1. Write the balanced chemical equation:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

2. Determine the limiting reactant:

* Calculate moles of H₂O produced from 2.40 mol C₂H₂:

- (2.40 mol C₂H₂) * (2 mol H₂O / 2 mol C₂H₂) = 2.40 mol H₂O

* Calculate moles of H₂O produced from 7.40 mol O₂:

- (7.40 mol O₂) * (2 mol H₂O / 5 mol O₂) = 2.96 mol H₂O

* The limiting reactant is C₂H₂ because it produces the least amount of H₂O.

3. Calculate the mass of H₂O produced:

* Use the moles of H₂O produced from the limiting reactant (2.40 mol):

- (2.40 mol H₂O) * (18.015 g H₂O / 1 mol H₂O) = 43.24 g H₂O

Therefore, 43.24 grams of H₂O can be produced by the reaction.