The water is formed from oxygen gas and...hydrogen gas, I'm assuming? It would have been nice for the question to have been a bit more explicit (not blaming you, of course).
Assuming that's the case, our chemical reaction would be:
2H₂(g) + O₂(g) → 2H₂O(l).
We are told that 1 mol of a gas has a volume of 24.0 dm³ at RTP. We can use this relation to determine the number of moles of O₂ gas that reacts given its initial volume, 33.5 dm³.
33.5 dm³ O₂(g)/24.0 dm³/mol = 1.396 mol O₂(g).
Since we are not given any information about H₂(g), or any other reactant for that matter, I am assuming that the O₂(g) is the limiting reactant. According to the equation, the stoichiometric ratio between O₂ and H₂O is 1:2. That is, for every one mole of O₂ that is consumed, two moles of H₂O are formed (i.e., the number of moles of H₂O formed is double the number moles of O₂).
Since 1.396 moles of O₂ reacts, 2(1.396) = 2.792 moles of H₂O are produced. To convert moles of water to grams, we multiply the number of moles of H₂O by the molar mass of H₂O:
(2.792 moles H₂O)(18.015 g/mol) = 50.3 g H₂O.
So, approximately 50.3 grams of water are formed from 33.5 dm³ of oxygen gas at RTP.
