (1/3) = 0.313 mole of NBr3, but there is more NBr3 present than that, so NBr3 is in excess and H2O is the limiting reactant.
(0.938 mol H2O) x (81.0 kJ / 3 mol H2O) = +25.3 kJ