ΔH = -730 kJ
Treat the energy as if it were a product in the equation. You get
CO + H₂ + O₂ → CO₂ + H₂O + 525 kJ
Then you do an ordinary stoichiometry calculation.
Step 1. Convert grams of water to moles of water.
Moles of H₂O = 25 g H₂O × (1 mol H₂O/18.02 g H₂O) = 1.39 mol H₂O
Step 2. Calculate the energy change
ΔH = 1.39 mol water × (-525 kJ/1 mol H₂O) = -730 kJ
