Co + h_2 + o_2 --> co_2 + h_2_o: delta h= 525 kj

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