Final answer:
To form 22.9 grams of oxygen gas, 824.06 grams of water are needed.
Step-by-step explanation:
To determine the mass of water needed to form 22.9 grams of oxygen gas, we need to use the stoichiometry of the chemical equation. From the balanced equation, 2H₂(g) + O₂(g) → 2H₂O(g), we can see that 1 mole of oxygen gas reacts with 2 moles of water. Therefore, we can set up a conversion factor using the molar mass of water to calculate the mass of water needed:
22.9 g O₂ * (2 moles H₂O / 1 mole O₂) * (18.02 g H₂O / 1 mole H₂O) = 824.06 g H₂O
Therefore, 824.06 grams of water are needed to form 22.9 grams of oxygen gas.